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3
Algebra
3.1 Rational expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 BAa1: Reduce to lowest terms . . . . . . . . . . . . . . . . . . . . . .
3.1.2 BAa2: Add, subtract, multiply, and divide . . . . . . . . . . . . . . .
3.1.3 BAa3: Find and simplify products of the form
(ax + b)(c/x + d). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.4 BAa4: Evaluate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.5 BAa5: Simplify complex fractions . . . . . . . . . . . . . . . . . . . .
3.2 Factoring quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 BAb1: Factor a quadratic trinomial in two variables . . . . . . . . . .
3.2.2 BAb3: Factor a perfect square trinomial in more than one variable .
3.2.3 BAb4: Recognize relationships among coefficients of a factorable
trinomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Operations on Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 BAe1: Simplify polynomial expressions involving products of binomials in one or more variables . . . . . . . . . . . . . . . . . . . . . .
3.3.2 BAe2: Factor a difference of squares . . . . . . . . . . . . . . . . . . .
3.3.3 BAe4: Divide a third degree by a first or second degree to obtain the
quotient and remainder . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.4 BAe5: Raise a binomial to the second or third power . . . . . . . . .
3.4 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.1 BAc2: Find the slope of a line from an equation . . . . . . . . . . . .
3.4.2 BAc3: Find the slope of a line from the coordinates of two points. . .
3.4.3 BAc4: Find the equation of a line from the coordinates of two points.
3.4.4 BAc5: Find the equation of a line given 1 point and the slope. . . . .
3.4.5 BAc6: Determine and/or plot the intercepts of a line, given an equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.6 BAc9: Given an equation determine if a point is above or below the
graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Quadratic/Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 BAd1: Quadratics of the form ax2 + b = 0 or (ax + b)2 + c = 0. . . . .
3.5.2 BAd2: Quadratics of the form ax2 +bx+c = 0 with a 6= 1 by factoring
over the integers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
3
3
3
5
7
8
9
11
11
13
15
18
18
19
20
22
24
24
26
28
31
33
36
38
38
40
3.5.3
BAd3: Quadratics using the quadratic formula, where the discriminant might be positive, negative or zero. . . . . . . . . . . . . . . . . 43
3.5.4 BAd4: Equations quadratic in u, where u = x2 , x3 , .... . . . . . . . . . 46
3.5.5 BAd5: Rational equations leading to quadratic equations. . . . . . . 49
3.5.6 BAd6: Rational equations leading to linear equations. . . . . . . . . . 52
3.5.7 BAd7: Solve literal equations involving rational expressions. . . . . . 54
3.6 Radicals and Fractionial exponents . . . . . . . . . . . . . . . . . . . . . . . . 56
3.6.1 BBa4: Express radicals in terms of fractional exponents and vice versa. 56
3.6.2 BBa1: Evaluate numerical expressions involving radicals and/or
fractional exponents. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.6.3 BBa2: Simplify radical expressions. . . . . . . . . . . . . . . . . . . . . 58
3.6.4 BBa3: Rationalize the numerator or denominator of a quotient involving radicals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
3.6.5 BBa5: Multiply radical expressions (monomials, binomials). . . . . . 61
3.6.6 BBa6: Simplify and combine like radicals. . . . . . . . . . . . . . . . . 62
3.6.7 BBa7: Simplify expressions involving radicals by changing to fractional exponents and using exponent laws. . . . . . . . . . . . . . . . 63
3.7 Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.7.1 BBb2: Use reciprocal rule to eliminate denominator negative exponents. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.7.2 BBb3: Simplify negative powers of sums and differences. . . . . . . . 66
3.7.3 BBb4: Evaluate or simplify expressions involving negative exponents. 67
3.8 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.8.1 BBc1: Evaluate numerical expressions involving absolute values. . . 69
3.8.2 BBc2: Recognize the graph of y = |x| and its variations. . . . . . . . . 70
3.8.3 BBc3: Solve equations of the type |ax + b| + c = d. . . . . . . . . . . . 74
3.9 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.9.1 BBd1: Solve or graph linear inequalities in one variable. . . . . . . . 76
3.9.2 BBe1: Solve consistent 2 x 2 systems by substitution or elimination. . 83
3.9.3 BBe2: Interpret 2 x 2 systems graphically . . . . . . . . . . . . . . . . . 87
3.9.4 BBe3: Recognize inconsistent or dependent systems algebraically or
graphically. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
3.10 Verbal Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
3.10.1 BBf1: Work/rate. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
3.10.2 BBf2: Distance/rate/time. . . . . . . . . . . . . . . . . . . . . . . . . . 97
3.10.3 BBf3: Mixture. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
3.10.4 BBf4: Number (reciprocal, digit, consecutive, integer, etc.). . . . . . . 103
3.10.5 BBf5: Area and perimeter. . . . . . . . . . . . . . . . . . . . . . . . . . 107
3.10.6 BBf6: Other miscellaneous problems. . . . . . . . . . . . . . . . . . . 110
3.10.7 BBf5: Area and perimeter. . . . . . . . . . . . . . . . . . . . . . . . . . 113
2
Chapter 3
Algebra
3.1
Rational expressions
3.1.1 BAa1: Reduce to lowest terms
We need to recall the rule for simplifying fractions. To do so, we need to factor out common factors and then we can divide out common factors. Doing it any other way is not
supported by the rules of arithmetic.
1.
x(x + 2)
x(2x + 4)
The numerator is factored. We need to factor out the denominator. We see that the
second factor contains common factor 2, so we factor it out and divide all common
factors x(x + 2), assuming x(x + 2) 6= 0:
=
=
2.
x(x + 2)
2x(x + 2)
1
x(x + 2) 1
× = .
x(x + 2) 2
2
x2 (x2 + 2)
x(2x2 + 4)
We factor out 2 from the denominator, then we divide out all common factors on the
second step. We can do it only if x 6= 0
=
x2 (x2 + 2)
2x(x2 + 2)
x(x2 + 2) 1
=
× x
x(x2 + 2) 2
1
= x.
2
3
3.1. RATIONAL EXPRESSIONS
3.
CHAPTER 3. ALGEBRA
(x + 9)(x − 2)
x(3x + 27)
We factor 3 from the denominator. Then we divide out all common factors, shown
on separate fractions and divide out them (assuming x − 9 6= 0).
=
=
4.
(x + 9)(x − 2)
x[3(x + 9)]
x+9 x−2
×
x+9
3x
x−2
.
=
3x
a(3x − 2)
(x + 3)(6 − 9x)
We observe if we reverse the signs in the parenthesis and factor out 3, we may have
common factors to divide out. We do so and have:
=
a(3x − 2)
(x + 3)(−3)(3x − 2)
We divide out (3x − 2) assuming that (3x − 2) 6= 0.
=−
5.
a
.
3(x + 3)
x(x2 + 2x + 1)
x(2x + 2)
We can factor the numerator further seeing that x2 + 2x + 1 = (x + 1)2 . There is only
one factor we can factor out from the denominator - the number 2. On the second
step, we divide out all common factors.
=
=
We assume x(x + 1) 6= 0.
x(x + 1)2
x2(x + 1)
x(x + 1) 1
× (x + 1)
x(x + 1) 2
1
= (x + 1).
2
4
CHAPTER 3. ALGEBRA
3.1. RATIONAL EXPRESSIONS
3.1.2 BAa2: Add, subtract, multiply, and divide
To add or subtract rational expression, we need to find the Least Common Denominator.
We multiply all expressions by the appropriate factors and then we add or subtract the
resulting numerators.
1.
x
(x + 2)
+
x(x + 4) 2
The LCD is 2x(x + 4).
=
2
(x + 2)
×
x(x + 4) 2
=
x x(x + 4)
×
2 x(x + 4)
x3 + 4x2 + 2x + 4
.
2x(x + 4)
(x + 3)
x
− =
x(x + 4) 2
LCD = 2x(x + 4)
=
(x + 3)2
x[x(x + 4)]
−
2x(x + 4)
2x(x + 4)
=
=
3.
+
2(x + 2) + x[x(x + 4)]
2x(x + 4)
=
2.
2x + 6 − x3 − 4x2
2x(x + 4)
−x3 − 4x2 + 2x + 6
.
2x(x + 4)
(x + 3)
x
×
x(x + 4) 2 + x
=
(x + 3)x
x(x + 4)(x + 2)
We divide out a common factor x, assuming (x 6= 0).
=
4.
x+3
.
(x + 4)(x + 2)
(x + 3)
x
÷
=
x(x + 4) 2 + x
=
(2 + x)
(x + 3)
×
=
x(x + 4)
x
=
(x + 3)(x + 2)
.
x2 (x + 4)
5
3.1. RATIONAL EXPRESSIONS
5.
CHAPTER 3. ALGEBRA
(x − 3)
x
−
x(x2 + 4) 2 + x
LCD = x(x2 + 4)(2 + x)
=
(x − 3)(2 + x)
x([x(x2 + 4)]
−
x(x2 + 4)(2 + x) x(x2 + 4)(2 + x)
x2 − x − 6 − x4 − 4x2
=
x(x + 2)(x2 + 4)
=
−x4 − 3x2 − x − 6
x(x + 2)(x2 + 4)
=−
x4 + 3x2 + x + 6
.
x(x + 2)(x2 + 4)
Note we put the minus sign in front of the fraction and all minuses were replaced
by pluses.
6
CHAPTER 3. ALGEBRA
3.1. RATIONAL EXPRESSIONS
3.1.3 BAa3: Find and simplify products of the form
(ax + b)(c/x + d).
We need to multiply expressions containing two terms. A convenient way of remembering how to do it is using FOIL method. We will show it in detail for # 1.
2
+5
1. (3x + 2)
x
2
2
+ (3x × 5) + 2 ×
+ (2 × 5)
= 3x ×
x
x
4
+ 10
x
4
= 15x + 16 + .
x
= 6 + 15x +
2. (x + 2)
1
−2
x
2
−4
x
2
= −2x − 3 + .
x
= 1 − 2x +
3. (5x − 7)
4
+5
3x
=
4. (−x + 2)
7
−5
x
20
28
+ 25x −
− 35
3
3x
85 28
− .
= 25x −
3
3x
14
− 10
x
14
= 5x − 17 + .
x
= −7 + 5x +
5. (7x − 2)
−2
+ 15
x
4
− 30
x
4
= 105x − 44 + .
x
= −14 + 105x +
7
3.1. RATIONAL EXPRESSIONS
CHAPTER 3. ALGEBRA
3.1.4 BAa4: Evaluate
1. If x = 2, then
(2 − x)
x−5
2 − (+2)
(+2) − 5
=
2. If x = −2, then
3. If x = 5, then
(2 − x)
x−5
(2 − x)
x2 − 5
=
0
= 0.
−3
=
2 − (−2)
(−2) − 5
=
4
4
=− .
−7
7
=
2 − (+5)
(+5)2 − 5
=
4. If x = −7, then
(2x2 − x)
x−5
=
5. If x = −3, then
(2 − x2 )
x2 − 5x
−3
3
=− .
20
20
105
35
2(−7)2 − (−7)
=
=− .
(−7) − 5
−12
4
=
=
2 − (−3)2
(−3)2 − 5(−3)
2−9
7
=− .
9 + 15
24
Note the use of parentheses for negative numbers. This way you will not forget to change
the sign if you need to.
8
CHAPTER 3. ALGEBRA
3.1. RATIONAL EXPRESSIONS
3.1.5 BAa5: Simplify complex fractions
We can look at the numerator of the complex fraction as a number, which we try to divide by the denominator fraction. Dividing by a fraction means that we multiply by the
reciprocal of that fraction. This step is shown only in # 1.
1. Reduce:
(x+2)(x+5)
x+3
(x+2)(x+5)
x+4
=
(x + 2)(x + 5) (x + 2)(x + 5)
÷
x+3
x+4
=
(x + 2)(x + 5)
x+4
×
x+3
(x + 2)(x + 5)
=
2. Reduce:
x+4
.
x+3
(x2 +2)(x+5)
x+3
x+2
(x+4)(x+1)
=
x+2
(x2 + 2)(x + 5)
÷
x+3
(x + 4)(x + 1)
=
(x2 + 2)(x + 5) (x + 4)(x + 1)
×
x+3
x+2
=
(x2 + 2)(x + 5)(x + 4)(x + 1)
.
(x + 3)(x + 2)
3. Reduce:
=
x(x+2)(x2 +5)
x+3
x+2
x(x+4)
x+2
x(x + 2)(x2 + 5)
÷
x+3
x(x + 4)
We factor out the numerator of the numerator fraction.
=
x(x + 2)(x2 + 5) x(x + 4)
×
x+3
x+2
=
x2 (x2 + 5)(x + 4)
.
x+3
9
3.1. RATIONAL EXPRESSIONS
CHAPTER 3. ALGEBRA
4. Reduce:
(x2 +2x)(x+5)
x+3
x+2
x2 +4x
x+2
(x2 + 2x)(x + 5)
÷ 2
x+3
x + 4x
x(x + 2)(x + 5) x(x + 4)
=
×
x+3
x+2
x2 (x + 5)(x + 4)
.
=
x+3
=
5. Reduce:
(x−2)(x+5)
x+3
−x+2
x+4
(x − 2)(x + 5) −x + 2
÷
x+3
x+4
(x − 2)(x + 5)
x+4
=
×
x+3
−x + 2
−(−x + 2)(x + 5)(x + 4)
=
(x + 3)(−x + 2)
=
=−
(x + 5)(x + 4)
.
x+3
10
CHAPTER 3. ALGEBRA
3.2
3.2. FACTORING QUADRATICS
Factoring quadratics
3.2.1 BAb1: Factor a quadratic trinomial in two variables
To factor quadratics, one way is to use FOIL method. We look for factors of the format
(A1 a + B1 b)(A2 a + B2 b). Evaluating this, we obtain:
(A1 a + B1 b)(A2 a + B2 b) = A1 A2 a2 + (A1 B2 + A2 B1 )ab + B1 B2 b2
We organize the information in a table. We will show the process in #1.
1. a2 + 5ab + 6b2
Looking at the above expression we can see A1 A2 = 1. One natural guess is A1 =
A2 = 1.
We replace A1 and A2 in the sum A1 B2 + A2 B1 = 1B2 + 1B1 = B2 + B1 . The sum
should be equal to 5.
We finally look at the product B1 B2 . It must be equal to 6. On the first step, we factor
product B1 B2 = 6. The sum is (+5), so we need to look only at positive factors. The
possibilities for the factors are (+6, +1) and (+3, +2). The first pair of factors has
sum (+7) and second one has sum (+5). We see the factors must be (+3) and (+2).
All the information is organized in the table below. In the left column are shown the
two factors of the product in the top row. Their sum is shown in the right column.
Product=6
+1, +6
+2, +3
Sum of factors
7 NO!
5 YES!
= a2 + 5ab + 6b2 = 1 × 1 × a2 + (1 × 2 + 1 × 3)ab + 2 × 3 × b2
= a2 + 2ab + 3ab + 6b2
= a(a + 2b) + 3b(a + 2b)
= (a + 2b)(a + 3b).
2. a2 − 5ab + 6b2
The only difference between # 2 and # 1 is that the sum B1 + B2 = −5, i.e. negative
five. It means that the factors are negative. The table shows the values:
Product=6
−2, −3
Sum of factors
−5 YES!
= a2 − 2ab − 3ab + 6b2
We combine first and second terms, and third and forth terms; then factor:
= a(a − 2b) − 3b(a − 2b)
= (a − 2b)(a − 3b).
11
3.2. FACTORING QUADRATICS
CHAPTER 3. ALGEBRA
3. a2 + 4ab − 5b2 =
Product=−5
+1, −5
−1, +5
Sum of factors
−5 NO!
+5 YES!
= a2 + 5ab − ab − 5b2
= a(a + 5b) − b(a + 5b)
= (a + 5b)(a − b).
4. 6a2 + 5ab + b2
We rearrange this in the following fashion
= b2 + 5ab + 6a2
We can now to use the table from #1
Product=6
+1, +6
+2, +3
Sum of factors
7 NO!
5 YES!
= b2 + 2ab + 3ab + 6a2
= b(b + 2a) + 3a(b + 3a)
= (b + 2a)(b + 3a).
5. a2 + 12ab + 20b2
Product=20
1, 20
2, 10
Sum of factors
21 NO!
12 YES!
= a2 + 2ab + 10ab + 20b2
= a(a + 2b) + 10b(a + 2b)
= (a + 2b)(a + 10b).
12
CHAPTER 3. ALGEBRA
3.2. FACTORING QUADRATICS
3.2.2 BAb3: Factor a perfect square trinomial in more than one variable
We need to recognize the perfect square formula:
A2 + 2AB + B 2 = (A + B)2
and
A2 − 2AB + B 2 = (A − B)2 = (B − A)2
The last identity resembles the property that the opposite numbers have one and the same
square.
Example: (−3)2 = 9 = 32
1. a2 − 6ab + 9b2
We see there are differences of numbers squared:
A2 = a2 ,
B 2 = 9b2 = (3b)2 ,
−2AB = −6ab = −2 × a(3b).
Then:
= a2 − 2a(3b) + (3b)2
= (a − 3b)2 .
We can check the proper factoring by remultiplying:
(a − 3b)2 = a2 − 2a × (3b) + (−3b)2 = a2 − 6ab + 9b2 .
2. 4a2 − 12ab + 9b2
A2 = (2a)2 ,
B 2 = 9b2 = (3b)2 ,
−2AB = −12ab = −2(2a)(3b).
Then:
= (2a)2 − 2(2a)(3b) + (3b)2
= (2a − 3b)2 .
3. 25a2 + 20ab + 4b2 =
A2 = (5a)2 ,
B 2 = 4b2 = (2b)2 ,
2AB = 20ab = 2(5a)(2b).
Then:
= (5a)2 + 2(5a)(2b) + (2b)2
= (5a + 2b)2 .
13
3.2. FACTORING QUADRATICS
CHAPTER 3. ALGEBRA
4. a2 − 2ab−1 + b−2
A2 = a2 ,
B 2 = b−2 = (b−1 )2 ,
−2AB = −2ab−1 .
Then:
= a2 − 2ab−1 + b−2
= (a − b−1 )2 .
This is a very important example. We see the square here is b−2 = (b−1 )2 and a2 . The
equivalent way you could see this is a2 − 2 × a/b + 1/b2 . We can say:
a
1
a2 − 2ab−1 + b−2 = a2 − 2 + 2 .
b b
Then:
1
(a − b−1 )2 = (a − )2 .
b
5. 36a2 + 36ab + 9b2
A2 = 36a2 = (6a)2 ,
B 2 = 9b2 = (3b)2 ,
2AB = 36ab = 2(6a)(3b).
Then:
= (6a)2 + 2(6a)(3b) + (3b)2
= (6a + 3b)2 .
Note! Parentheses are used in all the expressions above. This is first, so we do not get
confused, and second, to show not only the variable but also the whole expression in
parenthesis is squared. In the cross product, it shows the separated factors to you.
In all cases, if you feel you could be confused by notation, use parenthesis for indicating what arithmetic operation you need to do first.
14
CHAPTER 3. ALGEBRA
3.2. FACTORING QUADRATICS
3.2.3 BAb4: Recognize relationships among coefficients of a factorable
trinomial
Find a value of b so that the quadratic factors with integer coefficients, and give the factorization. There may be more than one correct answer. You only need give one correct
answer.
I will give you all possible answers. For #2 and #4, there are infinitely many of them. The
way these could be computed will be shown in a note at the end of the section for the
interested reader. The other items have a finite number of solutions.
1. x2 + (5 + b)x + 12.
We can use FOIL to find all the answers it this case. We look for whole numbers
which have product 12 and sum (5 + b). So, we factor 12 and show all groups of two
factors we may have: (1, 12), (2, 6), (3, 4); we could have negative factoring as well:
(−1, −12), (−2, −6), (−3, −4). For each pair of factors, we can compute the sum of
the factors, it must be (5 + b). We then will have an equation we can solve for b:
5 + b = 1 + 12,
b = 8.
In a table, we show all possible values for b. If your answer is not among them,
check your work.
Sum=5 + b
13
8
7
-13
-8
-7
Product=12
1, 12
2, 6
3, 4
-1, -12
-2, -6
-3, -4
b
8
3
2
-18
-13
-12
2. x2 + 7x + 2b.
We use FOIL method again. Let us have factors
x2 + 7x + 2b = (x + u)(x + v) = x2 + vx + ux + uv = x2 + (u + v)x + uv.
We choose one of the factors to be u = 3. Then, using the fact that the left and the
right side are simply two representations of the same trimoial, so u + v = 7 and
uv = 2b. We find v first.
v = 7 − u = 7 − 3 = 4.
Second, we find b.
2b = uv = 3 × 4 = 12,
15
3.2. FACTORING QUADRATICS
CHAPTER 3. ALGEBRA
b = 6.
One solution is x2 + 7x + 2b = (x + 3)(x + 4) if b = 6.
3. x2 + (7 − b)x + 24.
We use the same (FOIL) method as in #1. The table below shows the values of b for
which the trinomial is factored.
Sum=7 − b
25
14
11
10
-25
-14
-11
-10
Product=24
1, 24
2, 12
3, 8
4, 6
-1, -24
-2, -12
-3, -8
-4, -6
b
-18
-7
-4
-3
32
21
18
17
Example for the first row we have b = (−18). Then:
x2 + [7 − (−18)]x + 24 = x2 + 25x + 24 = (x + 1)(x + 24).
4. x2 + x + 9b.
We applay the same method as in #2:
x2 + x + 9b = (x + u)(x + v) = x2 + vx + ux + uv = x2 + (u + v)x + uv.
We can choose u = 2. Then, v = −1. We find then that uv = 9b, or
b=
uv
2 × (−1)
2
=
=− .
9
9
9
There is no restriction on b to be integer. We check:
x2 + x + 9 ×
−2
= x2 + x − 2 = (x + 2)(x − 1).
9
Note, that even b is not an integer, the quadratic trinomial still have integer coefficients.
5. x2 + (1 − b)x + 15
In the table is shown the solution using the same idea.
16
CHAPTER 3. ALGEBRA
3.2. FACTORING QUADRATICS
Sum=1 − b
16
8
-16
-8
Product=15
1, 15
3, 5
-1, -15
-3, -5
b
-15
-7
17
9
Check (using the last row):
x2 + (1 − 9)x + 15 = x2 − 8x + 15 = (x − 3)(x − 5).
Finding the solutions for #2: we can choose u ∈ N , then from u + v = 7 we find
v = 7 − u and using FOIL, we compute:
u(7 − u) = 2b,
b=
We have:
u(7 − u)
.
2
u(7 − u)
2
2
= x + [u + (7 − u)]x + u(7 − u) = (x + u)[x + (7 − u)].
x2 + 7x + 2 ×
17
3.3. OPERATIONS ON POLYNOMIALS
3.3
CHAPTER 3. ALGEBRA
Operations on Polynomials
3.3.1 BAe1: Simplify polynomial expressions involving products of binomials in one or more variables
1. (x − 1)(x + 2) + x(x − 3)
We multiply the two binomials using FOIL and expand the second product. On the
next step, we combine all the like terms.
= x2 + 2x − x − 2 + x2 − 3x
= (1 + 1)x2 + (2 − 1 − 3)x − 2
= 2x2 − 2x − 2.
2. (x − 1)(x + 2) + x(x − 3)
= x2 + 2x − x − 2 + x2 − 3x
= (1 + 1)x2 + (2 − 1 − 3)x − 2
= 2x2 − 2x − 2.
3. (x − 1)(x2 + 2) + x(x2 − 3)
This is again FOIL method. The only difference is one of the binomials is quadratic.
We will not have like terms expanding this product. However, when expanding
second product, we may find like terms:
= x3 + 2x − x2 − 2 + x3 − 3x
= (1 + 1)x3 − x2 + (2 − 3)x − 2
= 2x3 − x2 − x − 2.
4. (x − 2)(x + 2) − x(x − 3)
You may notice we have a product of sum and difference. There is a formula which
you’d like to remember: (A + B)(A − B) = A2 − B 2 . Here, A = x; B = 2. Check out
the FOIL method used below and see that the crossproducts subtract to 0.
= x2 + 2x − 2x − 4 − x2 + 3x
= 3x − 4.
5. (2x − 1)(x + 7) − x(x − 3)
= 2x2 + 14x − x − 7 − x2 + 3x
= (2 − 1)x2 + (14 − 1 + 3)x − 7
= x2 + 16x − 7.
18
CHAPTER 3. ALGEBRA
3.3. OPERATIONS ON POLYNOMIALS
3.3.2 BAe2: Factor a difference of squares
We will use the formula A2 − B 2 = (A − B)(A + B). I will give you the factors in each case.
Note, that in some cases, we can factor out a common factor which is not a binomial.
1. 25x2 − 81y 2
A2 = 25x2 = (5x)2 ,
B 2 = 81y 2 = (9y)2 .
We use parenthesis to change the order of arithmetic operations, first we do multiplication , and, then raise to a power.
= (5x)2 − (9y)2
= (5x − 9y)(5x + 9y).
2. 4x2 − 9y 2
A2 = 4x2 = (2x)2 ,
B 2 = 9y 2 = (3y)2 .
= (2x)2 − (3y)2
= (2x − 3y)(2x + 3y).
3. 16x2 − 144y 2
A2 = 16x2 = 42 (x)2 ,
B 2 = 144y 2 = 16(3y)2 .
= 16x2 − 16(3y)2
= 16[x2 − (3y)2 ]
= 16(x − 3y)(x + 3y).
4. 25x2 − 36y 2
A2 = 25x2 = (5x)2 ,
B 2 = 36y 2 = (6y)2 .
= (5x)2 − (6y)2
= (5x − 6y)(5x + 6y).
5. y 4 x2 − 16a2 b6
A2 = y 4 x2 = (y 2 x)2 ,
B 2 = 16a2 b6 = (4ab3 )2 .
In this example we have higher powers than two. We see the square could be a
power of a variable. Here, A = y 2 x, and B = 4ab3 .
= (y 2 x)2 − (4ab3 )2
= (y 2 x − 4ab3 )(y 2 x − 4ab3 ).
19
3.3. OPERATIONS ON POLYNOMIALS
CHAPTER 3. ALGEBRA
3.3.3 BAe4: Divide a third degree by a first or second degree to obtain
the quotient and remainder
The division of polynomials is very similar to long division of numbers.
Note when we have missing terms, in either dividend and the divisor, we have to show
these one way or another. First, we can write them with coefficient zero or the other way
is to leave an empty space in the dividend when writing it.
The division in detail (including the way one can check his answer) is shown #1 and
#2
1. (x3 − 3x + 4) ÷ (x2 + 2x)
The division:
x−2
x2 + 2x)x3 + 0x2 − 3x + 4
−(x3 + 2x2 )
−2x2 − 3x
−(−2x2 − 4x)
x + 4 reminder
=x−2+
x+4
.
z 2 + 2x
We check the division by multiplication. We multiply the answer and the dividend
and then we add the reminder. If we worked correctly, we will obtain the divisor.
answer × divisor + reminder = dividend
(x − 2)(x2 + 2x) + x + 4 = x3 + 2x2 − 2x2 − 4x + x + 4
= x3 − 3x + 4.
TRUE!
2. (3x3 + 3x2 + 4) ÷ (x2 − x + 2)
The division:
3x + 6
x2 − x + 2)3x3 + 3x2 + 0x + 4
−(3x3 − 3x2 + 6x)
6x2 − 6x + 4
−(6x2 − 6x + 12)
−8 reminder
= 3x + 6 +
20
−8
.
x2 − x + 2
CHAPTER 3. ALGEBRA
3.3. OPERATIONS ON POLYNOMIALS
Check:
answer × divisor + reminder = dividend
(3x + 6)(x2 − x + 2) − 8
= 3x3 − 3x2 + 6x + 6x2 − 6x + 12 − 8
= 3x3 + 3x2 + 4.
3. (x3 + 3x + 4) ÷ (x + 2) =
= x2 − 2x + 7 +
TRUE!
−10
.
x+2
4. (x3 + 4x2 − 3x + 4) ÷ (x − 3) =
= x2 + 7x + 18 +
58
.
x−3
5. (2x3 − 3x2 + 5x − 4) ÷ (x2 − 2) =
= 2x − 3 +
9x − 10
.
x2 − 2
Once more, here is the way to check the answer. We multiply the quotient and the divisor
and add a reminder (example #5):
(x2 − 2)(2x − 3) + 9x − 10 =
= 2x3 − 3x2 − 4x + 6 + 9x − 10 =
= 2x3 − 3x2 + 5x − 4.
We got the original expression ( the dividend), so that we worked properly.
21
3.3. OPERATIONS ON POLYNOMIALS
CHAPTER 3. ALGEBRA
3.3.4 BAe5: Raise a binomial to the second or third power
We can use the following formulas for raising to second and third powers:
(A + B)2 = A2 + 2AB + B 2 .
Using the formula for the sum, and the fact that −B = +(−B) we find
(A − B)2 = [A + (−B)]2 = A2 + 2A(−B) + (−B)2 = A2 − 2AB + B 2 .
The same way we compute the third power of the sum:
(A + B)3 = A3 + 3A2 B + 3AB 2 + B 3 .
Changing B to −B we obtain
(A − B)3 = [A + (−B)]3 .
= A3 + 3A2 (−B) + 3A(−B 2 ) + (−B)3 == A3 − 3A2 B + 3AB 2 − B 3 .
Note that squares are shown with + plus sign regardless whether they are with plus
or minus sign in the sum. The only difference when we square a difference is that the
crossproduct has negative sign.
Note, that in general we do not need two of the formulas: difference squared and difference cubed because these can be derived from the sum squared and the sum cubed by
appropriate change of signs.
1. (x + 1)2
A = x;
B = 1.
= x2 + (2 × x × 1) + 1 = x2 + 2x + 1.
(x + 1)3
= x3 + 3 × x2 × 1 + 3 × x × 12 + 13 = x3 + 3x2 + 3x + 1.
2. (x + 2)2
A = x;
B = 2.
(x + 2)3
= x2 + 2 × x × 2 + 22 = x2 + 4x + 4.
= x3 + 3 × x2 × 2 + 3 × x × 22 + 23 = x3 + 6x2 + 12x + 8.
22
CHAPTER 3. ALGEBRA
3.3. OPERATIONS ON POLYNOMIALS
3. (2x − 1)2
A = 2x;
B = −1.
= (2x)2 + 2 × (2x × (−1) + (−1)2 = 4x2 − 4x + 1.
(2x − 1)3
= (2x)3 + 3 × (2x)2 × (−1)1 + 3 × (2x) × (−1)2 + (−1)3 = 8x3 − 12x2 + 6x − 1.
4. (a + 2b)2
A = a;
B = 2b.
= a2 + 2a(2b) + (2b)2 = a2 + 4ab + 4b2 .
(a + 2b)3
= a3 + 3a2 (2b) + 3a(2b)2 + (2b)3 = a3 + 6a2 b + 12ab2 + 8b3 .
5. (x − 3)2
A = x;
B = 3.
= x2 + 2 × x × (−3) + (−3)2 = x2 − 6x + 9.
(x − 3)3
= x3 + 3 × x2 × (−3) + 3 × x × (−3)2 + (−3)3 = x3 − 9x2 + 27x − 27.
23
3.4. GRAPHS
3.4
CHAPTER 3. ALGEBRA
Graphs
3.4.1 BAc2: Find the slope of a line from an equation
The slope of a line is a fraction
slope of a line =
coeff x
.
coeff y
Example: for the equation 3y + 4x = 12 we subtract 4x from both sides.
3y + 4x − 4x = 12 − 4x,
3y = −4x + 12.
The slope is
slope of a line =
coeff x
−4
=
.
coeff y
3
We can divide by 3:
y=
12
−4
−4
x+
=
+ 4.
3
3
3
The slope is
4
slope = (− )/1.
3
1. The slope of the line y = 5x + 7 is 5.
5
y = x + 7.
1
2. The slope of the line 7y = −2x + 3 is −2/7
We divide both sides of the equation by 7:
2
3
y =− x+ .
7
7
3. The slope of the line 9y − 4x + 7 = 0 is 4/9.
We subtract from both sides (−4x + 7). Next, we divide the equation by 9.
9y = 4x − 7,
7
4
y = x− .
9
9
24
CHAPTER 3. ALGEBRA
3.4. GRAPHS
4. The slope of the line 3(y − 4) = 5(x + 7) is 5/3.
We divide by 3. Then, we add 4 to both sides and open the parenthesis.
5
y − 4 + 4 = (x + 7) + 4,
3
35
5
+ 4,
y = x+
3
3
47
5
x+ .
3
3
5. The slope of the line 12y + 4x = 7 is −1/3.
We solve for y:
12y = −4x + 7,
4
7
y =− x+ ,
12
12
7
1
y =− x+ .
3
12
25
3.4. GRAPHS
CHAPTER 3. ALGEBRA
3.4.2 BAc3: Find the slope of a line from the coordinates of two points.
We usually denote the slope with m. If we have to denote more than one slope, we could
use a subscript (example: m1 ). We use the formula
slope = m =
yR − yQ
.
xR − xQ
where yQ means the y coordinate of Q and all other accordingly. In the formula, we can
swap the place of the points P and Q (check!).
1. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 7). The slope of the
line that passes through Q and R
is
m=
yR − yQ
,
xR − xQ
m=
7−4
,
4 − (−3)
3
m= .
7
2. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 6). The slope of the
line that passes through Q and R is
m=
6−4
,
4 − (−3)
2
m= .
7
3. The coordinates of R are (−3, 4/3) and the coordinates of S are (4, 2/3). The slope of
the line that passes through R and S is
mRS =
2/3 − 4/3
,
4 − (−3)
2/3
2 1
=− × ,
7
3 7
2
mRS = − .
21
mRS = −
26
CHAPTER 3. ALGEBRA
3.4. GRAPHS
4. The coordinates of S are (5/7, 4) and the coordinates of T are (4, −6). The slope of
the line that passes through S and T is
mST =
23
−6 − 4
= (−10) ÷ ,
4 − 5/7
7
mST = (−10) ×
mST = −
23
,
7
70
.
23
5. The coordinates of T are (−3, −4) and the coordinates of U are (4, −4). The slope of
the line that passes through T and U is
mT U =
(−4) − (−4)
,
4 − (−3)
mT U = 0.
We see that if the y coordinates are the same, the slope is zero.
Note if the x coordinates of the two points are the same, then the slope is undefined!
This is because division by zero cannot be defined.
27
3.4. GRAPHS
CHAPTER 3. ALGEBRA
3.4.3 BAc4: Find the equation of a line from the coordinates of two
points.
Given the coordinates of two points, write an equation for the line in both point-slope
and slope intercept form if the line has a slope. If the line does not have a slope, write an
equation in the form y = C.
We have a two phase process.
First, we find the slope of the line. We can use the formulas from the previous section.
Second, we write point-slope form of the equation. It is
y − y0 = m(x − x0 ).
We denote here y0 and x0 corresponding coordinates of either of the given points. To find
intercept-slope form, we solve for y. We could open the parenthesis and the final form is
y = mx − mx0 + y0 = mx + (y0 − mx0 ).
1. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 7). Equations of the
line through Q and R are:
For point-slope form, we first compute the slope:
m=
3
7−4
= .
4 − (−3)
7
Now, we use the formula with point Q (−3, 4):
y − 4 = m[x − (−3)],
3
y − 4 = (x + 3).
7
For finding intercept-slope form, we solve for y:
3
3
+ 4,
y = x+ 3×
7
7
37
3
y = x+ .
7
7
28
CHAPTER 3. ALGEBRA
3.4. GRAPHS
2. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 6). Equations of the
line through Q and R are:
6−4
2
m=
= .
4 − (−3)
7
We use point R (4,6) for point slope form:
2
y − 6 = [x − (−3)],
7
2
y − 6 = (x + 3).
7
The intercept-slope form is:
6
2
y = x + + 6,
7
7
2
48
y = x+ .
7
7
3. The coordinates of R are (−3, 4/3) and the coordinates of S are
(4, 2/3).
Equations of the line through R and S are:
2
= − 21
.
m = 2/3−4/3
4−(−3)
The point slope form with S (4, 2/3):
y−
The intercept-slope form:
2
2
= − (x − 4).
3
21
8
2
2
x+
+ ,
21
21 3
22
2
y =− x+ .
21
21
y=−
4. The coordinates of T are (−3, 4) and the coordinates of U are
(−3, −4). The first coordinates are equal,therefore the slope is undefined. The line
has equation
x = −3.
5. The coordinates of S are (5/7, 4) and the coordinates of T are (4, −6). The equations
of the line through S and T are:
m=
The point slope form with S (4, 2/3)
−6 − 4
70
=− .
4 − 5/7
23
y − (−6) = −
29
70
(x − 4).
23
3.4. GRAPHS
The intercept-slope form:
CHAPTER 3. ALGEBRA
280
70
x+
− 6,
23
23
142
70
.
y =− x+
23
23
y=−
30
CHAPTER 3. ALGEBRA
3.4. GRAPHS
3.4.4 BAc5: Find the equation of a line given 1 point and the slope.
Given the slope of a line and the coordinates of a point on the line, write an equation for
the line in both point-slope and slope intercept form.
Recall the formulas for the point-slope form and the intercept-slope form:
y − y0 = m(x − x0 ),
y = mx − mx0 + y0 .
In the preceding, (x0 , y0 ) are the coordinates of the given point, the first equation is pointslope form, the second is the intercept-slope form.
1. A line passes through (3, −4) and has slope −3. The point-slope equation is
y − (−4) = −3(x − 3).
The intercept-slope form of the equation is
y = −3x + 9 − 4,
y = −3x + 5.
2. A line passes through (7, −4) and has slope −3/7. The point-slope equation is
3
y − (−4) = − (x − 7).
7
The intercept-slope form of the equation is
3
y = − x + 3 − 4,
7
3
y = − x − 1.
7
3. A line passes through (6, −11) and has slope −1/3. The point-slope equation is
1
y − (−11) = − (x − 6).
3
The intercept-slope form of the equation is
1
y = − x + 2 − 11,
3
1
y = − x − 9.
3
31
3.4. GRAPHS
CHAPTER 3. ALGEBRA
4. A line passes through (2/5, 6) and has slope 5. The point-slope equation is
2
y − 6 = 5(x − ).
5
The intercept-slope form of the equation is
y − 6 = 5x − 2,
y = 5x + 4.
5. A line passes through (3, 4/9) and has slope 5/9. The point-slope equation is
y−
5
4
= (x − 3).
9
9
The intercept-slope form of the equation is
5
15 4
y = x−
+ ,
9
9
9
5
11
y = x− .
9
9
32
CHAPTER 3. ALGEBRA
3.4. GRAPHS
3.4.5 BAc6: Determine and/or plot the intercepts of a line, given an
equation.
Given an equation for a line, give the horizontal and vertical intercepts of the line.
To find the intercept means we need to find points such that they are on the line, and
one of them has first coordinate zero (0, y1 ) and the other one has second coordinate
zero(x1 , 0). In practice, we replace y with zero and solve for x - x-intercept and x with
zero and solve for y -this is y-intercept.
y-intercept
We set
x=0 and
solve for y
0
c=
+
by
+
ax
(0,y1 )
(x1,0)
(0,0)
x-intercept
We set
y=0 and
solve for x
The other way is to plot the line, and then determine the intercepts. For this, we choose
any two x coordinates, plug then into the equation of the line and solve for y. We plot the
points, and find the horizontal and vertical intercepts.
1. An equation for a line is −3x + 4y = 12.
If y = 0, then
−3x + 4 × 0 = 12,
x=
The horizontal intercept is (−4, 0).
12
= −4.
−3
If x = 0, then
−3 × 0 + 4y = 12,
y = 3.
The vertical intercept is (0, 3).
33
3.4. GRAPHS
CHAPTER 3. ALGEBRA
2. An equation for a line is 5x + 4y = 20.
If y = 0, then
5x + 4 × 0 = 20,
x = 4.
The horizontal intercept is (4, 0) .
If x = 0, then
5 × 0 + 4y = 20,
y = 5.
The vertical intercept is (0, 5).
3. An equation for a line is −3x + 4y = 25.
If y = 0, then
−3x + 4 × 0 = 25,
25
25
x=
=− .
−3
3
The horizontal intercept is (−25/3, 0).
If x = 0, then
−3 × 0 + 4y = 25,
25
y= .
4
The vertical intercept is (0, 25/4).
4. An equation for a line is −3(x − 2) + 4(y − 7) = 12.
If y = 0 then
−3(x − 2) + 4(0 − 7) = 12,
−3x + 6 − 28 = 12,
−3x − 22 = 12,
−3x = 34
34
x=− .
3
The horizontal intercept is (−34/3, 0).
34
CHAPTER 3. ALGEBRA
3.4. GRAPHS
If x = 0, then
−3(0 − 2) + 4(y − 7) = 12,
6 + 4y − 28 = 12,
4y − 22 = 12,
4y = 34,
17
34
= .
y=
4
2
The vertical intercept is (0, 17/2).
5. An equation for a line is x + 4y = 13.
If y = 0, then
x + 4 × 0 = 13,
x = 13.
The horizontal intercept is (13, 0).
If x = 0, then
0 + 4y = 13,
13
y= .
4
The vertical intercept is (0, 13/4).
35
3.4. GRAPHS
CHAPTER 3. ALGEBRA
3.4.6 BAc9: Given an equation determine if a point is above or below
the graph.
Given an equation for a curve in the Cartesian plane and the coordinates of a point in the
plane, determine if the point is above or below the curve.
Example: An equation for a line is 3x + 4y = 12. The point with coordinates (4, 5) is above
this line because the point (4, 0) is on the line and 0 < 5.
In general, we evaluate the function at the point x. We compare the value of the function
to the y coordinate of the point.
Geometrically, we plot the function and a vertical line at horizontal coordinate x0 . If computed value of the function is bigger than the second coordinate of the point, then the
point is below the graph.
I will denote <>= the process of comparing of two numbers. The expression y(1)
means that I evaluate the function y at the point when x = 1.
1. Is the point (0, 4) above, below, or on the curve y = x3 + 2x − 1?
We compute the value of y at x = 0, and compare it with second coordinate of the
point specified - 4.
Is y(0) = 03 + 2 × 0 − 1 = −1 <>= 4?
y(0) = −1 < 4.
We see that y(0) is less than 4. It means that the point (0, 4) is above the graph.
2. Is the point (−3, 1) above, below, or on the curve y = −2x2 + 2x − 1?
Is y(−3) = −2(−3)2 + 2(−3) − 1 = −18 − 6 − 1 <>= 1?
y(−3) < 1.
The point (−3, 1) is above the graph y = −2x2 + 2x − 1 because y(−3) = −25 < 1.
3. Is the point (1, 1) above, below, or on the curve y = x3 + 2x − 2?
Is y(1) = 13 + 2 × 1 − 2 <>= 0?
y(1) = 1.
The point (1, 1) is on the graph y = x3 + 2x − 2 because y(1) = 1.
4. Is the point (3, −4) above, below, or on the curve y = x3 + 2x − 100?
Is y(3) = 33 + 2 × 3 − 100 <>= −4?
y(3) < −4.
The point (3, −4) is above the graph because y(3) = −67 < −4.
36
CHAPTER 3. ALGEBRA
3.4. GRAPHS
√
5. Is the point (16, 50) above,
below, or on the curve y = 5x + 2 x − 1?
√
Is y(16) = 5 × 16 + 2 16 − 1 = 80 + 8 − 1 <>= 50?
y(16) > 50.
√
The point (16, 50) is below the graph y = 5x + 2 x − 1 because y(16) = 87 > 50.
37
3.5. QUADRATIC/RATIONAL EQUATIONS
3.5
CHAPTER 3. ALGEBRA
Quadratic/Rational Equations
3.5.1 BAd1: Quadratics of the form ax2 + b = 0 or (ax + b)2 + c = 0.
Solve a quadratic equation where the square has been completed.
It is much easier to solve a quadratic equation once the square has been completed. We
simply move the free term on the other side and take square root from both sides of the
equation. We have two equations with one and the same left side and opposite right sides.
1. Solve 4x2 − 36 = 0 for x. We add the free term 36 to both sides of the equation:
4x2 − 36 + 36 = 0 + 36.
We solve for x:
36
= 9,
4
√
|x| = 9 = 3.
x2 =
Recall that when taking square root we compute the principal square root, i.e. the
square root without its sign. There are two numbers with absolute value of 3: x1 = 3
and x2 = −3. The solutions are: x1 = +3 and x2 = −3.
2. Solve 4(x − 12)2 − 36 = 0 for x.
4(x − 12)2 = 36,
36
(x − 12)2 =
= 9,
4
√
√
x − 12 = + 9 = +3 or x − 12 = − 9 = −3.
We solve these equations and find
x − 12 = 3,
x1 = 15.
x − 12 = −3,
x2 = −9.
The solutions are x1 = 15 and x2 = 9.
3. Solve −3a2 + 36 = 0 for a.
−3a2 = −36,
−36
a2 =
= 12,
−3
38
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
√
√
a1 = + 12 = +2 3,
√
√
a2 = − 12 = −2 3.
√
√
The solutions are a1 = +2 3 and a2 = −2 3. Note, that the solutions may not be
rational numbers so that the FOIL device is not always useful.
4. Solve 5(x + 3)2 − 25 = 0 for x.
5(x + 3)2 = 25,
25
= 5,
(x + 3)2 =
5
√
√
x + 3 = + 5 or x + 3 = − 5.
√
x1 = −3 + 5,
√
x2 = −3 − 5
√
√
The solutions are x1 = −3 + 5 and x2 = −3 − 5
5. Solve 2(x + 9)2 − 162 = 0 for x.
2(x + 9)2 = 162,
162
(x + 9)2 =
= 81,
2
√
√
x + 9 = + 81 = +9 or x + 9 = − 81 = −9.
x1 = −9 + 9 = 0,
x2 = −9 − 9 = −18.
The solutions are x1 = 0 and x2 = −18.
39
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
3.5.2 BAd2: Quadratics of the form ax2 + bx + c = 0 with a 6= 1 by
factoring over the integers.
Solve a quadratic equation where the quadratic expression factors over the integers.
To solve quadratic equation we use zero product principle: the product of two numbers
is zero if and only if at least one of the factors is zero. For quadratics, to find the factors
of the quadratic expression and apply the principle means to make each factor equal to 0.
If we have rational solutions to the equation, we can use FOIL method for factoring. The
details will be shown with #1.
Be advised this is a method of trials, so you must be consistent and try until you find
the answer or exhaust all possibilities. Check your work carefully to avoid making errors.
1. Solve 2x2 + 5x − 3 = 0 for x.
FOIL method for factoring. We need two First terms with product (+2): the only combination is (+2, +1). We need two Last terms with product (−3): they could be
(+1, −3) and (−1, +3). We compute the sum Outer terms product plus Inner terms
product until we have a sum equal to (+5).
In the table, FTPr means First terms product, LTPr means Last terms product and
OIpSum means Outer and Inner terms product sum. Outer terms are shown in outer
positions, and Inner terms in inner. The information is organized in the following
table. For example the first row represents the quadratic in factors product [(+1)x +
(+1)] × [(+2)x + (−3)] - the terms come in the order they are in FOIL.
FTPr=2
+1, +2
+1, +2
+2, +1
+2, +1
LTPr=-3
+1, -3
-1, +3
+1, -3
-1, +3
OIpSum=+5
-1
+1
-5
+5
We see we have the sum we need in the last row. Then, the equation factors
0 = 2x2 + 5x − 3 = (2x − 1)(x + 3).
We now can solve
2x − 1 = 0 and x + 3 = 0.
The first one gives the solution x =
1
2
and second x = (−3).
For all the following problems, only the table, resulting equations and solutions will
be shown.
2. Solve 4x2 + 4x − 3 = 0 for x.
40
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
FTPr=4
+1, +4
+1, +4
+4, +1
+4, +1
+2, +2
+2, +2
LTPr=-3
+1, -3
-1, +3
+1, -3
-1, +3
+1, -3
-1, +3
OIpSum=+4
+1
-1
-11
+11
-4
+4
The equivalent forms of the equation are
0 = 4x2 + 4x − 3 = (2x − 1)(2x + 3).
The solutions are x =
1
2
and x = − 23 .
3. Solve 6a2 + 13a + 6 = 0 for a.
FTPr=6
+1, +6
+1, +6
+1, +6
+1, +6
+2, +3
+2, +3
LTPr=6
+2, +3
+3, +2
+6, +1
+1, +6
+1, +6
+3, +2
OIpSum=+13
+15
+20
+37
+12
+15
+13
The equivalent forms of the equation are
0 = 6a2 + 13a + 6 = (2a + 3)(3a + 2).
The solutions of the equation are a1 = − 32 and a2 = − 32 .
4. Solve 6x2 − x − 15 = 0 for x.
FTPr=6
+1, +6
+1, +6
+2, +3
+2, +3
LTPr=-15
-5, +3
+3, -5
-5, +3
+3, -5
OIpSum=-1
-27
+13
-11
-1
The equation is
0 = 6x2 − x − 15 = (2x + 3)(3x − 5).
The solutions of the equation are x1 = − 32 and x2 = 35 .
41
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
5. Solve 6x2 + 13x − 15 = 0 for x. We can use the table above to solve # 5 as well. The
table is reduced to only one row:
FTPr=6
+1, +6
LTPr=-15
+3, -5
OIpSum=+13
+13
The equation is
0 = 6x2 + 13x − 15 = (x + 3)(6x − 5).
The solutions of the equation are x1 = −3 and x2 = 56 .
42
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
3.5.3 BAd3: Quadratics using the quadratic formula, where the discriminant might be positive, negative or zero.
Solve a quadratic equation where the roots may not be rational numbers. Either we complete the square or use the quadratic formula.
We use the quadratic formula. It can be used in every case. If the equation is
ax2 + bx + c = 0
and we solve for x, the solutions of the equation are
x1,2 =
−b ±
√
b2 − 4ac
.
2a
This means there are two solutions:
x1 =
−b +
and
√
b2 − 4ac
2a
√
b2 − 4ac
.
2a
The other general way for solving quadratic equations is by completing the square. We
ilustrate it in the following problem.
x2 =
−b −
1. 2x2 + 5x − 4 = 0.
We add 4 to both sides of the equation.
2x2 + 5x = 4.
We divide both sides by 2.
5
x2 + x = 2.
2
Here is why this technique is called ”completing the square”. On the left side we
want to have a perfect square. The first term is x. We know the crossproduct is 25 x.
This is the double product of the first and the second term. We know that the first
one is x and determine the second one (called below b):
5
2 × x × b = x.
2
We solve for b: b = 54 . Therefore we need to add to both sides the square of b,i.e. ( 45 )2 :
5
x +2× x+
4
2
2
2
5
5
=2+
.
4
4
43
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
We factor the square on the left.
5
x+
4
2
=
57
.
16
We take square roots from both sides and have:
r
5
57
x+ =+
4
16
or
r
5
57
.
x+ =−
4
16
We have plus or minus because there are two numbers with square
(+3)2 = (−3)2 = 9. We solve for x:
√
5
57
,
x1 = − +
4
4
√
5
57
x2 = − −
.
4
4
p
57/16, just as
2. 4x2 − 4x − 5 = 0.
We will use the same method for this equation.
4x2 − 4x = 5,
(2x)2 − 2(2x) × 1 + 12 = 5 + 1,
(2x + 1)2 = 6.
√
2x + 1 = + 6
or
The solutions are x1 = (−1 +
√
√
2x + 1 = − 6.
6)/2 and x2 = (−1 −
√
6)/2.
3. 6a2 + 13a + 7 = 0:
Here we will use the quadratic formula. Since a = 6, b = 13 and c = 7 we use the
formulas:
√
−b + b2 − 4ac
x1 =
2a
and
√
−b − b2 − 4ac
.
x2 =
2a
x1 =
−(+13) +
p
(+13)2 − 4 × (+6) × (+7)
,
2×6
44
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
−13 ±
√
169 − 168
,
12
−13 + 1
x1 =
= −1.
12
x1 =
For x2 we have
7
−13 − 1
=− .
12
6
Then, the solutions are x1 = −1 and x2 = −7/6.
x2 =
4. x2 + 2x + 15 = 0.
−2 + 22 − 4 × 1 × 15
,
2×1
√
−2 + −56
x1 =
.
2
√
Recall the fact that −1 = i is an imaginary number which we assume has square
equal to (−1). We can write then
p
√
√ √
√
−56 = (−1)56 = −1 56 = 56i.
√
−2 + 56i
x1 =
.
2
We can factor out 2:
√
√
−2 − 22 × 14i
−2 + 56i
=
2
2
√
√
2(−1 + 14i
= −1 + 14i.
=
2
so the solution is
√
x1 = −1 + 14.
x1 =
The same way
√
x2 = −1 −
√
14i.
√
The solutions are x1 = −1 + i 14 and x2 = −1 − i 14.
5. 6x2 + 13x + 15 = 0.
x1 =
√
132 − 4 × 6 × 15
,
2×6
√
−13 + −191
x1 =
,
12
√
−13 + i 191
x1 =
.
12
−13 +
The same way:
√
−13 − i 191
x2 =
.
12
√
√
The solutions are x1 = (−13 + i 191)/12 and x2 = (−13 − i 191)/12.
45
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
3.5.4 BAd4: Equations quadratic in u, where u = x2 , x3 , ....
Solve an equation which is quadratic in a power of x.
To solve an equation which is quadratic in a power of x, we make use of the property of
exponents as in example shown below:
(x2 )3 = (x3 )2 = x2×3 = x6 .
The property states if we have a power of a variable (say x), we can break the exponent
down into two factors and when computing this way, the result is the same. You can
check for yourself (22 )3 = (23 )2 = 26 .
We can use this rule for rational exponents as well. Recall the denominator of the exponent is the root which we take to compute the number.
1. x2/3 + 3x1/3 + 2 = 0.
We see that 23 = 2 × 31 . We can simplify the equation by letting
x1/3 = u.
This is our substitution. Then, we have
u2 = x2/3 .
The equation becomes
u2 + 3u + 2 = 0.
We see we can solve this equation easily (u+1)(u+2) = 0. The solutions are u1 = −1
and u2 = −2.
The equation is not solved yet! We need to go back and use the substitution to find x:
x1/3 = −1.
We find
x1 = −1.
The other solution u2 = −2 will give us
1
x 3 = −2,
x2 = −8.
2. x6 + 3x3 + 2 = 0. We substitute
x3 = v.
The equation becomes
x6 + 3x3 + 2 = (x3 )2 + 3(x3 ) + 2 = v 2 + 3v + 2 = 0.
46
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
The last right equation is solved above. We have the solutions v1 = −1 and v2 = −2.
We now solve
x3 = −1,
x3 + 1 = 0.
We factor this equation and find (x + 1)(x2 − x + 1) = 0. Now, we use zero product
principle and find
x + 1 = 0,
x1 = −1.
Next, we solve
x2 − x + 1 = 0.
Using the quadratic formula, we find two complex number roots:
p
1 + (−1)2 − 4 × 1 × 1
x2 =
,
2
√
3
1
.
x2 = + i
2
2
The other root is
√
1
3
x3 = − i
.
2
2
For the second solution of u we solve the same way:
x3 = −2,
x3 + 2 = 0,
√
√
√
3
3
3
(x + 2)(x2 − 2x + ( 2)2 ) = 0.
This equation has three other roots:
√
3
x4 = − 2.
√
√
To ease the computations, we first compute the discriminant of (x2 − 3 2x+( 3 2)2 ) =
0:
p
√
√
3
3
D = (− 2)2 − 4 × 1 × ( 2)2 ) = −3 3 (22 ).
The remaining solutions are
√
3
x5 =
q p
2 + −3 3 (22 )
2
,
√
√
2
3
3
x5 =
+i 2
2
2
and
√
√
3
√
2
3
3
−i 2
.
x6 =
2
2
√
6
3
3/2,
All the solutions
of
the
equation
x
+
3x
+
2
=
0
are
x
=
−1,
x
=
1/2
+
i
1
2
√
√
√
√
√
√
x3 = 1/2 − i 3/2, x4 = − 3 2, x5 = 3 2(1/2 + i 3/2) and x6 = 3 2(1/2 − i 3/2).
√
3
47
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
3. 2x4 − 3x2 + 1 = 0.
We substitute x2 = u.
2u2 − 3u + 1 = 0.
We use FOIL and find (2u − 1)(u − 1) = 0. Thus, u1 = 12 and u2 = 1. We find all the
solutions for x now:
1
x2 = ,
2
r
√
1
2
x1,2 = ±
=±
.
2
2
√
√
The first two solutions are x1 = 2/2 and x2 = − 2/2. Finally, we solve
x2 = 1,
x3,4 = ±1.
The other two solutions are x3 = 1 and x4 = −1.
4. x + 5 + 4x−1 = 0.
This equation is not in a quadratic form for any power of x. But, we can transform
it in this form by multiplying it by x:
(x + 5 + 4x−1 )x = 0 × x
x2 + 5x + 4 = 0
This is a regular quadratic equation and we solve it for x:
(x + 4)(x + 1) = 0
The solutions are x1 = −4 and x2 = −1.
5. x2 − 2 + x−2 = 0.
We use the same idea for this equation (but multiplying by x2 ):
x4 − 2x2 + 1 = 0
We substitute x2 = u:
u2 − 2u + 1 = 0
This equation is (u − 1)2 = 0. Therefore, there is only one repeated solution u1,2 = 1.
We solve for x now and find x1 = 1 and x2 = −1. The original equation is not a
polynomial equation so that it has only two solutions as stated above.
48
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
3.5.5 BAd5: Rational equations leading to quadratic equations.
Solve a rational equation which can be reduced to a quadratic equation.
We need to transform the rational equation to quadratic equation by using a process
known as ”clearing fractions”.
We multiply the equation by the Least Common Denominator of all rational fractions in the
equation. This is not all. We need to list all restrictions, i.e., all values of x for which the
equation is defined. Recall we cannot divide by zero.
1.
2x − 5
= x + 2.
x+1
We see that the LCD = x + 1. This denominator must not be equal to zero, so we
need x 6= −1. We multiply now the equation by (x + 1):
2x − 5 = (x + 2)(x + 1).
We simplify it and solve for x:
2x − 5 = x2 + 3x + 2,
x2 + x + 7 = 0.
We use the quadratic formula:
x1 =
−1 +
x2 =
−1 −
and
√
√
12 − 4 × 1 × 7
2
12 − 4 × 1 × 7
.
2
There are two complex number solutions:
√
3 3
1
x1 = − + i
2
2
and
2.
√
1
3 3
x2 = − − i
.
2
2
2x − 5
x+2
=
.
x+1
x−1
LCD = (x + 1)(x − 1).
We restrict x:
x 6= −1 and x 6= +1.
We can clear the fractions now:
(2x − 5)(x − 1) = (x + 2)(x + 1),
49
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
2x2 − 7x + 5 = x2 + 3x + 2,
x2 − 10x + 3 = 0.
We use quadratic formula:
x1 =
10 +
p
(−10)2 − 4 × 1 × 3
2
10 −
p
(−10)2 − 4 × 1 × 3
.
2
and
x2 =
The solutions are
x1 = 5 +
and
x2 = 5 −
3.
√
√
22
22.
2x − 5
= 3x + 2.
x+1
LCD = x + 1, then x 6= −1.
Clearing fractions:
2x − 5 = (3x + 2)(x + 1),
2x − 5 = 3x2 + 5x + 2,
3x2 + 3x + 7 = 0.
The solutions are
x1 =
−3 +
x2 =
−3 −
and
√
√
32 − 4 × 3 × 7
2×3
32 − 4 × 3 × 7
,
2×3
i.e.
√
75
1
x1 = − + i
2
6
and
4.
√
75
1
x2 = − − i
.
2
6
1
1
+
= 3.
x+1 x−1
LCD = (x + 1)(x − 1), thus x 6= −1 and x 6= +1.
Clearing fractions:
1
1
+
(x + 1)(x − 1) = 3(x + 1)(x − 1).
x+1 x−1
50
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
We open the parenthesis on the left and remove the common factors:
(x − 1) + (x + 1) = 3(x2 − 1).
We solve for x:
3x2 − 3 = 2x,
3x2 − 2x − 3 = 0.
The solutions are:
x1 =
2+
p
2−
p
(−2)2 − 4 × 3 × (−3)
,
2×3
and
x1 =
i.e.
(−2)2 − 4 × 3 × (−3)
2×3
√
1
10
x1 = +
3
3
and
√
1
10
.
x2 = −
3
3
3x − 5
= 2x − 1.
x−1
LCD = x − 1, thus x 6= 1.
5. Solve
Clearing fractions:
3x − 5 = (2x − 1)(x − 1),
3x − 5 = 2x2 − 3x + 1,
2x2 − 6x + 6 = 0,
x2 − 3x + 3 = 0.
The solutions are:
x1 =
3+
p
(−3)2 − 4 × 1 × 3
2×1
3−
p
(−3)2 − 4 × 1 × 3
,
2×1
and
x2 =
i.e.
√
3
3
x1 = + i
2
2
and
√
3
3
.
x2 = − i
2
2
51
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
3.5.6 BAd6: Rational equations leading to linear equations.
Solve a rational equation that reduces to a linear equation.
There is no difference between solving rational equations leading to quadratic equations
and those leading to linear equations. We simply have only one solution.
1.
2x2 + 5
= x + 2.
2x + 1
LCD = 2x + 1, thus x 6= −1/2.
Clearing fractions:
2x2 + 5 = (x + 2)(2x + 1),
2x2 + 5 = 2x2 + 5x + 2.
We solve for x:
5x = 3,
3
x= .
5
2.
x+2
2x − 5
=
.
2x + 1
x−1
LCD = (2x + 1)(x − 1), thus x 6= −1/2 and x 6= 1.
Clearing fractions and solving:
(2x − 5)(x − 1) = (2x + 1)(x + 2),
2x2 − 7x + 5 = 2x2 + 5x + 2,
−12x = −3,
1
x= .
4
3.
3x2 − 5
= 3x + 2.
x+1
LCD = x + 1, thus x 6= −1.
Clearing fractions and solving:
3x2 − 5 = (3x + 2)(x + 1),
3x2 − 5 = 3x2 + 5x + 2,
5x = −7,
7
x=− .
5
52
CHAPTER 3. ALGEBRA
4.
3.5. QUADRATIC/RATIONAL EQUATIONS
1
x
+
= 1.
x+1 x−1
LCD = (x + 1)(x − 1), thus x 6= ±1.
We clear fractions and solve:
1
x
+
(x + 1)(x − 1) = 1 × (x + 1)(x − 1),
x+1 x−1
x(x − 1) + 1 × (x + 1) = x2 − 1,
x2 − x + x + 1 = x2 − 1.
We have now
+1 = −1 FALSE!
We see this equation does not have any solution.
5.
4x2 − 5x + 7
= 2x − 1.
2x − 1
LCD = 2x − 1, thus x 6= 1/2.
Clearing fractions and solving:
4x2 − 5x + 7 = (2x − 1)(2x − 1),
4x2 − 5x + 7 = 4x2 − 4x + 1,
−x = −6,
x = 6.
53
3.5. QUADRATIC/RATIONAL EQUATIONS
CHAPTER 3. ALGEBRA
3.5.7 BAd7: Solve literal equations involving rational expressions.
Solve a rational equation in several variables for a specific variable.
We use the same method when solving literal equations as when we use numbers. The
only difference is we keep the expression with all variables. Keep in mind variables represent numbers and, if we cannot do something with numbers, we cannot do it with
variables too (example: dividing by zero is not defined so we cannot allow to have denominators equal to zero).
Solve each of the following equations for b.
1.
1 1
+ = 1.
b a
We isolate on the first step 1/b by subtracting 1/a from both sides and simplify:
1
a−1
1
=1− =
.
b
a
a
We take the reciprocal of both sides (after restricting a 6= 1 since 1/b cannot be zero):
b=
2.
a
.
a−1
a c
+ = 1.
b d
To isolate b on one side, we first subtract the fraction c/d from both sides and simplify:
a
c
d−c
=1− =
.
b
d
d
db
We multiply both sides by
( after restricting c 6= d):
d−c
d
c−d
d
a
×b
=
×
b.
b
c−d
d
c−d
Dividing out common factors will give us
b=
3.
ad
.
d−c
1 1
1
+ =
.
b a
a+b
We can clear fractions as we have it done before, restricting a 6= 0 b 6= 0 and a 6= −b.
We can multiply by ab(a + b):
a(a + b) + b(a + b) = ab,
54
CHAPTER 3. ALGEBRA
3.5. QUADRATIC/RATIONAL EQUATIONS
a2 + ab + ab + b2 = ab.
We subtract ab from both sides of the equation and see that we have a quadratic
equation for b. We solve it for b using quadratic formula:
b1,2
Then,
4.
b2 + ab + a2 = 0,
√
−a ± a2 − 4 × 1 × a2
=
.
2×1
√
3|a|
a
,
b1 = − + i
2
2
√
3|a|
a
.
b2 = − − i
2
2
a b
+ = 1.
b a
We clear fractions after restricting a 6= 0 and b 6= 0:
b
a
ab + = 1ab,
b
a
a2 + b2 = ab.
We subtract ab from both sides of the equation. This is a quadratic equation for b.
We solve it using quadratic formula:
b1,2
Then,
5.
b2 − ab + a2 = 0,
p
−(−a) ± (−a)2 − 4 × 1 × a2
=
.
2×1
√
a
3|a|
,
b1 = + i
2
2
√
3|a|
a
b2 = − i
.
2
2
1 1
− = 2.
b a
The restrictions are a 6= 0 and b 6= 0. We can proceed as in #1: we isolate 1/b on one
side of the equation, and simplify.
1
2a + 1
1
=2+ =
.
b
a
a
We cannot have the left side zero, so we restrict the right side not to be equal to zero
a 6= −1/2. We then take the reciprocal of both sides:
b=
55
a
.
2a + 1
3.6. RADICALS AND FRACTIONIAL EXPONENTS
3.6
CHAPTER 3. ALGEBRA
Radicals and Fractionial exponents
3.6.1 BBa4: Express radicals in terms of fractional exponents and vice
versa.
Rewrite each expression using
the alternate
form for expressing roots. Do not simplify.
√
√
n
n
m
1/n
Recall that A is actually A and ( B) is B m/n :
A1/n =
√
n
A
and
√
n
( B)m = B m/n .
1.
√
36
m = 1 and n = 2.
= 361/2 .
2.
√
3
−27
m = 1 and n = 3.
= (−27)1/3 .
3. 1003/2
m = 3 and n = 2.
4. 255/2
n = 2 and m = 1.
5. (−32)2/5
m = 2 and n = 5.
√
= ( 100)3 .
√
= ( 25)5 .
√
= ( 5 −32)2 .
56
CHAPTER 3. ALGEBRA
3.6. RADICALS AND FRACTIONIAL EXPONENTS
3.6.2 BBa1: Evaluate numerical expressions involving radicals and/or
fractional exponents.
Simplify numerical expressions involving roots in either radical or fractional exponent
form.
1.
√
36
= 6.
2.
√
3
−27
= −3.
3. 1003/2
√
= ( 100)3 = 103 = 1000.
4. 255/2
√
= ( 25)5 = 55 = 3125.
5. (−32)2/5
√
= ( 5 −32)2 = (−2)2 = 4.
57
3.6. RADICALS AND FRACTIONIAL EXPONENTS
CHAPTER 3. ALGEBRA
3.6.3 BBa2: Simplify radical expressions.
Reduce and combine radicals in each radical expression. You may assume all variables
represent positive quantities.
Recall a perfect square is a number or expression which is a square of a rational number
or another expression. Example 4 = 22 and a4 = (a2 )2 . The same way we can define
perfect third power, fourth power and so on. In general, perfect n-th power is a number
or expression which is an n-th power of a number or expression.
1.
p
3
16x5 y 3 /z 6
We see that 16 = 23 × 2 and x5 = x3 x2 . The other expressions y 3 and z 6 are perfect
third powers (also known as perfect cube).
p
= 3 (23 x3 y 3 /z 3 )(2x2 )
p
3√
3
= 3 (2xy/z 2 ) 2x2
√
2xy
2xy √
3
3
= ( 2 )3/3 2x2 = 2 2x2 .
z
z
2.
3.
4.
5.
p
4
81x5 y 9 /z 6
√
45 +
√
80 −
p
3
27x7 y 3 /z 8
p
12x4 y 3
p
4
(34 x4 y 8 /z 4 )(xy/z 2 )
p
= [(3xy 2 )4 ]1/4 4 xy/z 2
r
3xy 2 4 xy
.
= 2
z
z2
=
√
20
√
√
√
32 × 5 + 42 × 5 − 22 × 5
√
√
√
√
= 3 5 + 4 5 − 2 5 = 5 5.
=
p
3
(33 x6 y 3 /z 6 )(x/z 2 )
p
= [(3x2 y/z 2 )3 ]1/3 3 x/z 2
r
3x2 y 3 x
.
= 2
z
z2
=
p
= (22 x4 y 2 )(3y)
1/2 p
= (2x4 y)2
3y
p
= 2x2 y 3y.
58
CHAPTER 3. ALGEBRA
3.6. RADICALS AND FRACTIONIAL EXPONENTS
3.6.4 BBa3: Rationalize the numerator or denominator of a quotient involving radicals.
To rationalize the numerator or denominator of an expression, we use the following formula:
(A + B)(A − B) = A2 − B 2 .
We will multiply by an expression of one, containing the conjugate of the radical number.
The conjugate of A + B is A − B, and the conjugate of A − B is A + B.
√
√
x− y
1.
(x − y)
√
√
√
√
The conjugate of√( x − y) is ( x + y). Therefore, we need to multiply with ex√
x+ y
pression of one √x+√y :
√
√
√
√
x− y
x+ y
×√
=
√
x−y
x+ y
√
√
( x)2 − ( y)2
√
=
√
(x − y)( x + y)
x−y
1
1
=
×√
√ =√
√ , if x 6= y.
x−y
x+ y
x+ y
√
√
x+ y
2.
x+y
√
√
The conjugate is x − y.
√
√
√
√
x+ y
x− y
=
×√
√
x+y
x− y
=
(x − y)
√
√ , if x 6= y.
(x + y)( x − y)
√
√
x+2− x−2
3.
4
√
√
The conjugate is x + 2 + x − 2.
√
√
√
√
x+2− x−2
x+2+ x−2
√
×√
=
4
x+2+ x−2
√
√
( x + 2)2 − ( x − 2)2
√
√
=
4( x + 2 + x − 2)
x+2−x+2
√
= √
4( x + 2 + x − 2)
1
√
.
=√
x+2+ x−2
59
3.6. RADICALS AND FRACTIONIAL EXPONENTS
CHAPTER 3. ALGEBRA
3x
4. √
√
x2 + 1 − y
We need now the√conjugate of the denominator:
√
The conjugate is x2 + 1 + y.
√
√
x2 + 1 + y
3x
=√
√ ×√ 2
√
x2 + 1 − y
x +1+ y
√
√
3x( x2 + 1 + y)
=
if x2 + 1 6= y.
2
x +1−y
5
5. √
√
x− y
√
√
The conjugate is x + y.
√
√
x+ y
5
=√
√
√ ×√
x− y
x+ y
√
√
5( x + y)
=
, if x 6= y.
x−y
60
CHAPTER 3. ALGEBRA
3.6. RADICALS AND FRACTIONIAL EXPONENTS
3.6.5 BBa5: Multiply radical expressions (monomials, binomials).
Multiply the radical expression. You may assume all variables represent positive quantities. Combine like terms, but do not simplify otherwise.
√
√
1. 2x + 3y 3
p
= 3(2x + 3y).
p
√
2. 4 81x5 y 9 4 7xyz
p
p
= 4 (81x5 y 9 )(7xyz) = 4 567x6 y 10 z.
√ √
3. 45 10
√
√
= 45 × 10 = 450.
p
√
4. 3 27x7 y 3 3 x + y
p
= 3 (27x7 y 3 )(x + y).
p
p
5. 3 12x4 y 3 3 5x2 y 3
p
= 3 (12x4 y 3 )(5x2 y 3 ).
61
3.6. RADICALS AND FRACTIONIAL EXPONENTS
CHAPTER 3. ALGEBRA
3.6.6 BBa6: Simplify and combine like radicals.
Simplify the radical expression. You may assume all variables represent positive quantities. Combine like terms and extract all possible roots.
Recall that a square root of a number, is a number which square is the given number. We
call a number or expression a perfect square if it is a square of another rational number or
expression. We simplify by extracting the largest possible perfect square.
The same way is defined perfect cube — third power of a number, perfect forth and so
on...
1.
2.
3.
√ √
54 3
p
p
4
81x5 y 9 4 7xy 2 z
√ √
45 20
√
√
= 54 × 3 = 2 × 33 × 3
√
√ √
√
= 81 × 2 = 81 2 = 9 2.
p
4
(81x4 y 8 )(7x2 y 3 z)
p
1p
= [(34 xy 2 )4 ] 4 4 7x2 y 3 z = 3xy 2 4 7x2 y 3 z.
=
=
4.
5.
p
√
3
27x7 y 3 3 3x + 6y
p
p
3
12x4 y 5 3 5x2 y 4
√
√
32 × 5 × 5 × 22 = 32 × 52 × 22
√
= 302 = 30.
p
= 3 (33 x6 y 3 )x(3x + 6y)
1/3 p
3
= (3x2 y)3
x(3x + 6y)
p
= 3x2 y 3 x(3x + 6y).
p
3
60x4+2 y 5+4
√
1/3 √
3
3
60 = x2 y 3 60.
= (x2 y 3 )3
=
62
CHAPTER 3. ALGEBRA
3.6. RADICALS AND FRACTIONIAL EXPONENTS
3.6.7 BBa7: Simplify expressions involving radicals by changing to fractional exponents and using exponent laws.
Use fractional exponents to combine like terms.
Recall the index of the radical is the denominator of the exponent fraction. We multiply
exponents with one and the same base by adding the exponents. We use the same rules
for adding fractions as usual.
1.
p
√
4
x5 y 3 xy
= (x5 y)1/4 (xy)1/3 = x5/4+1/3 y 1/4+1/3
= x19/12 y 7/12 .
2.
3.
4.
5.
p
p
3
x4 y 2 2 2xy 3
p
p
5
x4 y 2 3 x3 y 5
p
p
2
x4 y 2 4 2xy 3
p
p
3
x5 y 4 6 x5 y
= (x4 y 2 )1/3 (2xy 3 )1/2
= x4/3+1/2 y 2/3+3/2 21/2 = x11/6 y 13/6 21/2 .
= (x4 y 2 )1/5 (x3 y 5 )1/3
= x4/5+1 y 2/5+5/3 = x9/5 y 31/15 .
= (x4 y 2 )1/2 (2xy 3 )1/4
= 21/4 x2+1/4 y 1+3/4 = 21/4 x9/4 y 7/4 .
= (x5 y 4 )1/3 (x5 y)1/6
= x5/3+5/6 y 4/3+1/6 = x5/2 y 3/2 .
63
3.7. NEGATIVE EXPONENTS
3.7
CHAPTER 3. ALGEBRA
Negative Exponents
3.7.1 BBb2: Use reciprocal rule to eliminate denominator negative exponents.
Write each expression so it has no negative exponents.
Recall for any real number a and and any integer n we have
a−n =
1
.
an
3−2 =
1
32
Examples:
and
1
= 53 .
5−3
It turns out removing the negativity of the exponent means to raise to the opposite positive exponent the reciprocal of the number-base.
All the laws of exponents hold when using negative exponents.
1.
x3 y 4
x−3 y −1
= x3 y 4 (x3 y 1 ) = x3+3 y 4+1 = x6 y 5 .
We see, in this example, that the reciprocal of a number could be expressed as the
number raised to power (−1).
2.
6x2 y −5
3x−3 y −7
= 2x2 y −5 (x3 y 7 ) = 2x2+3 y −5+7 = 2x5 y 2 .
We see the laws of exponents hold when the exponents are negative.
3.
4.
w2 y 3
x−4 y −2
= w2 y 3 (x4 y 2 ) = x4 y 5 w2 .
x3/2 y 4
x−1/2 y −2
= x3/2 y 4 (x1/2 y 2 ) = x(3/2)+(1/2) y 6 = x2 y 6 .
Note the laws hold even when the exponents are fractions.
64
CHAPTER 3. ALGEBRA
5.
x−3/2 y 1/2
x−3 y −1
3.7. NEGATIVE EXPONENTS
= x−3/2 y 1/2 (x3 y) = x(−3/2)+3 y (1/2)+1 = x3/2 y 3/2 .
In conclusion, example #2 underlines the use of exponent laws with negative numbers
and #4 — fractional exponents. In all cases, when the base is the same, we multiply by
adding exponents and divide by subtracting the exponents.
65
3.7. NEGATIVE EXPONENTS
CHAPTER 3. ALGEBRA
3.7.2 BBb3: Simplify negative powers of sums and differences.
Write each expression so that it has no negative exponents.
We will see the base could not only be a number but it could be an expression containing addition and subtraction.
To remove the negativity of the base, we exchange the base with its reciprocal. Example:
−2
4
2
1
1
=
= 2 .
4
4
1. (x + 2)−1
=
2. (x − 2)−2
=
1
.
x+2
1
.
(x − 2)2
3. (x + y + 2)−2
=
1
.
(x + y + 2)2
4. (3x + 2)−2 =
=
1
.
(3x + 2)2
=
1
.
(x − 1)3
5. (x − 1)−3
66
CHAPTER 3. ALGEBRA
3.7. NEGATIVE EXPONENTS
3.7.3 BBb4: Evaluate or simplify expressions involving negative exponents.
1. If a = 2 and b = 5 then a−2 + b
= 2−2 + 5 =
1
21
1
+5=5 = .
2
2
4
4
2. If a = −2 and b = 5 then a−3 + b−1
= (−2)−3 + 5−1 =
1
1
+ 1
3
(−2)
5
−5 + 8
3
1 1
= .
=− + =
8 5
8×5
40
3. Combine like terms and eliminate the negative exponents:
a+b
+ b−1
a−1
=
=
a+b
1/a + 1/b
a+b
(b + a)/(ab)
Recall that dividing a number to another number is actually multiplication with the
reciprocal of the second number:
= (a + b) ×
ab
= ab, if a 6= −b.
a+b
67
3.7. NEGATIVE EXPONENTS
CHAPTER 3. ALGEBRA
4. Combine like terms and eliminate the negative exponents:
2a2 + b
a−2 + b−1
We use the same method:
2a2 + b
=
1/a2 + 1/b
2a2 + b
=
(b + a2 )/(a2 b)
=
(2a2 + b)a2 b
.
a2 + b
5. Combine like terms and eliminate the negative exponents:
2a2 b
a−2 + b−3
=
=
=
2a2 b
1/a2 + 1/b3
2a2 b
(b3 + a2 )/(a2 b3 )
2a4 b4
(2a2 b)(a2 b3 )
=
.
b 3 + a2
a2 + b 3
68
CHAPTER 3. ALGEBRA
3.8
3.8. ABSOLUTE VALUE
Absolute Value
3.8.1 BBc1: Evaluate numerical expressions involving absolute values.
Give the value of an expression involving absolute value.
Example: If x = 2 then |x − 1| − |3 − x| = 0.
Recall the absolute value of a number is the number without its sign. So, | + 2| = 2 and
| − 3| = 3.
1. If x = 5, then |3x − x2 | − |3 − 2x|
= |3 × 5 − 52 | − |3 − 2 × 5|
= | − 10| − | − 7|
= 10 − 7 = 3.
2. If x = −5, then −|x − x2 | + |3 + 2x|
= −|(−5) − (−5)2 | + |3 + 2 × (−5)|
= −| − 5 − 25| + |3 − 10|
= −30 + 7 = −23.
3. If x = 0, then |3x − x2 | − | − 3 − 2x|
= |3 × 0 − 02 | − | − 3 − 2 × 0|
4. If x = −3, then
√
= 0 − 3 = −3.
x2 − |x|
=
p
(−3)2 − | − 3|
Recall the square root is only the positive square root of a number, so
√
9 = 3, therefore
= 3 − 3 = 0.
5. If x = −1, then |3 − 4x2 | + |8 − 2x| + x
= |3 − 4(−1)2 | + |8 − 2(−1)| + (−1)
= |3 − 4| + |8 + 2| − 1 = 1 + 10 − 1 = 10.
69
p
(−3)2 =
3.8. ABSOLUTE VALUE
CHAPTER 3. ALGEBRA
3.8.2 BBc2: Recognize the graph of y = |x| and its variations.
Example: Which of the following is the graph of y = |x| + 1?
EXAMPLE EXPLANATION: We can consider the most basic absolute value graph.
What we need is another expression, whose graph we know. This is the linear expression.
The absolute value is the number without its sign. We can define:
|x| = −x when x < 0
y=
|-3|= -(-3)=3
|x|
When x is negative,
then absolute value of
x is the opposite: -x
and
|x| = +x when x ≥ 0.
|2| = 2
y=
|x|
When x is positive,
absolute value of x is
the number itself.
Both are parts of a line but not the whole line. The graph of absolute value function is
given below.
It appears as the latin letter V. All functions with absolute value have this basic shape.
The other characteristics of this graph are its vertex and axis of symmetry. The vertex for
this graph is point (0, 0) and axis of symmetry is x = 0.
70
CHAPTER 3. ALGEBRA
3.8. ABSOLUTE VALUE
Basic shape of y = |x|
a)
b)
c)
10
–10 –8
–6
–4
d)
10
10
10
8
8
6
y
4
6
6
6
4
4
4
2
2
2
–2 0
–2
–2 0
–2
–2 0
–2
2
4 x 6
8
–10 –8
10
–6
–4
8
2
4 x 6
8
10
–10 –8
–6
–4
8
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–4
–4
–6
–6
–6
–6
–8
–8
–8
–8
–10
–10
–10
–10
2
4 x 6
8
10
2
4 x 6
8
10
Only graph b) has desired shape.
Solution: b
1. Circle the graph of y = |3x| + 1.
a)
b)
–10 –8
–6
–4
c)
d)
10
10
10
8
8
8
6
6
y
4
6
6
4
4
4
2
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
10
8
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–4
–4
–6
–6
–6
–6
–8
–8
–8
–8
–10
–10
–10
–10
There are three graphs with desired shape. We consider a), c) and d). Note the function is symmetric about x = 0. This means if we replace a number, say x = 2, with
its opposite, x = −2 the function value is the same. Indeed, 6 = |3 × 2| = f (2) and
6 = |3 × (−2)| = f (−2)|. Only graphs a) and d) are symmetric about x = 0 so we only
consider them. The vertices are at (0, 10) for a) and (0, 1) for d). Only graph d) has the
vertex on the same place as the function y = |3x| + 1.
Solution: d.
71
3.8. ABSOLUTE VALUE
CHAPTER 3. ALGEBRA
2. Circle the graph of y = −|3x| + 10.
a)
b)
c)
10
8
–10 –8
–6
10
8
10
8
8
6
6
6
6
4
4
4
4
2
2
2
–2 0
–2
–4
d)
10
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–4
–4
–6
–6
–6
–6
–8
–8
–8
–8
–10
–10
–10
–10
2
One way to find the proper graph is to compute a few points using the formula given
and compare them with corresponding points from the graphs. We see now:
f (0) = −|3 × 0| + 10 = 10 - it matches graph a).
We compare it with the other graphs (instructive): for b) the graph shows f (0) = 4 - it
is not the same value (10), the same way f (0) = −8 for graph c) and f (0) = 1 for d).
Therefore, the graph is a).
Solution: a.
3. Circle the graph of y = |3 − x| + 1.
a)
b)
–10 –8
–6
–4
c)
d)
10
10
10
8
8
8
10
8
6
6
6
6
4
4
4
4
2
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–4
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–4
–2 0
–2
2
4 x 6
8
10
–4
–6
–6
–6
–6
–8
–8
–8
–8
–10
–10
–10
–10
We find the vertex and axis of symmetry. We need to find what the axis of symmetry
is for this function. The lowest point of the function is when the absolute value has its
minimum. The minimum value for the absolute value function is 0. For this function it
can be obtained when x = 3, i.e. we solve |3 − x| = 0. We find that x = 3. This is our axis
of symmetry.
Looking at the picture we see the graphs a) and c) are symmetric about x = 0. Graph b) is
not symmetric so it is not a solution. We are left with choice d). This graph is symmetric
about x = 3.
Solution: d.
4. Circle the graph of y = |2x + 4| − 3.
a)
b)
–10 –8
–6
–4
c)
d)
10
10
10
8
8
8
10
8
6
6
6
6
4
4
4
4
2
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–4
–4
–6
–6
–6
–6
–8
–8
–8
–8
–10
–10
–10
–10
2
4 x 6
8
10
We find the minimum value the function can have: it is when the absolute value has its
minimum 0. That minimum is when |2x + 4| = 0 or x = −1/2. Then, f (−1/2) = −3. The
only graph with minimum value of f (x) = −3 is b).
72
4 x 6
8
10
CHAPTER 3. ALGEBRA
3.8. ABSOLUTE VALUE
Note we can use any other point to find the graph. For example,we compute the value
of all functions at x = −2 and compare them with the analytical value which is y(−2) =
|2 × (−2) + 4| − 3 = −3. Graph a) shows value of (+3), graph b) - (−3) and graph c) - (+6).
Graph d) is undefined for the point x = (−2). The only graph which matches is b).
Solution: b.
5. Circle the graph of y = |3(x − 5)| − 7.
a)
b)
c)
10
8
–10 –8
–6
–4
d)
10
10
8
6
6
4
4
2
2
2
–2 0
–2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
10
8
6
4
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
8
6
4
2
2
4 x 6
8
10
–10 –8
–6
–4
–2 0
–2
–4
–4
–4
–4
–6
–6
–6
–6
–8
–8
–8
–8
–10
–10
–10
–10
2
4 x 6
8
10
We need to consider graphs b), c) and d).We again will use the minimum value of the
function. It is (−7). The only function with minimum value of (−7) is c) so it is the
solution.
Solution: c.
73
3.8. ABSOLUTE VALUE
CHAPTER 3. ALGEBRA
3.8.3 BBc3: Solve equations of the type |ax + b| + c = d.
Solve each equation for x if possible. If it is not possible, state why.
1. For solving |2x + 4| + 3 = 9 we first isolate the absolute value on one side of the
equality:
|2x + 4| = 6
We use the fact that absolute value is the number without its sign. Therefore, we
need to solve two equations:
2x + 4 = 6
and
2x + 4 = −6
(we use the fact | + 6| = | − 6| = 6). The solutions are:
2x = 6 − 4,
x1 = 1.
2x = −6 − 4,
x2 = −5.
The solutions are x1 = 1 and x2 = −5.
2. |5x + 3| − 3 = 8.
|5x + 3| = 8 + 3 = 11.
The equations are: 5x + 3 = 11 and 5x + 3 = −11.
5x + 3 = 11,
5x = 8,
8
x1 = .
5
5x + 3 = −11,
5x = −14.
x=−
The solutions are x1 = 8/5 and x2 = −14/5.
74
14
.
5
CHAPTER 3. ALGEBRA
3.8. ABSOLUTE VALUE
3. |x − 4| + 3 = 1.
We isolate the absolute value expression:
|x − 4| = 1 − 3 = −2.
We have on the left side a positive number, and on the right side — negative. These
two numbers could never be equal. Therefore, this equation does not have a solution.
4. |3x − 4| + 7 = 11.
|3x − 4| = 11 − 4.
The equations are 3x − 4 = 4 and 3x − 4 = −4.
3x − 4 = 4
3x = 8,
8
x1 = .
3
3x − 4 = −4,
3x = 0,
x2 = 0.
The solutions are x1 = 8/3 and x2 = 0.
5. |4 − 3x| − 3 = 8.
|4 − 3x| = 8 + 3 = 11.
The equations are 4 − 3x = 11 and 4 − 3x = −11.
4 − 3x = 11,
3x = −7,
7
x1 = − .
3
4 − 3x = −11,
3x = 15,
x2 = 3.
The solutions are x1 = −7/3 and x2 = 3.
75
3.9. LINEAR INEQUALITIES
3.9
CHAPTER 3. ALGEBRA
Linear Inequalities
3.9.1 BBd1: Solve or graph linear inequalities in one variable.
Solve each inequality algebraically and graphically. Give your answer algebraically and
graph it on a number line.
Example: If 2x + 5 ≤ 9 then x ≤ 2.
1. 3x + 4 < 7.
On the first step we add (−4) to both sides of the inequality:
3x + 4 + (−4) < 7 + (−4) = 3.
On the second step we multiply with the reciprocal of coefficient in front of x:
3x ×
1
1
<3× ,
3
3
x < 1.
x
-
+
1
We solve all the inequalities the same way. To make it easier, we always transfer x to
the side it will have a positive coefficient. When we multiply by its reciprocal, this
number is positive, it does not change the way of the inequality: We have greater or
equal, or less or equal ...
2. 3x + 7 ≤ −4x + 8.
We add this time (4x − 7) to both sides.
3x + 7 + (4x − 7) ≤ −4x + 8 + (4x − 7),
7x ×
1
1
≤1× ,
7
7
1
x≤ .
7
76
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
xx
-
x
xx
xx
1/7
+
3. 9 − x ≥ 5x + 3.
9 − x + (x − 3) ≥ 5x + 3 + (x − 3),
1
1
6 × ≥ 6x × ,
6
6
1 ≥ x.
xx
-
x
xx
xx
1
+
4. 12x + 4 < 4x − 8.
12x + 4 + (−4x − 4) < 4x − 8 + (−4x − 4),
1
1
8x × < −12 × ,
8
8
3
x<− .
2
X +
-3/2
5. 11x + 3 > 19x − 5.
11x + 3 + (−11x + 5) > 19x − 5 + (−11x + 5),
8×
1
1
> 8x × ,
8
8
1 > x.
77
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
x
1
78
+
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
BBd2: Solve or graph linear inequalities in two variables.
Solve each inequality for the indicated variable and graph your solution.
When solving an equation or inequality in one variable, we have zero, one or more
numbers as solutions. We can look at these numbers as points on the number line and we
can plot them. In this case, we can think of them as a set of solutions.
Analogously, we can think of the solutions of inequality in two variables as the set containing all ordered pairs of numbers which make the inequality true sentence. When plotting the solution, we have it as a region in (x, y) plane. When solving linear inequality in
two variables, the solution set contains half of plane.
1. Solve for y and graph your solution: 2x − 3y < 4x − 9.
2x − 3y + (3y − 4x + 9) < 4x − 9 + (3y − 4x + 9),
(−2x + 9) ×
1
1
< 3y × ,
3
3
2
− x + 3 < y,
3
2
y + x − 3 > 0.
3
We see we have the line y + 2/3x − 3 = 0 as a border between two regions in x, y
plane — y + 2/3 − 3 > 0 and y + 2/3x − 3 < 0. To find which one is bigger, we choose
any point which is not on the line. One good test point is (0, 0), if it is not on the
line. We check:
2
0 + × 0 − 3?0.
3
The answer is that this point is in the half plane y + 2/3x − 3 < 0. The solution of the
inequality is the other half of the plane. The line is dashed because we have strict
inequality and therefore, the border (the line) is not part of the solution.
THE TOP HALF OF PLANE
IS THE SOLUTION.
y
Y=
-2
/3X
+3
X
79
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
2. Solve for x and graph your solution: 5x − 3y ≤ 4x − 9.
5x − 3y − 4x + 9 ≤ 4x − 9 − 4x + 9,
x − 3y + 9 ≤ 0,
and solved for x
x ≤ 3y − 9.
We check x ≤ 3y − 9, using again (0, 0) and find:
0≤3×0−9
FALSE!
therefore, the half of plane containing (0, 0) is not the solution of the problem. The
dividing line is part of the solution because we have a less or equal sign.
THE SOLUTION IS THE BOTTOM
PART OF THE PLANE (UNDER THE LINE)
Y
X
-3
y=1/3x
3. Solve for y and graph your solution: 2x − 3y > x − y + 99.
2x − 3y + (−2x + y) > x − y + 99 + (−2x + y),
−2y > −x + 99,
99
y < 1/2x − .
2
We check with point (0, 0):
99
1
0? × 0 − ,
2
2
99
0>− .
2
We see that ordered pair (0, 0) is not a solution of the inequality, therefore that part
of the plane is not the solution set. The border line is dashed because it is not part
of the solution.
80
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
y
/2
-99
2x
=1/
y
x
BOTTOM HALF OF THE PLANE IS THE
SOLUTION.
4. Solve for x and graph your solution: 2x + 13y < 4x − 9.
2x + 13y + (−2x + 9) < 4x − 9 + (−2x + 9),
1
1
< 2x × ,
2
2
9
13
y + < x.
2
2
We use the ordered pair (x, y) = (0, 0) and check if it makes the above inequality a
true sentence. It is false because
(13y + 9) ×
13/2 × 0 + 9/2 > 0 = x
but we need the inequality the opposite way.
Note here and in #5 we have x as a function of y so that the axes changed their
places.
THE SOLUTION IS TOP PART OF THE PLANE.
x
/2
+9
y
3/2
1
x=
81
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
5. Solve for x and graph your solution: 2x − 3y ≥ 4x − 9.
2x − 3y + (−2x + 9) ≥ 4x − 9 + (−2x + 9),
−3y + 9 ≥ 4x − 9 − 2x + 9 = 2x,
1
1
(−3y + 9) × ≥ 2x × ,
2
2
9
3
− y + ≥ x.
2
2
As usual, we check using a pair of numbers (x, y) = (0, 1):
3
9
− × 1 + ?0,
2
2
3 9
+ = 3 ≥ 0.
2 2
This pair makes the inequality a true sentence, so this is the solution half of plane.
x=
-3
/2y
+9
/2
The solution is bottom half of the plane
82
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
3.9.2 BBe1: Solve consistent 2 x 2 systems by substitution or elimination.
Solve a pair of equations algebraically. Check your answers.
1. {4x + 5y = 11, 6x + 11y = −5}
We want to eliminate variable x. For this, we need to multiply the first equation by
3, and the second by 2:
Note I use vertical line ”|” to denote that the operation given on right is performed
on both sides of the equation — here we multiply, for example, the first equation
with 3 — all coefficients are multiplied with this number. This notation will be used
in many other plases.
4x + 5y = 11
6x + 11y = −5
| × (+3),
| × (−2).
+12x + 15y = 33,
−12x − 22y = 10.
We add these equations and then solve for y:
+12x − 12x + 15y − 22y = 33 + 10,
−7y = 43
1
|× −
7
,
43
.
7
We use the same way to find x, only this time we multiply the first equation with 11
and the second with (−5):
y=−
+44x + 55y = 121,
−30x − 55y = 25.
14x = 146,
83
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
x=
146
73
= .
14
7
We now check our answer:
292 − 215
43
77
73
=
+5× −
=
= 11,
4×
7
7
7
7
73
43
35
438 − 473
6×
+ 11 × −
=−
= −5.
=
7
7
7
7
2. {2x + 5y = 0, 6x + 11y = 5}
We will apply for this example the substitution method. We use the first equation to
express x :
2x = −5y,
5
x = − y.
2
Now, we substitute that value in the second equation and solve for y:
5
6x + 11y = 6 × − y + 11y = 5,
2
−15y + 11y = 5,
−4y = 5,
5
y=− .
4
On the next step, we use the expression for x (found before) and calculate its value:
5
25
5
5
x=− y=− × −
=
2
2
4
8
We check our work:
5
25 − 25
25
+5× −
=
= 0,
2×
8
4
4
25
75 − 55
5
6×
=
+ 11 × −
= 5.
8
4
4
3. {x + 5y = 8, x + 11y = −5}
We see both equations contain the variable x with coefficient of one, so this system is
suitable for solving using the substitution method. We express x in the first equation
in terms of y. On the next step, we substitute this value into the second equation and
solve it for remaining variable, in this case, y.
x = 8 − 5y,
84
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
x + 11y = (8 − 5y) + 11y = −5.
We add (−8) to each side.
6y + 8 = −5
| + (−8),
6y = −13,
13
y=− .
6
We now can return to the first equation x = 8 − 5y and compute x:
13
x = 8 − 5y = 8 − 5 × −
,
6
x=
113
48 − (−65)
=
.
6
6
We now check the solution:
113
13
48
113 − 65
x + 5y =
+ 5× −
=
= 8.
=
6
6
6
6
113
113 − 143
13
−30
x + 11y =
=
+ 11 × −
=
= −5.
6
6
6
6
4. {4x − 5y = −2, 6x + 10y = −5}
We will mix the methods in this example. We will use the elimination method on
the first step to eliminate y. For this we multiply first equation by 2 and add it to
second equation.
4x − 5y = −2
8x − 10y = −4,
6x + 10y = −5,
14x = −9,
9
x=− .
14
| × 2,
Now, we will substitute the value of x we found in one of the equations and solve it
for the second variable, y:
4x − 5y = −2 | + (−4x),
1
−5y = −2 − 4x |× − ,
5
1
y = − (−2 − 4x).
5
Note that −1/5 = (−1)(1/5) and −2 − 4x = (−1)(2 + 4x), therefore we can get rid
of these minus ones by multiplying them together. The process for computing the
85
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
value of y is shown below.
y=
y=
y=
y=
y=
1
(2 + 4x),
5
1
9
2+4× −
,
5
14
1 2 × 14 − 4 × 9
×
,
5
14
1 × −18
,
5 × 14
4
− .
35
We check our answer:
4x − 5y =
6x + 10y =
4
9
− 5 × − 45
= −2,
4 × − 14
4
9
+ 10 × − 35
=
6 × − 14
−5.
5. {3x + 5y = 1, 6x + 11y = −5}
We multiply the first equation by (−2) and add it to the second one. We substitute
the value of y in the first equation and solve it for x:
3x + 5y =
1| × (−2),
−6x − 10y =
−2,
6x + 11y =
−5,
y=
−7,
3x =
1 − 5y,
3x = 1 − 5 × (−7),
3x =
36,
x=
12.
We check our work:
3x + 5y = 3 × 12 + 5 × (−7) = 1,
6x + 11y = 6 × 12 + 11 × (−7) = −5.
It may seem easier to solve a system using the substitution method and comparing it
with the elimination method but it is not. For example, what if we have a system with
4 equations and decide to use this method? On the first step, we express one variable
in terms of other three. Then, we substitute the result in the other three equations and
simplify. We need to repeat the process again —it means the method is not used for large
systems.
86
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
3.9.3 BBe2: Interpret 2 x 2 systems graphically .
Graph each of the following systems. Put the coordinates of the point of intersection on
your graph.
Example: {4x + 5y = 1, 6x + 11y = 5}.
Solution: x = −1 and y = 1, so (−1, 1) is the point of intersection. Graph each of the
following systems. Put the coordinates of the point of intersection on your graph.
Explanation ( applied to all of the exercises below):
Recall that solution set for an equation in two variables can be drawn in a plane. For a
linear equation, it is a line in the plane. If we have two lines, they (a) can be parallel, (b)
can have an intersection point or (c) — can be one and the same line. Then, the solution
set of a system in two variables and two equations is the set of all points, which make
each of the equations true sentence - i.e. all points in the plane which belong to both lines.
Graphically this is the intersection point of the two lines (see the graph below).
Intersection point (-1,1)
6x+11y=5
(-1,1)
4x
+5
y=
1
87
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
1. {4x + 5y = −11, 6x + 11y = 5}.
Intersection point about (-10, 7)
6x+11y=5
(-73/7, 43/7)
4x+5y=-11
Graphing helps us to find out whether the solution exists. The precision, however,
is not high. The analytic method shows that x = −73/7 and y = 43/7.
We check:
43
73
292 215
77
+5×
4× −
=−
+
= − = −11,
7
7
7
7
7
438 473
35
43
73
=−
+
=
= 5.
+ 11 ×
6× −
7
7
7
7
7
88
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
2. {2x + 5y = 0, 6x + 11y = −5}
6x + 11y = -5
2x + 5y = 0
(-25/8, 5/4)
We check:
25
5
25 25
2× −
+5× =− +
= 0,
8
4
4
4
5
75 55
25
+ 11 × = − +
6× −
= −4.
8
4
4
4
3. {x + 5y = −8, x + 11y = 5}
(-113/6, 13/6)
x+
x+
5y
=
11y
=5
-8
We check:
48
113 65
13
+
= − = −8,
=−
6
6
6
6
113
113 143
30
13
−×
=−
+ 11 ×
+
=
= 5.
6
6
6
6
6
113
+5×
−
6
4. {4x − 5y = 2, 6x + 10y = 5}
89
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
6x
+1
4x
We check:
0y
-
=5
=
5y
2
(9/14, 4/35)
9
18 4
− 5 × 435 =
− = 2,
14
7
7
27 8
9
+ 10 × 435 =
+ = 5.
6×
14
7
7
4×
5. {3x + 5y = −1, 6x + 11y = 5}
3x + 5y = -1
(-12 , 7)
6x +11y = 5
We check:
3 × (−12) + 5 × 7 = −36 + 35 = −1,
6 × (−12) + 11 × 7 = −72 + 77 = 5.
90
CHAPTER 3. ALGEBRA
3.9. LINEAR INEQUALITIES
3.9.4 BBe3: Recognize inconsistent or dependent systems algebraically
or graphically.
Indicate whether each system represents either a pair of parallel lines, the same line, or a
pair of lines that intersect.
Example: Indicate whether each system represents either a pair of parallel lines, the
same line, or a pair of lines that intersect.
{4x + 5y = 1, 8x + 10y = 5}.
Solution of the solved example:
We rearrange both equations in slope-intercept form, i.e. solve them for y:
4x + 5y = −11,
5y = −4x − 11,
4
11
y =− x− .
5
5
8x + 10y = 5,
10y = −8x + 5,
5
8
y=− + ,
10 10
1
4
y =− x+ .
5
2
We look at the expressions for y. What we see is both have slope −4/5 and different
y-intercepts. This means the lines are parallel.
1. {4x + 5y = −11, 6x + 11y = 5}.
We can solve the problem graphically. We plot the two lines and observe whether
they are parallel, one and the same line or intersect each other. The graph below
shows that the lines intersect each other.
4x
+5
y=
6x
+1
11
1y
=5
Rearranged equations show the same:
4
11
y =− x+ ,
5
5
5
6
y =− x+ .
11
11
91
3.9. LINEAR INEQUALITIES
CHAPTER 3. ALGEBRA
Slopes are −4/5 and −6/11.
For the remaining systems I will show only the rearranged equations and the concusion.
2. {x + 5y = 10, 2x + 10y = 20}.
x = −5y + 10,
x = −5y + 10.
This system is one and the same line.
3. {−x + 11y = −8, x − 11y = 5}.
1
8
x− ,
11
11
5
1
y = x+ .
11
11
y=
This is a system of parallel lines.
4. {4x = 2 + 2y, 8x − 4y = 5}.
1
1
x= y+ ,
2
2
1
5
x= y+ .
2
2
This is a system of parallel lines.
5. {3x + 5y = −1, 6x + 11y = 5}.
3
1
y =− x− ,
5
5
5
6
y =− x+ .
11
11
This is a system of intersecting lines.
92
CHAPTER 3. ALGEBRA
3.10
3.10. VERBAL PROBLEMS
Verbal Problems
3.10.1 BBf1: Work/rate.
Solve each of the following problems involving rates and work.
Example: Two bricklayers are to build a wall. The first can lay 50 bricks per hour and
the second can lay 70 bricks per hour. How long will it take them to construct the wall if
it contains 1200 bricks?
Solution:
Suppose only the first builder works on the wall. He will lay 50 bricks in one hour. If he
gets a helper, i.e. the second builder comes to work, both will lay 120 bricks in one hour.
For two hours they lay 240 bricks, for 3— 360 and so on. The builders need to lay 1200
bricks. To find out the time, we divide the total number of bricks to the number of bricks
which the two builders can lay per hour. This way we find:
T IM E =
Total number of bricks
Number bricks laid per hour
1200
= 10
120
The two builders need 10 hours to construct a wall containing 1200 bricks.
t=
Definition:
Rate is the amount of work done in one unit time by one or more people.
1. It takes one bricklayer 20 hours to lay 1200 bricks, and it takes another bricklayer 16
hours to lay 1200 bricks. Working together, how long will it take them to lay 2400
bricks?
We look at the formula above. We know the total amount of work to be done but we
do not know the rate at which it could be done. Therefore, we need to find the rate
of the two builders together. If the first builder can lay 1200 bricks in 20 hours, this
rate can be found by dividing the number of bricks built by the time for doing so:
RAT E =
Total number of bricks
.
Time for building
We denote the first builder rate by RAT E1 and the second builder rate by RAT E2 .
RAT E1 =
1200
= 60 bricks per hour.
20
93
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
The first builder rate is 60 bricks per hour. We find the second builder’s rate the
same way:
1200
RAT E2 =
= 75 bricks per hour.
16
The rate at which the two builders work is the sum of the rates of the two builders:
RAT E = RAT E1 + RAT E2 = 60 + 75 = 135 bricks per hour.
The two builders working together have a rate of 135 bricks per hour. They need to
lay 2400 bricks. We use now the formula from above:
T IM E =
Total number of bricks
,
RAT E
2400
2
= 17 .
135
3
These two builders need 17 hours and 40 minutes to lay 2400 bricks.
t=
2. My hot water faucet fills my bathtub in 20 minutes and my cold water faucet fills
my bathtub in 16 minutes. If I run both at the same time, how long will it take to fill
the bathtub?
We take another look at the rate definition. We do not know the volume of the bath.
Let us consider it 1 unit(i.e. it could be 1 f t3 , 1 m3 or we can make another unit for
this particular problem - it is 1 bath). Then, the rate for filling the bath is:
RATE =
1 bath
Time to fill one bath
We will denote RAT E1 the rate hot water faucet fills the bath and RAT E2 the rate
cold water faucet fills the bath. Then:
RAT E1 =
1
bath per minute,
20
RAT E2 =
1
bath per minute.
16
If we open both faucets the rate is:
RAT E = RAT E1 + RAT E2 =
1
9
1
+
=
bath per minute.
20 16
80
We can now use the formula for computing time:
T IM E =
1 bath
1
=
.
RAT E
9/80
If I open both hot and cold water faucets, I can fill the bath for 8 98 minutes, approximately 8 minutes and 53 seconds.
94
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
3. One secretary can type 120 words per minute, while another can type 130 words
per minute. Working together, how long would it take them to type a document
containing 100,000 words?
The two secretaries working together type 250 words per minute. The time for
typing 100,000 words could be found by the same formula:
T IM E =
100, 000
Number of words
=
= 400.
RAT E
250
The two secretaries working together can type in 100,000 words for 400 minutes or
6 hours and 40 minutes.
4. It takes one groundskeeper an hour to mow a football field, and it takes another an
hour and twenty minutes to mow the same field. Working together, how long does
it take to mow the field?
This is an another variation for the above problem. We have one field. The rates for
mowing the field are ( in minutes):
RAT E1 =
1
fields per minute,
60
1
fields per minute.
60 + 20
The rate for the two grounkeepers is
RAT E2 =
RAT E = RAT E1 + RAT E2 =
1
1
7
+
=
fields per minute.
60 80
240
The time for mowing is:
T IM E =
1
7
240
= 34
2
7
The two groundkeepers can mow the field for 34 27 minutes, approximately 34 minutes
and 17 seconds (if nobody stops to get a sip of water).
5. My neighbor and I want to fill a swimming pool. His hose delivers 15 gallons per
minute, and mine delivers 12 gallons per minute. Alas, the pool leaks 3 gallons per
minute. When full, the pool contains 10,000 gallons of water. How long will it take
us to fill the pool?
Let’s see what happens after one minute if I and my neighbor start filling the swimming pool. We will have in the pool 12 gallons of water coming from my hose, 15
95
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
gallons of water coming from my neighbors hose, but there will be 3 gallons less because of a leek. Effectively, there will be 24 gallons in the pool. If we denote RAT E1
— the rate of my hose, RAT E2 — rate of my neighbor’s hose and RAT E3 the rate at
which the pool leaks, we can write:
RAT E = RAT E1 + RAT E2 − RAT E3 = 24 gallons per minute.
Now, we can compute the time for filling the swimming pool:
T IM E =
2
10000
= 416
24
3
We need 416 32 minutes to fill the swimming pool or 416 minutes and 40 seconds.
96
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
3.10.2 BBf2: Distance/rate/time.
Solve each of the following problems involving distance, rate and time.
Example: If I can run 7.5 miles per hour for two hours, how long will it take me to run
10 miles?
We make use of the formula connecting distance, speed and time:
distance = speed × time.
In this example the distance is 10 miles, the speed is 7.5 miles per hour, and we do not
know the time. We use the expression for time :
time =
distance
10
1
=
=1
speed
7.5
3
The time for running 10 miles is 1 31 hours.
1. A car travels 10 miles in 8 minutes. What is its average speed in miles per hour?
There are two ways to solve this problem. We can compute what is the average
speed per minute and then, knowing that one hour has 60 minutes, we multiply
this number by 60, and thus we find the average speed per hour. Here are the
computations:
Rate per minute (denoted by RPmin):
RP min =
5
10
= miles per minute.
8
4
We compute the speed:
speed = 60 × RP min = 60 ×
5
= 75 mph.
4
The speed is 75 mph.
The other way is to express 8 minutes in hours:
2
8
hours =
hours.
60
15
We can now compute the average speed using the main formula directly:
8 minutes =
speed =
distance
10
=
time
2/15
speed = 75 mph.
NOTE: We always need to use one and the same units for all different variables in
the formula, i.e. if one variable is in miles, the other need to be in miles per unit
time, the units time must be the same and so on.
97
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
2. A car travels 10 miles at an average speed of 45 miles per hour. How many minutes
does this take?
We make use of the same formula (speed×time=distance):
time =
distance
10
2
=
= ho.
speed
45
9
Because we need the time in minutes we multiply the found time in hours by 60,
the number of minutes in 1 hour:
time (min) = time (ho) × 60 =
40
1
2
× 60 =
= 13 min.
9
3
3
Above, we denote [time(min)] the time for traveling 10 mi in minutes and [time(ho)]
the same in hours.
It takes 13 31 minutes for a car with average speed 45 mph to travel 10 miles.
3. It takes 11 minutes for a leaf to drift 1/4 mile down a river. How fast is the river
flowing in feet per second?
We will express the time in seconds and the distance in feet, then the answer we find
will be in feet per second. There are 5280 feet in a mile.
speed × time = distance,
speed =
1/4 × 5280
11 × 60
1320
= 20 feet/sec.
660
The speed of the current is 20 feet per second.
=
4. A river is flowing at 2 feet per second. How long will it take for a boat to drift 100
feet? Since all data is in desired units, we write:
time =
distance
speed
100
= 50 sec.
2
The boat drifts 100 feed for 50 seconds.
=
98
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
5. Jackie Joyner-Kersee once ran 800 meters in 2 minutes and 17.61 seconds. What was
her average speed in kilometers per hour?
We express 800 meters in kilometers 800 m= 0.8 km. We express 2 min 17.61 sec in
hours: 2 min 17.61 = (2 × 60 + 17.61)/3600 ho. We now compute the average speed:
speed =
=
distance
time
0.8
= 20.929 km/ho.
137.61/3600
Jackie Joyner-Kersee speed was 20.929 km/ho.
99
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
3.10.3 BBf3: Mixture.
Solve each of the following problems involving rates and work.
Example: I have one gallon of a 12% alcohol solution and one gallon of a 20% alcohol
solution. How much of each should I mix together to get one gallon of a 15% alcohol
solution?
Example solution:
Recall when we have two unknown quantities, we need two different equations to find
the amounts. First, we need to have one gallon of the solution. If we denote with x the
quantity of 12% solution and with y the quantity of 20% solution, we know when we add
them together, their sum is one gallon solution. We translate it:
x + y = 1.
We need 15% solution. If we have 1 gallon of 12% solution, then the amount of alcohol
it contains is 12% × 1 gallons, but we must take another quantity — x gallons. In this
quantity, we have 12% × x gallons of alcohol. The same way we determine in y gallons
20% solution we have 20% × y gallons of alcohol. We know when we add these together,
we need to have 1 gallon 15% solution. The alcohol in it is 15% × 1 gallons. The equatioin
is
.12x + .2y = .15.
Solving above equations together, we find x = 0.625 and y = 0.375.
We need 0.625 gallons 12% solution and 0.375 gallons 20% solution.
Remember, the check of the answer is part of the solution:
0.625 + 0.375 = 1.000 liters,
12% × 0.625 + 20% × 0.375
= 0.075 + 0.075 = 0.150 = 15% × 1liters.
1. My radiator contains 4 gallons of a 35% antifreeze solution. How much should I
replace with pure antifreeze to get 4 gallons of a 50% antifreeze solution?
We start with 4 gallons of 35% solution. We take out x gallons of it and put back x
gallons of 100% solution. The result is 4 gallons 50% solution. From the picture, we
can visualize the equation:
(4 × 0.35) − (x × 0.35) + x = 4 × 0.5.
100
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
4 gal 35 %
solution
x gal 35 %
solution
_
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+
x gal 100 %
solution
x with x gall
x
x
x
x 100 %
xx
x
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xx
x
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x
x
x
x
x
x
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We replace x
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solution
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xx
4 gal 50%
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We solve the resulting equation and find x =
12
13
gallons.
2. Ten gallons of a 5% alchohol solution is mixed with fifteen gallons of a 23% alcohol
solution. What is the concentration of alcohol in the resulting solution?
10 gall 5 %
solution
+
xxxx
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15 gall 23 %
solution
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=
25 gall x %
solution
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From the graph we see that total amount of solution is 25 gallons. We see also the
alcohol in it is (5% × 10) + (23% × 15). Recall the concentration is computed by
dividing the amount of alcohol by the amount of the solution:
concentration =
amount of alcohol
× 100,
amount solution
101
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
concentration =
0.05 × 10 + 0.23 × 15
× 100 = 15.8%.
10 + 15
3. Peanuts cost $1.50 per pound, and walnuts cost $3.25 a pound. I want to blend
them to get 10 pounds of nuts that sells for $2.50 per pound. How many pounds of
peanuts and how many pounds of walnuts should I use?
We denote with x the amount of peanuts and y the amount of walnuts. The information is organized in the chart below.
Quantity
Price
Cost
Peanuts
x
$1.50
$1.50x
Walnuts
y
$3.25
$3.25y
Mix
10
$2.50
$2.50 × 10
We can translate the chart into equations: the quantities must add to 10 (x + y = 10)
and the costs must match (1.5x + 3.25y = 10 × 2.5). We find then we need to use 30/7
pounds of peanuts and 40/7 pounds of walnuts.
4. Cinnamon sugar is made by blending sugar, at $0.40 per pound with cinnamon at
$40.00 per pound. Cinnamon sugar sells for $10.00 per pound. How much cinnamon is in each pound?
We can use the same kind of chart to solve this problem. The information is given
below.
Quantity
Price
Cost
Cinnamon
x
$40.00
$40.00x
Bl. sugar
y
$0.40
$0.40y
Cinn. shugar
1
$10
$10 × 1
We write down the equations as in #3:
x + y = 1,
total amount40x + 0.4y = 10 × 1
cost equation.
After solving the equations, we find we need 8/11 pounds of blended sugar and
3/11 pounds of cinnamon.
5. Fuel for a two cycle engine is made by blending gasoline with oil. The gasoline costs
$2.20 per gallon and the oil costs $4.00 per gallon. How much should one gallon of
the mixture cost if it contains 32 times as much gasoline as oil?
We know here the ratio in which we blend the gasoline and oil. If we use 32 gallons
of gasoline and one gallon of oil, we will have 33 gallons of mixture. The cost of
price of this mixture is 32 × $2.20 + $4.00 = $74.4. We divide this dollar amount by
the number of gallons and find the cost per gallon: $2.26.
102
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
3.10.4 BBf4: Number (reciprocal, digit, consecutive, integer, etc.).
Solve each of the following problems involving numbers.
Example: The sum of three consecutive integers is 24. What are these integers?
Example solution:
Let denote the first integer with x, then the next consecutive integer is (x + 1) and the next
is (x + 2). We know the sum of the three of them is 24. The equation is:
x + (x + 1) + (x + 2) = 24.
We solve the equation for x and find x = 7.
Answer: 7, 8 and 9.
Check:
7 + 8 + 9 = 24.
1. 3 more than twice a number is 17. What is the number?
We denote the number with x. We translate into mathematical language: three more
than twice is (3 + 2x) and the answer is 17:
3 + 2x = 17.
After solving equation we find that the number is 7.
Answer: 7.
Check:
3 + 2 × 7 = 17.
2. A real number larger than 1 is added to its reciprocal and the result is 29/10. What
is this number?
We denote the number with x. We know the sum of the number (x) and its reciprocal
(1/x) is 29/10. The equation is
1
29
x+ = .
x
10
We multiply this equation with x and the resulting equation is quadratic:
29
x + 1 = 0,
10
10x2 − 29x + 10 = 0.
x2 −
We solve it and find x1 = 5/2 and x2 = 2/5. These two numbers are reciprocals so
only one is bigger than one.
Answer is x = 5/2.
Check:
5
1
5 2
+
= +
2 5/2
2 5
4
29
25
+
= .
=
10 10
10
103
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
3. The sum of three consecutive odd integers is 375. What are these integers?
We denote the smallest integer with x, therefore next integer is (x + 2) and the third
one is (x + 4). We have the equation:
x + (x + 2) + (x + 4) = 375.
We solve the equation for x and find x = 123.
The integers are 123, 125 and 127.
Check:
123 + 125 + 127 = 375.
4. The product of two integers is 21 and their sum is 10. What are these integers?
We denote one of the numbers with x and the other with y. We know their sum is
10 so one equation is x + y = 10. The product is 21 so the equation is xy = 21. The
system is:
x + y = 10,
xy = 21.
The way to solve this nonlinear system is by expressing y in the first equation:
y = 10 − x.
We substitute y in the second equation and solve.
x(10 − x) = 21,
10x − x2 − 21 = 0,
x2 − 10x + 21 = 0.
We find the solutions of this equation are 3 and 7. They are the numbers with a sum
of 10 and product of 21. Thus, the system of equations has two solutions but there
is only one answer to the question since 3 + 7 = 7 + 3 and 3 × 7 = 7 × 3.
Answer: 3 and 7. The explanation above cleans up the check of the answer.
5. The product of two real numbers is 3 and their sum is 5. What is the larger of these
two real numbers?
We use the same method as in #4. I will write down the system:
x + y = 5,
xy = 3.
We express y from the first equation:
y = 5 − x.
104
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
Using the second equation we write:
3 = xy = x(5 − x) = 5x − x2 ,
x2 − 5x + 3 = 0.
This equation has irrational solutions. The discriminant is D = (−5)2 − 4 × 1 × 3 =
25 − 12 = 13. The larger number is
x=
The other number is
y=
We check:
x+y =
5+
5+
√
13
2
5−
√
13
√
13
2
.
5+
√
13
= 5,
2
√
√
√
5 + 13 5 − 13
52 − ( 13)2
25 − 13
xy =
×
=
=
= 3.
2
2
4
4
2
+
.
6. Twice one real number plus six times another is 13 while the average of these numbers is 4. What are these numbers?
We denote first number with x and second with y. We use the information now,translating
it into equations: Twice the first number plus 6 times second is 13: 2x + 6y = 13. The
average is 4: x+y
= 4. The system:
2
2x + 6y = 13,
x+y
= 4.
2
We multiply the second equation by 4:
2x + 2y = 16.
We subtract that equation from the first one:
2x + 6y − (2x + 2y) = 13 − 16 = −3.
We solve for y:
4y = −3,
3
y=− .
4
The corresponding x is
x=
105
35
.
4
3.10. VERBAL PROBLEMS
We check:
CHAPTER 3. ALGEBRA
35/4 + (−3/4)
x+y
=
= 4,
2
2
35
3
35 9
2x + 6y = 2 ×
+6× −
− = 13.
=
4
4
2
2
106
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
3.10.5 BBf5: Area and perimeter.
Solve each of the following problems involving numbers.
Example: The area of a circle is 25π square feet. What is the circumference of this
circle?
Example solution:
Recall that the area A, of a circle is
A = πr2 ,
where r is the radius and the circumference, l, is
l = 2πr.
We find the radius r from the expression for the area: r = 5. We use now the second
formula and find that the circumference is 10π. Answer: 10π feet.
1. The circumference of a circle is 15 feet. What is the area of this circle?
We use the same formulas but in opposite way. From the circumference formula we
find the radius of the circle.
l = 2πr,
r=
l
15
=
.
2π
2π
Using the formula for the area we find:
A = πr2 = 3.14 × 2.392 = 17.95
The area is 17.92 square feet.
2. The perimeter of a square is 64 meters. What is the area of this square?
Recall the formulas for the perimeter p = 4 × s and the area is A = s2 , where s is
the length of a side of the square. We use the formula for the perimeter and find the
side of the square:
p = 4 × s,
s=
p
64
=
= 16.
4
4
Using the formula for the area we compute
A = s2 = 162 = 256.
The are of a square with perimeter 64 meters is 256 square meters.
107
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
3. The area of an equilateral triangle is 16 square inches. What is the perimeter of this
triangle?
Recall the formula of the area of a triangle is A = 21 aha . In this formula a is one side
of the triangle, and ha is the perpendicular to the same side. In equilateral triangle
is the same. We only need to find the length of the perpendicular. To do so, we use
the right angle triangle in which the legs are AH and HC, the hypotenuse is AC.We
know that AC = a and AH = a/2. We do not know the length of the third side. We
use the Pythagorean theorem and write:
AH 2 + HC 2 = AC 2 ,
a 2
+ h2a = a2 ,
2
a 2 3a2
2
2
,
=
ha = a −
2
4
√
3a
ha =
.
2
We can now use the formula of the area and the height we found.
√
1
1
3a
,
A = a × ha = a ×
2
2
2
√
3 2
A=
a.
4
We find a:
s
s
4A
4 × 16
√
a= √ =
≈ 6.08 inches.
3
3
The perimeter is the sum of the lengths of the sides, in this case — three times the
side:
p = 3 × a = 3 × 6.08 ≈ 18.24 inches.
The perimeter is 18.24 inches.
4. A rectangle is twice as wide as it is high, and the perimeter is 12 meters. What is the
area of this rectangle?
Let denote the high with x. Then, the wide is twice the high, i.e. (2x). The perimeter
is the double sum of the two : p = 2(x + 2x). It is 12. We write:
6x = 12
x=2
The high is 2 meters and the wide is 4 meters long. The area of the rectangle is the
product of the wide and high, and is 8 square meters.
The area of the rectangle is 8 square meters.
108
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
5. The area of a rectangle is six square inches and the perimeter is 10 inches. What are
the dimensions of this rectangle.
We denote the wide of the rectangle with x and the high with y. We know that the
perimeter is doubled sum of the sides and is 10: 2(x + y) = 10. The area is the
product of the sides and is 6: xy = 6. The system is
2(x + y) = 10
xy = 6
After solving the system we find x = 2 and y = 3.
The dimensions of the rectangle are 2 inches and 3 inches.
6. If we quadruple the area of a circle, by what factor do we increase its circumference?
Let A is the area of the initial circle and it is
A = πr2
Let denote r1 the radius of quadruple circle. We use the same formula:
4A = πr12
We substitute A in this expression and obtain:
4A = 4π(r2 ) = πr12
We solve the above expression for r1 :
r1 = 2r
We denote with l the circumference of the original circle and with l1 of the quadrupled one. We express now the circumference of that bigger circle in terms of the
smaller one:
l1 = 2πr1 = 2π(2r) = 2(2πr) = 2l
If we quadruple the ares of a circle, its circumference is doubled.
109
3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
3.10.6 BBf6: Other miscellaneous problems.
Solve each of the following problems involving distance, rate and time.
1. I own two cars. One gets 22 miles to the gallon of gasoline, and the other gets 32
miles to the gallon of gasoline. I drove 720 miles and used 30 gallons of gasoline.
How many miles was each car driven?
We denote with x the miles first car was driven and y — the other one respectively.
The first car (Car 1) runs 22 miles a gallon and the second one — 32. If the first car is
driven x miles, it will consume x/22 gallons gasoline. For the second one the value
is y/32 gallons. We can organize the information in the chart below:
Miles
Miles per gallon
Gas
Car 1
x
22
x/22
Car 2
y
32
y/32
Totals
720 mi
30 gal
There are two relations in this chart — one is the total distance traveled, and the
other is total gasoline used. On the first row, the distances traveled are x and y.
We add them anf the total is 720 miles. In the third row, we can add the gasoline
consumed by a car and the total is 30 gallons. The system is:
x + y = 720,
x
y
+
= 30.
22 32
After solving this system we find x = 528 and y = 192.
The car getting 22 miles a gallon traveled 528 miles and the other one traveled 192
miles.
2. Advance tickets to a play cost 7 dollars, and tickets at the door cost 10 dollars. If 300
tickets were sold and there were 2700 dollars of income, how many tickets of each
type were sold?
Let’s denote x the number of the tickets sold in advance and y — at the door. We
organize all the information in the chart below:
# of tickets
Price
Income
In advance
x
$7.00
$7.00x
110
At the door
y
$10.00
$10.00y
Total
300
—
$2700
CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
Again, we see two relations we can use to find out how many tickets of each kind
were sold — the number of the tickets and the incomes. The system is:
x + y = 300
# of tickets equation,
7x + 10y = 2700
income equation.
After solving the system we find x = 100 and y = 200.
There were 100 tickets sold in advance and 200 tickets sold at the door.
3. It takes 6 hours for a plane to fly from west to east from Los Angeles to New York
City, and it takes 8 hours for the plane to fly back, owing to the constant west to
east jet stream. If it is 3200 air miles between cities, how fast is the jet stream wind
blowing?
We make a drawing for better understanding the problem. It is given below:
N. Y. City
L. A.
plane
jet
time = 6 hours
distance = 3200 miles
plane
jet
time = 8 hours
The distance in both directions is 3200 miles. If we denote the speed of plane in still
air by pl and the speed of the jet stream by j, then from LA to New York the speed
of the plain is pl + j and the time for flying is 6 hours. In the opposite direction, the
speed is pl − j and the time is 8 hours. We have two relations:
6(pl + j) = 3200,
8(pl − j) = 3200.
We can easily solve this system. The solution is pl = 1400/3 and j = 200/3.
The speed of the jet stream is 200/3 mph or 66.67 mph.
4. One child is 5 times as old as another. The sum of their ages is 18 years. How old
will the younger child be in three years?
Let denote the age of the yonger child now by x. Then, the age of the other child is
5x. The sum of the two ages is 18. We have the equation:
5x + x = 6x = 18.
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3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
We solve the equation and find x = 3. After three years this child will be three years
older.
The age of the child after three years will be 6 years.
5. One leg of a right triangle is 3 times as long as the other. The area of the triangle is
6 square meters. What is the perimeter of this right triangle?
Recall the formula for the area A = 1/2 × ab of a right triangle, where a and b are the
legs. The shorter leg is a and the longer one is 3a. Using this information, we find a:
1
a × 3a = A = 6,
2
1 2
3a = 6.
2
We find a = 2 or a = −2. The length cannot be negative so we have only one
solution: a = 2. Then, the legs are 2 meters and 6 meters.
We need now to find the third side. We use the formula for the sides in the right
triangle: c2 = a2 + b2 . In our case, b = 3a, therefore
c 2 = a2 + b 2
= a2 + (3a)2 = 22 + 62 = 40.
√
We find c = 2 10. The length cannot be negative so we do not consider the negative
solution of the equation for the hypothenuse. The perimeter of the right triangle is
√
p = a + b + c = 2 + 6 + 2 10
√
= 8 + 2 10.
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CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
3.10.7 BBf5: Area and perimeter.
Solve each of the following problems involving numbers.
Example: The area of a circle is 25π square feet. What is the circumference of this
circle?
Example solution:
Recall that the area A, of a circle is
A = πr2 ,
where r is the radius and the circumference, l, is
l = 2πr.
We find the radius r from the expression for the area: r = 5. We use now the second
formula and find that the circumference is 10π. Answer: 10π feet.
1. The circumference of a circle is 15 feet. What is the area of this circle?
We use the same formulas but in opposite way. From the circumference formula we
find the radius of the circle.
l = 2πr,
r=
l
15
=
.
2π
2π
Using the formula for the area we find:
A = πr2 = 3.14 × 2.392 = 17.95
The area is 17.92 square feet.
2. The perimeter of a square is 64 meters. What is the area of this square?
Recall the formulas for the perimeter p = 4 × s and the area is A = s2 , where s is
the length of a side of the square. We use the formula for the perimeter and find the
side of the square:
p = 4 × s,
s=
p
64
=
= 16.
4
4
Using the formula for the area we compute
A = s2 = 162 = 256.
The are of a square with perimeter 64 meters is 256 square meters.
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3.10. VERBAL PROBLEMS
CHAPTER 3. ALGEBRA
3. The area of an equilateral triangle is 16 square inches. What is the perimeter of this
triangle?
Recall the formula of the area of a triangle is A = 21 aha . In this formula a is one side
of the triangle, and ha is the perpendicular to the same side. In equilateral triangle
is the same. We only need to find the length of the perpendicular. To do so, we use
the right angle triangle in which the legs are AH and HC, the hypotenuse is AC.We
know that AC = a and AH = a/2. We do not know the length of the third side. We
use the Pythagorean theorem and write:
AH 2 + HC 2 = AC 2 ,
a 2
+ h2a = a2 ,
2
a 2 3a2
2
2
,
=
ha = a −
2
4
√
3a
ha =
.
2
We can now use the formula of the area and the height we found.
√
1
1
3a
,
A = a × ha = a ×
2
2
2
√
3 2
A=
a.
4
We find a:
s
s
4A
4 × 16
√
a= √ =
≈ 6.08 inches.
3
3
The perimeter is the sum of the lengths of the sides, in this case — three times the
side:
p = 3 × a = 3 × 6.08 ≈ 18.24 inches.
The perimeter is 18.24 inches.
4. A rectangle is twice as wide as it is high, and the perimeter is 12 meters. What is the
area of this rectangle?
Let denote the high with x. Then, the wide is twice the high, i.e. (2x). The perimeter
is the double sum of the two : p = 2(x + 2x). It is 12. We write:
6x = 12
x=2
The high is 2 meters and the wide is 4 meters long. The area of the rectangle is the
product of the wide and high, and is 8 square meters.
The area of the rectangle is 8 square meters.
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CHAPTER 3. ALGEBRA
3.10. VERBAL PROBLEMS
5. The area of a rectangle is six square inches and the perimeter is 10 inches. What are
the dimensions of this rectangle.
We denote the wide of the rectangle with x and the high with y. We know that the
perimeter is doubled sum of the sides and is 10: 2(x + y) = 10. The area is the
product of the sides and is 6: xy = 6. The system is
2(x + y) = 10
xy = 6
After solving the system we find x = 2 and y = 3.
The dimensions of the rectangle are 2 inches and 3 inches.
6. If we quadruple the area of a circle, by what factor do we increase its circumference?
Let A is the area of the initial circle and it is
A = πr2
Let denote r1 the radius of quadruple circle. We use the same formula:
4A = πr12
We substitute A in this expression and obtain:
4A = 4π(r2 ) = πr12
We solve the above expression for r1 :
r1 = 2r
We denote with l the circumference of the original circle and with l1 of the quadrupled one. We express now the circumference of that bigger circle in terms of the
smaller one:
l1 = 2πr1 = 2π(2r) = 2(2πr) = 2l
If we quadruple the ares of a circle, its circumference is doubled.
115