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1
2
Arithmetic
1.1 Whole Numbers and Integers . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 ARa3: Prime factorization . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 ARa4: Greatest common divisor and least common multiple . . . . .
1.1.3 ARa6: Correct order of operations . . . . . . . . . . . . . . . . . . . .
1.2 Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 ARb1: Reducing fractions to lowest terms . . . . . . . . . . . . . . . .
1.2.2 ARb2: Adding, subtracting, multiplying and dividing fractions . . .
1.2.3 ARb3: Recognizing correct and incorrect cancellation . . . . . . . . .
1.2.4 ARb4: Addition, subtraction, multiplication and division of mixed
numerals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Decimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 ARc2: Arranging fractions and decimals . . . . . . . . . . . . . . . .
1.3.2 ARc4: Identify the magnitude . . . . . . . . . . . . . . . . . . . . . . .
1.3.3 ARc5: Convert between scientific and decimal notation . . . . . . . .
1.4 Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 ARd1: Estimate the quotient of two whole numbers . . . . . . . . . .
1.4.2 ARd2: Rounding a number to a given precision (nearest power of ten)
1.4.3 ARd4: Estimate the result of a computation using intervals . . . . . .
1.5 Ratio, proportion and percent . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 ARe1: Convert among equivalent forms: ratio, decimal and percent
1.5.2 ARe2: Find a given percent of a base b . . . . . . . . . . . . . . . . . .
1.5.3 ARe3: Find what percent of a is b . . . . . . . . . . . . . . . . . . . . .
1.5.4 ARe4: Convert from one measurement unit to another . . . . . . . .
1.5.5 ARe5: Solve a proportion given three terms . . . . . . . . . . . . . . .
1.5.6 ARe6: Solve a proportion given a verbal description. . . . . . . . . .
1.5.7 ARe7: Find the third, given two of the following: base amount,
percent of increase/decrease, resulting amount. . . . . . . . . . . . .
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Basic Algebra
31
2.1 Formulas and expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.1.1 AAa1: Evaluate expressions given numerical values for variables . . 31
2.1.2 AAa2: Find the numerical value of one variable in a formula given
numerical values of the other variables . . . . . . . . . . . . . . . . . 32
1
2.1.3
2.1.4
2.1.5
2.1.6
2.2
2.3
2.4
AAa3: Solve a formula for one variable in terms of the others . . .
AAa4: Simplify expressions by removing grouping symbols . . . .
AAa5: Simplify rational expressions . . . . . . . . . . . . . . . . . .
AAa6: Add, subtract, multiply or divide polynomial and rational
expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.7 AAa7: Convert from verbal statement to algebraic expression or
equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 AAb1: Solve ax + b = cx + d when a, b, c, d are integers . . . . . . .
2.2.2 AAb2: Solve ax + b = cx + d when a, b, c, d are rational numbers .
2.2.3 AAb3: Solve ax + b = cx + d when some of a, b, c, d are letters . . .
2.2.4 AAb4: Set up and solve verbal problems using linear equations. . .
2.2.5 AAb5: Solve ax + b = cx + d after removing grouping symbols. . .
Multiplication and Factoring . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 AAc1: Find products of the form (ax + b)(cx + d) and simplify . . .
2.3.2 AAc2: Find products of the form (ax + b)2 and simplify . . . . . . .
2.3.3 AAc3: Factor expressions containing a common monomial factor .
2.3.4 AAc4: Factor by grouping an expression like ax + ac + bx + bc . . .
2.3.5 AAc5: Factor expressions like x2 + bx + c . . . . . . . . . . . . . . .
Positive Integer Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 AAd1: Simplify exponents using the “add exponents” rule . . . . .
2.4.2 AAd2: Simplify expressions using “subtract exponents” rule . . . .
2.4.3 AAd3: Find power of power using “multiply exponents” rule . . .
2.4.4 AAd4: Find whole number power of a rational number . . . . . . .
A Glossary
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2
Chapter 1
Arithmetic
1.1
Whole Numbers and Integers
1.1.1 ARa3: Prime factorization
1. One way to find the prime factorization of a positive integer is using factors tree.
This is shown below.
360
9
40
10
3
4
5
3
2
2
2
Here is one tree. There are many variations. The prime factors are doublecircled.
Answer: the prime factorization of 360 is
360 = 23 × 32 × 5.
3
1.1. WHOLE NUMBERS AND INTEGERS
CHAPTER 1. ARITHMETIC
2. The other way to find prime factorization is to start dividing the number given by
the primes we know it is divisible by. Below, we see the process for number 1260.
The number times a prime factor is found is shown in parenthesis.
1260 ÷ 2 = 630
630 ÷ 2 = 315
315 ÷ 5 = 630
63 ÷ 3 = 21
21 ÷ 3 = 7
7÷7=1
prime factor 2
prime factor 2
prime factor 5
prime factor 3
prime factor 3
prime factor 7
Answer: the prime factorization of 1260 is
1260 = 22 × 32 × 5 × 7.
3.
9900 = 22 × 32 × 52 × 11
4.
23400 = 23 × 32 × 52 × 13
5.
4200 = 23 × 3 × 52 × 7
4
(1)
(2)
(1)
(1)
(2)
(1)
CHAPTER 1. ARITHMETIC
1.1. WHOLE NUMBERS AND INTEGERS
1.1.2 ARa4: Greatest common divisor and least common multiple
1. We first factor both numbers into primes: 9 = 32 and 15 = 3 × 5. The number 9 has
two factors of 3; the number 15 has one factor of 3 and one factor of 5.
The greatest common divisor (GCD) is the product of the factors found in both numbers, raised to the lowest power in all numbers. In this example, we find only one
common factor — 3. Its lowest power is 1, so the GCD is 3.
The least common multiple (LCM) is the product of all factors in both numbers, each
raised to the maximum power in all numbers. We have two factors of 3 (found in
number 9) and one factor of 3 (found in 15). In 15, we find factor 5, raised to first
power. Finally, LCM is 32 × 5 = 45.
Answer: for 9 and 15 the greatest common divisor is GCD = 3
and the least common multiple is LCM = 45.
2.
14 = 2 × 7.
21 = 3 × 7.
Factors found in both numbers — only 7. Its power in both numbers is 1.
Answer: for the numbers 14 and 21 the greatest common divisor is GCD = 7
and the least common multiple is LCM = 2 × 3 × 7 = 42.
3.
12 = 22 × 3.
15 = 3 × 5.
Factors found in both numbers — only 3. Its power in both numbers is 1.
Answer: for the numbers 12 and 15 the greatest common divisor is GCD = 3
and the least common multiple is LCM = 22 × 3 × 5 = 60.
4.
8 = 23 .
12 = 22 × 3.
Factors found in both numbers — only 2. Its lowest power is 2.
Answer: for the numbers 8 and 12 the greatest common divisor is GCM = 22 = 4
and the least common multiple is LCM = 23 × 3 = 24.
5
1.1. WHOLE NUMBERS AND INTEGERS
CHAPTER 1. ARITHMETIC
5.
22 = 2 × 11.
55 = 5 × 11.
Factors found in both numbers — 11. Its power in both numbers is 1.
Answer: for the numbers 22 and 55 the greatest common divisor is GCM = 11
and the least common multiple is LCM = 2 × 5 × 11 = 110.
6
CHAPTER 1. ARITHMETIC
1.1. WHOLE NUMBERS AND INTEGERS
1.1.3 ARa6: Correct order of operations
1. First, we multiply or divide and second, we add or subtract.
5 + 2 × 4 = 5 + (2 × 4)
= 5 + 8 = 13.
2. We multiply and divide on the first step. If we have more than one multiplication,
we do all of them from left-to-right.
5 × 2 + 3 × 8 = (5 × 2) + (3 × 8)
= 10 + 24 = 34.
3.
9 × 3 − 3 × 8 = (9 × 3) − (3 × 8)
= 27 − 24 = 3.
4. We multiply on the first step. Then we use the left-to-right rule on addition and
subtraction.
9 + 3 − 3 × 8 = 9 + 3 − (3 × 8)
= 9 + 3 − 24
= 12 − 24 = −12.
5. Multiplication and division are operations of the same type, so we use the left-toright rule. We subtract on second step:
9 × 3 − 6 ÷ 3 = (9 × 3) − (6 ÷ 3)
= 27 − 2 = 25.
7
1.2. FRACTIONS
1.2
CHAPTER 1. ARITHMETIC
Fractions
1.2.1 ARb1: Reducing fractions to lowest terms
We factor the numbers, then find the GCD, factor it out and reduce the fraction by dividing that factor out.
1. For 12 and 18, GCD = 6:
2. For 54 and 12, GCD = 6:
3. For 98 and 21, GCD = 7:
12
22 × 3
=
18
2 × 32
2
6×2
= .
=
6×3
3
54
2 × 33
= 2
12
2 ×3
6×9
9
=
= .
6×2
2
2 × 72
98
=
21
3×7
7 × 14
14
=
= .
7×3
3
4. For 37 and 303, GCD = 1 so this fraction cannot be reduced further:
37
37
=
.
303
303
5. For 55 and 77, GCD = 11:
11 × 5
55
=
77
11 × 7
5
= .
7
8
CHAPTER 1. ARITHMETIC
1.2. FRACTIONS
1.2.2 ARb2: Adding, subtracting, multiplying and dividing fractions
When adding or subtracting, first we find the LCM and extend the fractions by multiplying them by the appropriate expression of one to obtain a common denominator.
We multiply fractions by multiplying the numerators and this is the new numerator, and
we do so with the denominators obtaining the new denominator.
We divide fractions by multiplying the first fraction by the reciprocal of the second one.
1. For numbers 2 and 3, LCM = 6:
1 1
1×2 1×3
+ =
+
3 2
3×2 2×3
=
5
2 3
+ = .
6 6
6
1×1
1 1
× =
3 2
3×2
1
= .
6
2. For numbers 9 and 2, LCM = 18:
2×2 3×9
2 3
− =
−
9 2
9×2 2×9
=
27
23
4
−
=− .
18 18
18
2 2
2 3
÷ = ×
9 2
9 3
2×2
4
=
= .
9×3
27
3. For numbers 7 and 14, LCM = 14:
3
1
3×2
1
+
=
+
7 14
7 × 2 14
1
7
6
+
=
14 14
14
1×7
1
=
= .
2×7
2
=
3
1
3×1
×
=
7 14
7 × 14
3
= .
98
9
1.2. FRACTIONS
CHAPTER 1. ARITHMETIC
4. For numbers 12 and 3, LCM = 12:
1
5
1×4
5
− =
−
12 3
12 3 × 4
=
5
4
1
−
= .
12 12
12
1
5
3
5
÷ =
×
12 3
12 1
5×3
5
5×3
=
= .
=
12 × 1
4×3
4
5.
14
5 × 14
5×7×2
5
×
=
=
16 15
16 × 15
2×2×2×2×5×3
10 × 7
7
=
= .
10 × 24
24
14
5
15
5
÷
=
×
16 15
16 14
5 × 15
75
=
=
.
16 × 14
224
10
CHAPTER 1. ARITHMETIC
1.2. FRACTIONS
1.2.3 ARb3: Recognizing correct and incorrect cancellation
Decide if each of the following statements is true or false. If the statement is false, correct
the right hand side to give a correct statement.
1.
2+1
4+2
=
.
2+8
1+4
TRUE.
2.
1+2+3
= 5.
2
FALSE. The sum in the numerator is 6. We can then factor out 2. The correct answer
is 3.
1+2+3
6
= = 3.
2
2
3.
9 + 12
= 1.
21
TRUE.
4.
3×7
7
=
.
303
101
.
TRUE.
5.
3
5+5
= .
20
2
FALSE. When we add 5 + 5, we have 10. We can factor out 10, so the answer is 1/2.
5+5
10
1
=
= .
20
20
2
11
1.2. FRACTIONS
CHAPTER 1. ARITHMETIC
1.2.4 ARb4: Addition, subtraction, multiplication and division of mixed
numerals
1. When we add or subtract mixed numerals, first we add or subtract the fractions and
then we add or subtract the whole number part.
2
2
3
13
+ (1 + 5)
+
1 +5 =
18
3
18 3
13 12
+
+ (1 + 5)
=
18 18
=
7
25
+6=7 .
18
18
2. Note in the solution below that you can borrow an integer and transform it into a
fraction with the given denominator so that both fraction and the integer have the
same sign. In this case, we borrow a negative one.
7
1
1
7
2 −5 =
+2−5
−
13
3
13 3
21 13
−3
−
=
39 39
=
8
39
8
−3=
−
−2
39
39 39
31
31
= − − 2 = −2 .
39
39
3. We cannot easily multiply or divide mixed numerals directly. It is easiest to transform them first into improper fractions. Then, we can use rules we already know
for proper fractions.
11
1
3 ×5 =
18
3
54 11
+
18 18
×
65 16
×
18
3
5 × 13 × 24
= 2
3 ×2×3
=
5 × 13 × 23
33
7
520
= 19 .
=
27
27
=
12
15 1
+
3
3
CHAPTER 1. ARITHMETIC
4.
1.2. FRACTIONS
1
2
2 ÷5 =
7
3
14 1
+
7
7
÷
15 2
+
3
3
45 2
+
9
9
15 17
÷
7
3
3
15
×
=
7
17
45
15 × 3
=
.
=
7 × 17
119
=
5.
2
3
9 ÷5 =
5
9
45 3
+
5
5
÷
48 47
÷
5
9
9
48
×
=
9
47
48 × 9
=
5 × 47
197
432
=1
.
=
235
235
=
13
1.3. DECIMALS
1.3
CHAPTER 1. ARITHMETIC
Decimals
1.3.1 ARc2: Arranging fractions and decimals
There are many ways to compare fractions. The easiest one is to find the LCD (Least
Common Denominator) of the two fractions and convert the fractions to that denominator.
Then we can compare the numerators — the fraction with bigger numerator is larger.
This algorithm is more difficult when comparing more than two fractions. However, if a
fraction (a) is bigger than the the bigger of other two (let’s say b < c, and c < a), we do
not need to compare the fractions a and b.
1. We start with 5/3 and 3/2. Their LCD is 6, the equivalent fractions are 10/6 and
15/6 respectively. Therefore, 3/2 > 5/3.
We now compare 2/3 and 5/3. The denominators are the same, therefore 2/3 < 5/3.
We compare 2/3 and 3/5 — LCD=15. The fractions are 10/15 and 9/15 . It follows
then that 2/3 > 3/5.
We compare 3/5 and 6/11 — LCD=55. The fractions are 33/55 and 30/55. Our conclusion is 3/5 > 6/11.
NOTE , THAT WE CAN COMPARE THE DENOMINATORS IF THE NUMERATORS ARE
THE SAME ( WHEN COMPARING POSITIVE NUMBERS ). T HEN , THE FRACTION WITH
smaller DENOMINATOR IS larger!
EXAMPLE :
3/5 = 6/10 > 6/11.
Arranged in increasing order the numbers are:
6/11 < 3/5 < 2/3 < 5/3 < 3/2.
2. To compare two positive numbers, in decimal notation, we start at the left and compare corresponding digits. When two digits differ, the number with the larger digit
is the larger of the two numbers. To facilitate the comparison, extra zeros can be
written to the right of the last decimal place.
We start with 0.0010 and 0.00014. The first two digits in both numbers are zeros, the
third digit in first number is one, and in second is zero. Therefore, 0.0010 > 0.00014.
In the same way: 0.1002 > 0.0010, 0.0003 = 0.00030 > 0.00014, 0.0020 > 0.0010,
0.0020 < 0.1002, 0.0003 < 0.0010.
Arranged in increasing order the numbers are:
0.00014 < 0.0003 < 0.0010 < 0.0020 < 0.1002.
14
CHAPTER 1. ARITHMETIC
1.3. DECIMALS
3. 5/8 < 3/4;
4/9 < 5/8;
3/7 < 4/9;
3/8 < 3/7.
Arranged in increasing order the numbers are:
3/8 < 3/7 < 4/9 < 5/8 < 3/4.
4. 1.2423 > 1.0014;
1.0014 > .0038;
0.0038 > 0.0032;
0.0032 > 0.00038.
Arranged in increasing order the numbers are:
0.00038 < 0.0032 < 0.0038 < 1.0014 < 1.2423.
5. 3/7 > 3/9 = 1/3;
5/13 = 35/91 < 39/91 = 3/7;
5/13 > 5/15 = 1/3;
1/3 = 5/15 > 4/15;
7/9 = 49/63 > 27/63.
Arranged in increasing order the numbers are:
4/15 < 1/3 < 5/13 < 3/7 < 7/9.
15
1.3. DECIMALS
CHAPTER 1. ARITHMETIC
1.3.2 ARc4: Identify the magnitude
To identify the magnitude, we find the first digit which is not zero. There are two cases:
a) The leftmost nonzero digit is on the left of the decimal point. We count how
many digits are on the left from the decimal point. This number shows us the smallest
power of ten bigger than the number - our upper bound. The lower bound is the upper
bound divided by 10.
b) The leftmost nonzero digit is on the right side of the decimal point. We count
how many zeros there are between it and the decimal point. This number will show us
the upper power of ten. The lower bound is equal to the upper bound divided by 10.
1. The first number is 0.457. We are in case b). There are 0 zeros between first non zero
digit and the decimal point. It means that the upper bound is 100 . The lower bound
then is 10−1 :
10−1 ≤ 0.457 ≤ 100 = 1.
2. The second number is 0.0257. There is one zero. We write:
0.01 = 10−2 ≤ 0.0257 ≤ 10−1 = 0.1.
3. The number is 13.0457. We are in case a). We count two digits on the left of the
decimal point. Then, 102 is bigger than the number and 101 is smaller:
10 = 101 ≤ 13.0457 ≤ 102 = 100.
4.
0.0001 = 10−4 ≤ 0.000457 ≤ 10−3 = 0.001.
5.
103 ≤ 3000.457 ≤ 104 = 10000.
16
CHAPTER 1. ARITHMETIC
1.3. DECIMALS
1.3.3 ARc5: Convert between scientific and decimal notation
A number is in scientific notation if it is represent by the product of two factors, the first
is between one and ten, and the second is 10 raised to a whole power. In practice, we take
the decimal point, and move it as many places right as the number power of ten is (when
that power is positive), or, move it the same number of places to the left (when that power
is negative).
Recall that multiplying with powers of ten we simply shift the decimal point that
many places as many zeros we have to the right. The same with division, only in this case
we move decimal point to the left.
1. The first number we convert is 14.87. We can represent it as 14.87 = 1.487 × 10.
14.87 = 1.48710 = 1.487 × 101 .
1.528 × 10−1 =
2.
1.528
= 0.1528.
10
1.487
1.487
=
= 1.487 × 10−2 .
100
102
1.528 × 103 = 1.528 × 1000 = 1528.
0.01487 =
3.
1487 = 1.487 × 1000 = 1.487 × 103 .
1.528
= 0.001528.
1.528 × 10−3 =
103
4.
1140.027 = 1.140027 × 1000 = 1.140027 × 103 .
1.2806
1.2806 × 10−4 =
= 0.00012806.
104
5.
10040.87 = 1.004087 × 10000 = 1.004087 × 104 .
1.528 × 105 = 1.528 × 103 × 102 = 1528 × 100 = 152800.
17
1.4. ESTIMATION
1.4
CHAPTER 1. ARITHMETIC
Estimation
1.4.1 ARd1: Estimate the quotient of two whole numbers
1.
354
≤4
95
because 95 ≈ 100 (”≈” means about equal).
3≤
2.
5≤
586
≤6
106
because 106 ≈ 100.
3.
985
≤6
195
because 985/197 = 5 and 985/195 > 985/197 = 5.
This is difficult estimate. The estimate can be seen using 985 = 1000 − 15 = 5(200 −
3) = 5 × 197, therefore, 985/197 = 5, and the fact that if the denominator of a fraction
is bigger, the fraction is smaller (see ARc2).
5≤
4.
1234
≤ 23
55
because 22 × 55 = 2 × 5 × 11 × 11 = 10 × 121 = 1210 and 1234 = 1210 + 24.
22 ≤
5.
19 ≤
9874
≤ 20
501
because 501 ≈ 500.
18
CHAPTER 1. ARITHMETIC
1.4. ESTIMATION
1.4.2 ARd2: Rounding a number to a given precision (nearest power of
ten)
1. The number 781 is bigger than 500, so that it is closer to 103 than to 102 .
781 ≈ 103 .
2. This number, 1.781, is much closer to one (100 ) than to ten (101 ).
1.781 ≈ 100 .
3.
7.81 ≈ 101 .
4.
0.0781 ≈ 0.1 = 10−1 .
5. The number 22781 is closer to 104 = 10000 than to 105 = 100000. Its approximation
is 104 .
22781 ≈ 104 .
19
1.4. ESTIMATION
CHAPTER 1. ARITHMETIC
1.4.3 ARd4: Estimate the result of a computation using intervals
1. Estimate the sum of numbers between 50 and 60 and numbers between 100 and 115.
We add numbers which are between 50 and 60 (for the first one) and 100 and 115.
Then, the least numbers we could add are 50 and 100, and the largest ones are 60
and 115. Adding the numbers we found will give us the smallest and the biggest
number we could have: 50 + 100 = 150; 60 + 115 = 175. The answer is between
150 = 50 + 100 and 175 = 60 + 115.
2. Estimate the sum of numbers between 500 and 600 and numbers between 1000 and
2000.
The first number is between 500 and 600,the second is between 1000 and 2000. The
answer is between
1500 = 500 + 1000 and 2600 = 600 + 2000.
3. Estimate the product of numbers between 40 and 60 and numbers between 100 and
150.
The smallest numbers we would multiply are 40 (the first factor) and 100 (the second factor). Then, the product is at least 40 × 100 = 4000. The largest numbers we
could multiply are 60 (the first factor) and 150 (second one). Then, the largest number we can obtain is 60 × 150 = 9000. The answer is between
4000 = 40 × 100 and 9000 = 60 × 150.
4. Estimate the subtraction of numbers between 50 and 60 from numbers between 100
and 200.
When we subtract numbers, the logic is the opposite. We find the smallest difference
subtracting the largest subtrahend 1 from the smallest minuend. It is 100 − 60 = 40.
We find the biggest difference by subtracting the least subtrahend from the biggest
minuend. In our case: 200 − 50 = 150. The answer is between:
1
40 = 100 − 60 and 150 = 200 − 50.
The number we subtract from is minuend.The number we subtract is subtrahend. In the example above
100 is minuend and 60 is subtrahend.
20
CHAPTER 1. ARITHMETIC
1.4. ESTIMATION
5. Estimate the division of numbers between 50 and 60 by numbers between 5 and 10.
We find the least answer by dividing the smallest dividend 2 by the largest divisor.
This way, we found 50/10 = 5. We find the biggest answer by dividing the biggest
dividend by the least divisor. We compute 60/5 = 12. The answer is between:
5=
60
50
and 12 = .
10
5
2
The dividend is the number we divide and divisor is the number we divide by. In the example above, 50
is the dividend and 10 is the divisor.
21
1.5. RATIO, PROPORTION AND PERCENT
1.5
CHAPTER 1. ARITHMETIC
Ratio, proportion and percent
1.5.1 ARe1: Convert among equivalent forms: ratio, decimal and percent
1.
35% ≡ 0.35 ≡
35
.
100
72% ≡ 0.72 ≡
72
.
100
0.42 ≡ 42% ≡
42
.
100
2.
3.
4.
5.
12.5
1
=
≡ 0.125 ≡ 12.5%.
8
100
20
1
=
≡ 0.20 ≡ 20%.
5
100
22
CHAPTER 1. ARITHMETIC
1.5. RATIO, PROPORTION AND PERCENT
1.5.2 ARe2: Find a given percent of a base b
1. Percent means part of a hundred. 37% of 22 means
37
× 22.
100
37% of 22 =
37
× 22 = 8.14.
100
25% of 39 =
25
× 39 = 9.75.
100
2.
3.
37.5% of 64 =
37.5
× 64 = 24.
100
5.5% of 120 =
5.5
× 120 = 6.6.
100
4.
5.
11% of 55 =
11
× 55 = 6.05.
100
23
1.5. RATIO, PROPORTION AND PERCENT
CHAPTER 1. ARITHMETIC
1.5.3 ARe3: Find what percent of a is b
Percent is part of a hundred. What percent of a is b means
1. 13 is ? of 65:
2. 16 is ? of 64:
3. 5 is ? of 80:
4. 32 is ? of 256:
5. 11 is ? of 110:
13
× 100% = 20%.
65
16
× 100% = 25%.
64
5
× 100% = 6.25%.
80
32
× 100% = 12.5%.
256
11
× 100% = 10%.
110
24
b
a
× 100.
CHAPTER 1. ARITHMETIC
1.5. RATIO, PROPORTION AND PERCENT
1.5.4 ARe4: Convert from one measurement unit to another
In general, we usually use a conversion table (it is shown below). It is a well-known fact
that there are 1760 yards in one mile.
1 (statute) mile is
1 furlong is
1 chain is
1 fathom is
1 yard is
1 foot is
8 furlongs
10 chains
11 fathoms
2yards
3 feet
12 inches
1. In one yard, there are 3 feet, therefore 2.25 yards are 2.25 × 3 = 6.75 feet. We have 12
inches in one foot, and, in 6.75 feet there are 6.75 × 12 = 81 inches.
2. We transform 7920 feet into yards:
7920 ft
= 2640 yards.
3 ft/yd
Next, we use the fact that in one mile there are 1760 yards:
2640
= 1.5 statute miles.
1760
3. 15 inches are 15/12 feet:
15
= 1.25 feet.
12
4. We assume US gallons (one US gallon equals 8 pints, so one quarter USG has two
pints).
1.75 gal = 1.75 × 8 pints
= 14 pints =
14
quarts = 7 quarts.
2
Answer: 1.75 gallons are 7 quarts.
5. 3.5 feet are 3.5 × 12 inches:
3.5 × 12 = 42 inches.
25
1.5. RATIO, PROPORTION AND PERCENT
CHAPTER 1. ARITHMETIC
1.5.5 ARe5: Solve a proportion given three terms
1. We can think of proportions as two different representations of the same fraction. 5
is to 3 as x is to 10 means that if we divide 5 by 3, we will have the same answer as
x
. We solve this equation for x:
when we divide x by 10. With other words 35 = 10
5
x
= .
3
10
5
50
= .
3
3
We solve all the other examples the same way.
x = 10 ×
2.
3.
4.
5.
9
13
= .
5
x
x
5
= .
9
13
45
9×5
= .
x=
13
13
15
x
= .
3
10
9
3 × 15
= .
x=
10
2
16
12
= .
x
10
x
10
= .
12
16
15
12 × 10
= .
x=
16
2
15
19
= .
x
14
14
x
= .
19
15
19 × 14
266
x=
=
.
15
15
26
CHAPTER 1. ARITHMETIC
1.5. RATIO, PROPORTION AND PERCENT
1.5.6 ARe6: Solve a proportion given a verbal description.
1. Fred can drive 75 miles in 4 hours. How long does it take Fred to drive 100 miles if
he drives at the same rate?
We express the given information as a proportion. Our assumption is that Fred
drives at the same rate. We can say that 75 miles is to 4 hours as 100 miles is to x
hours. We have:
100
75
=
,
4
x
x=
4
16
× 100 = .
75
3
Fred can drive 100 miles in 16/3 hours (5 hours 20 minutes).
2. 5 pounds of salami costs $15.35. My dog ate $9.21 worth of salami. How many
pounds did she eat?
5 (pounds) is to $15.35 (cost for 5 pounds) as x (pounds) is to $9.21 :
5
x
=
,
15.35
9.21
x=
5
× 9.21 = 3.
15.35
The dog ate 3 pounds.
3. My bicycle tire makes 130 revolutions going around a circular track whose circumference is 950 feet. If it makes 195 revolutions going around a different circular track,
what is the circumference of this new track?
130 (revolutions) is to 950 (feet) as 195 (revolutions) is to x (feet).
130
195
=
,
950
x
x=
950
× 195 = 1425.
130
The circumference of the new track is 1425 feet.
27
1.5. RATIO, PROPORTION AND PERCENT
CHAPTER 1. ARITHMETIC
4. 60 feet of rope costs 24 dollars. How much does 100 feet of the same rope cost?
60 (feet) is to 24 (dollars) as is 100 (feet) to x (dollars).
100
60
=
,
24
x
x=
24
× 100 = 40.
60
100 feet of rope cost 40 dollars.
5. A recipe calls for 3 cups of flour and 2 cups of sugar. I only have 1 and 1/3 cups of
sugar. If I want to reduce the recipe, how much flour will I need to use?
3 (cups of flour) is to 2 (cups of sugar) as x (cups of flour) is to 1 13 (cups of sugar).
We can write the mixed numeral as a fraction 1 13 = 34 .
3
x
=
.
2
4/3
Recall also that dividing by a fraction is equivalent to multiplying with its reciprocal.
x=
3 4
× = 2.
2 3
The reduced recipe needs 2 cups of flour.
28
CHAPTER 1. ARITHMETIC
1.5. RATIO, PROPORTION AND PERCENT
1.5.7 ARe7: Find the third, given two of the following: base amount,
percent of increase/decrease, resulting amount.
1. Yesterday a shirt cost $10.00. Today it costs $8.25. What is the percentage decrease
in the price?
We find first the value of the decrease. It is $10.00 − $8.25 = $1.75. Then we divide
the decrease to the yesterday’s price and multiply that ratio by 100.
$10.00 − $8.25 = $1.75.
1.75
× 100 = 17.5.
10.00
The percentage decrease in a price of the shirt is 17.5%.
% decrease =
2. A store is having a sale, everything is 25 percent off. I bought a shirt for $8.00. How
much would the shirt have cost me if there was no sale?
25 percent off means that one must pay 75 percent of the regular price of the shirt.
We then say that 75% of the price is $8.00. Therefore, 75% is to $8.00 as 100% is to x
dollars.
100
75
=
.
8
x
800
2
x=
= 10 .
75
3
The shirt cost $10.67.
3. Today I weigh 166 pounds, which is 2% less than I weighed yesterday. How much
did I weigh yesterday?
166 pounds are 98% of my weight from yesterday. 98 percent is to 166 as 100 percent
is to x.
100
98
=
.
166
x
166
19
x=
× 100 = 169 .
98
49
19
My weight yesterday was 169 49 pounds.
4. Yesterday the temperature in my office was 85 degrees. Today the thermostat was
replaced and it is now 70 degrees. What is the percentage decrease in the temperature?
We use the same algorithm as in #1:
decrease = 85 − 70 = 15.
%decrease =
15
300
11
decrease
× 100 =
× 100 =
= 17 .
yesterday’s temperature
85
17
17
11
The percentage decrease in the temperature is 17 17
%.
29
1.5. RATIO, PROPORTION AND PERCENT
CHAPTER 1. ARITHMETIC
5. There are 50 gallons of water in a tank. 20 gallons of water are added. What is the
percentage increase in the water in the tank?
The initial quantity was 50 gallons. We add 20 gallons, which is the amount of the
increase. We say then:
20
× 100 = 40.
% increase =
50
The percentage of increase of the water in the tank is 40%.
30
Chapter 2
Basic Algebra
2.1
Formulas and expressions
2.1.1 AAa1: Evaluate expressions given numerical values for variables
1. If x = 2 and y = −3, then
2. If x = 1 and y = 2, then
2 + (−3)2
x + y2
=
= −11.
x+y
2 + (−3)
√
x + y3
1 + 23
=
= 1.
x+y
1+2
p
3. If x = 2 and y = −3, then (x − 1)(x − 2) + y = (2 − 1)(2 − 2) + (−3) = −3.
4. If x = 7 and y = 2/3, then
5. If x = 1 and y = 5, then
x
7
=
x + 3y
7+3×
2
3
7
= .
9
x4 + y 2
14 + 52
26
13
=
=
= .
x+y
1+5
6
3
31
2.1. FORMULAS AND EXPRESSIONS
CHAPTER 2. BASIC ALGEBRA
2.1.2 AAa2: Find the numerical value of one variable in a formula given
numerical values of the other variables
We, in general, first substitute and then simplify the expression solving it for the quantity
we do not know.
1. 2LW + 2HW + 2LH = 15. Find L if H = 2 and W = 4.
We substitute the known quantities and solve for L.
(2 × L × 4) + (2 × 2 × 4) + (2 × L × 2 = 15),
8L + 16 + 4L = 15,
12L = −1,
L=−
1
.
12
2. LW H = 30. Find L if H = 2 and W = 5.
LW H = 30,
L × 2 × 5 = 30,
L = 3.
3. Ax2 + Bx + C = 13. Find A if x = 2, B = 4 and C = −2.
Ax2 + Bx + C = A22 + (4 × 2) + (−2) = 13,
4A + 8 − 2 = 13,
4A + 6 = 13,
4A = 13 − 6 = 7,
7
A= .
4
32
CHAPTER 2. BASIC ALGEBRA
4.
2.1. FORMULAS AND EXPRESSIONS
A+B
= E. Find C if A = 2, B = 4, D = 9 and E = 7.
C +D
A+B
= E.
C +D
We substitute and solve for C.
A+B
= E,
C +D
2+4
= 7,
C +9
6 = (C + 9)7,
6 = 7C + 63,
7C = 6 − 63 = −57,
57
C=− .
7
5. A2 + B 2 = C 2 . Find B if A = 7 and C = 25 and B < 0.
A2 + B 2 = C 2 ,
72 + B 2 = 252 ,
B 2 = 625 − 49 = 576.
We can factor 576 to see whether it is a perfect square.
576 = 2 × 288 = 2 × 2 × 144 = 2 × 2 × 12 × 12 = 24 × 24 = 242 .
Knowing the fact that B < 0 we write the answer
√
B = − 576 = −24.
33
2.1. FORMULAS AND EXPRESSIONS
CHAPTER 2. BASIC ALGEBRA
2.1.3 AAa3: Solve a formula for one variable in terms of the others
In general we use commutative, distributive and associative laws and add or subtract
terms from each side of the identities in order to isolate the variable we need to solve for.
1. 2LW + 2HW + 2LH = S. Find L in terms of S, H, and W .
We subtract 2HW from both sides of the identity. Then, we factor out L. We finally
divide the equality by the factor 2W + 2H, assuming it is not equal to zero.
2LW + 2LH = S − 2HW,
L(2W + 2H) = S − 2HW,
S − 2HW
L=
.
2W + 2H
2. LW H = V . Find L in terms of W , H, and V . We simply divide V by WH, assuming
HW 6= 0 :
V
.
L=
WH
A+B
. Find A in terms of B, n, R, and T .
nR
First, we multiply the identity by (nR).
3. T =
T nR = A + B
Second, we subtract B from both sides of the identity.
nT R − B = A + B − B = A.
4.
A+B
= E. Find C in terms of A, B, D, and E.
C +D
(A + B) = E(C + D),
A+B
,
C +D =
E
A+B
− D,
C=
E
so long as E 6= 0.
5. A2 + B 2 = C 2 . Find B in terms of A and C if B < 0.
B 2 = C 2 − A2 .
Because B < 0, and the function square root of x is always non-negative, we must
have the negative of the square root.
√
B = − C 2 − A2 .
34
CHAPTER 2. BASIC ALGEBRA
2.1. FORMULAS AND EXPRESSIONS
2.1.4 AAa4: Simplify expressions by removing grouping symbols
The distributive law is in charge in the following examples. Do not forget to multiply all
the addends by the multiplier. Note, that brackets can be used as another set of parenthesis. Remember, we open the parenthesis from the inside out, i.e. the most inner pair
first.
1. 4(x − 5) + 3(x + 2)
= 4x − 20 + 3x + 6
= 7x − 14
= 7(x − 2).
2. 4[x − (x − 3)] + 3[x + (2 − 3x)]
= 4[x − x + 3] + 3[x + 2 − 3x]
We needed to change the signs of x and 3 to minus and plus accordingly. Then,
x − x = 0.
= 4 × (3) + 3((−2x + 2)
= 12 + 3[−2x + 2]
= 12 − 6x + 6 = 18 − 6x
= 6(3 − x).
3. 7(x + 15) + 3[4y − (x + 2)]
= 7x + 105 + 3[4y − x − 2]
= 7x + 105 + 12y − 3x − 6
= 4x + 12y + 99.
4. 9(y + x − 5) + 3[(x + 2) − y]
= 9y + 9x − 45 + 3[x + 2 − y]
= 9y + 9x − 45 + 3x + 6 − 3y
= 12x + 6y − 39
= 3(4x + 2y − 13).
5. −3(x − 5) + 4(x + 2)
= −3x + 15 + 4x + 8
= x + 23.
35
2.1. FORMULAS AND EXPRESSIONS
CHAPTER 2. BASIC ALGEBRA
2.1.5 AAa5: Simplify rational expressions
We factor out common factors and simplify by dividing out factors equivalent to expressions equal to one.
1.
5x + 10
+ 1)(x + 2)
The numerator has common factor 5 and we factor it out.
(x2
=
(x2
5(x + 2)
+ 1)(x + 2)
We see that the numerator and the denominator have common factor x + 2, so we
have a factor equal to one — (x + 2)/(x + 2), assuming x 6= −2.
=
2.
(a2 + ab)
a
x2
5
.
+1
We assume a 6= 0.
a(a + b)
a
= a + b.
=
3.
5(x + 2) + 10
4x + 16
We assume x 6= −4.
=
5(x + 2 + 2)
4(x + 4)
=
4.
xy + yz
ya + yb
5(x + 4)
4(x + 4)
5
= .
4
We assume y 6= 0.
y(x + z)
y(a + b)
x+z
.
=
a+b
=
5.
abc
ab + bc
We assume b 6= 0.
abc
b(a + c)
ac
=
.
a+c
=
36
CHAPTER 2. BASIC ALGEBRA
2.1. FORMULAS AND EXPRESSIONS
2.1.6 AAa6: Add, subtract, multiply or divide polynomial and rational
expressions
When adding rational expressions we first find the Least Common Denominator -(LCD). We
multiply each expression by the appropriate factor so all expressions will have one and
the same denominator.
Dividing fractions is actually multiplying by the reciprocal fraction of the divisor. We
must, in all cases to assume that all the denominator factors are not equal to zero.
1.
5
5x + 10
÷
+ 1)(x + 2) x + 2
As mentioned, we multiply by the reciprocal of the divisor instead of dividing. In
the first rational expression, we factor out 5 in the numerator. We assume x 6= −2.
(x2
=
(x2
5(x + 2)
x+2
×
+ 1)(x + 2)
5
We see that there are equal factors in the numerator and the denominator - 5 and
(x + 2). We divide out these factors and obtain:
=
x+2
.
x2 + 1
2. (5x2 + 4x + 9) + (3x3 − 2x + 11)
We open the parenthesis and combine like terms. We arrange the polynomial in
descending order of exponents.
= 3x3 + 5x2 + (4x − 2x) + (9 + 11)
= 3x3 + 5x2 + 2x + 20.
3. (−5x2 + 14x + 90) − (3x3 − 2x + 11)
We open the parenthesis as a first step. The sign before the second pair is negative.
It changes all the signs of the terms in these parenthesis:
= −5x2 + 14x + 90 − 3x3 + 2x − 11
= −3x3 − 5x2 + (14x + 2x) + (90 − 11)
= −3x3 − 5x2 + 16x + 79.
37
2.1. FORMULAS AND EXPRESSIONS
CHAPTER 2. BASIC ALGEBRA
4. (4x4 + 38x3 + 23x2 − 57x − 21) ÷ (x2 + 9x + 3)
We use a process similar to the long division of numbers.
4x2 + 2x − 7
x2 + 9x + 3)4x4 + 38x3 + 23x2 − 57x − 21
−(4x4 + 36x3 + 12x2 )
2x3 + 11x2 − 57x
−(2x3 + 18x2 + 6x)
−7x2 − 63x − 21
−(−7x2 − 63x − 21)
0
(4x4 + 38x3 + 23x2 − 57x − 21) ÷ (x2 + 9x + 3) = 4x2 + 2x − 7.
5. (5x2 + 4x + 9) × (3x3 − 2x + 11) =
When multiplying polynomials, we multiply each term in the first parenthesis with
each term in the second pair and combine like terms. In our case, before simplifying,
we must have 9 terms! This will help you to check your answer.
= (15x5 − 10x3 + 55x2 ) + (12x4 − 8x2 + 44x) + (27x3 − 18x + 99)
= 15x5 + 12x4 + (−10 + 27)x3 + (55 − 8)x2 + (44 − 18)x + 99
= 15x5 + 12x4 + 17x3 + 47x2 + 26x + 99.
Another way is to organize the multiplication as multiplication of numbers. Note,
in the process below how the terms are organized leaving space for missing terms.
3x3 − 2x + 11
5x2 + 4x + 9
27x3
− 18x + 99
4
2
12x
− 8x + 44x
15x5
10x3 + 55x2
15x5 + 12x4 + 17x3 + 47x2 + 26x + 99
38
CHAPTER 2. BASIC ALGEBRA
2.1. FORMULAS AND EXPRESSIONS
2.1.7 AAa7: Convert from verbal statement to algebraic expression or
equation
1. The sum of the squares of the lengths of the sides of a right triangle is equal to the
square of the length of the hypotenuse.
Let K and L be the lengths of the sides and let H be the length of the hypotenuse.
K 2 + L2 = H 2 .
2. The surface area of a rectangular solid is found by multiplying the length and the
width, multiplying the height and the width, and multiplying the height and the
length, adding these three quantities together, and doubling the resulting number.
Let S be the surface area, H be the height, W be the width, and L be the length.
S = (LW + HW + HL) × 2.
3. If you triple Fred’s age, you get the same number as if you double Fred’s age and
add 10 years.
Let A be Fred’s age.
3A = 2A + 10.
4. Five touchdowns, five extra points, and three field goals gives forty four points. A
touchdown is six points, a field goal is three points and an extra point is one point.
Let T be the number of touchdowns, E be the number of extra points, and F be the
number of field goals.
We multiply the number of touchdowns T by the points of each: 6, the number of
extra points E by the points of each: 1, and the field goals F by the points of each: 3.
We add all three products and the sum equals 44:
6T + E + 3F = 44.
5. In 36 minutes against the Bucks, Jermaine O’Neal sank eighteen of twenty eight two
point buckets and nineteen of twenty five free throws, for a total of fifty five points.
Let T be the number of two point buckets and F be the number of free throws. We
know that free throw is 1 point so that we multiply the number of free throws F by
one. We find the points of buckets sunk by multiplying their number by two - the
weight of each one. Then, we add these two numbers and their sum is 55:
2T + F = 55.
39
2.2. LINEAR EQUATIONS
2.2
CHAPTER 2. BASIC ALGEBRA
Linear Equations
2.2.1 AAb1: Solve ax + b = cx + d when a, b, c, d are integers
We need to isolate the unknown variable on one side of the equation and all other terms
on the other side.
1. 3x + 4 = 8x − 11:
We subtract 3x from both sides of the equation and add 11 in the same way.
3x − 3x + 4 + 11 = 8x − 3x − 11 + 11,
5x = 15.
We divide now both sides of this equation by 5:
5x
15
= ,
5
5
x = 3.
The last step in solving an equation is checking the answer. We do so by substituting
the solution in the original equation:
3 × 3 + 4 = 13,
8 × 3 − 11 = 13,
so 3 × 3 + 4 = 8 × 3 − 11.
The solution of the equation 3x + 4 = 8x − 11 is x = 3.
2. 4x − 9 = 2x + 1:
4x − 2x − 9 + 9 = 2x − 2x + 1 + 9,
2x = 10,
x = 5.
Check:
4 × 5 − 9 = 11,
2 × 5 + 1 = 11,
so 4 × 5 − 9 = 2 × 5 + 1.
40
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
3. 5x − 44 = −2x + 3:
7x = 47,
47
x= .
7
Check:
73
47
− 44 = − ,
7
7
73
−2 × 477 + 3 = − ,
7
5×
so 5 ×
4. −3x + 24 = −8x − 11:
47
− 44 = −2 × 477 + 3.
7
5x = −35,
x = −7.
Check:
−3 × (−7) + 24 = 45,
−8 × (−7) − 11 = 45,
so − 3 × (−7) + 24 = −8 × (−7) − 11.
5. 3x + 4 = −11:
3x = −15,
x = −5.
Check:
3 × (−5) + 4 = −11
so 3 × (−5) + 4 = −11
41
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
2.2.2 AAb2: Solve ax + b = cx + d when a, b, c, d are rational numbers
Solving linear equations with rational coefficients is pretty much the same as solving
linear equation with integer coefficients. The only difference is that, when simplifying
we need to find LCD of the numbers we need to add, and use the proper way of addition/subtraction of fractions and for multiplication/division as well.
NOTE! A common mistake students make is to omit x from one of the fractions. The
other mistake is not using parenthesis, when appropriate, in which case the fractions are
incorrectly added/subtracted or multiplied. The way fractions are written is important,
because sometimes students do not recognize where the fraction or expression ends. It is
also important for the student to check his/her work.
1.
2
1
1
2
x+ = x+ :
5
3
5
3
2
1
1 1
1
1
2 1
x− x+ − = x− x+ − ,
5
5
3 3
5
5
3 3
1
1
x= .
5
3
In above equation, if we divide both sides by 15 , we will have a coefficient of one in
front of unknown variable x. In other words, we multiply by the reciprocal number
of the coefficient of x.
5
x= .
3
Check:
2 5 1
× + = 1,
5 3 3
1 5 2
× + = 1,
5 3 3
1 5 2
2 5 1
so × + = × + .
5 3 3
5 3 3
2.
1
1
2
2
x+
= x+ .
7
14
5
5
1
1
1
1
1
2
1
2
x− x+
−
= x− x+ − ,
7
5
14 14
5
5
5 14
2 × 14
1×5
2×5 1×7
x=
−
−
,
7×5 5×7
5 × 14 14 × 5
23
3
x= .
35
70
42
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
We divide both sides by 3/35 (Recall, that is, we multiply by 35/3).
23 35
3
23
3
3
x=
=
÷
÷
× ,
35 35
70 35
70
3
x=
Check:
3.
2
1
1
1
x+ = x+ :
9
7
3
14
23
.
6
1
49
7
2 23
×
+
=
= ,
7
6
14
42
6
1 23 2
23 12
35
7
×
+ =
+
=
= ,
5
6
5
30 30
30
6
2 23
1
1 23 2
so ×
+
= ×
+ .
7
6
14
5
6
5
1 2
−
3 9
x=
1
1
− ,
7 14
1
1
x= ,
9
14
1
1
÷
,
x=
14
9
x=
Check:
4.
3
1
1
1
x+ =− x+ :
4
6
5
10
9
.
14
9
1
2
2
×
+ = ,
9 14 7
7
1
9
1
4
×
+
=
=
3 14 14
14
9
1
1
9
2
+ = ×
so ×
9 14 7
3 14
3 1
+
4 5
x=
+
1
1
− ,
10 6
19
2
x=− ,
20
30
2
20
x=− × ,
30 19
4
x=− .
57
43
2
,
7
1
.
14
2.2. LINEAR EQUATIONS
Check:
CHAPTER 2. BASIC ALGEBRA
1
4
1
1
6
19
13
3
+ =− + =−
× −
+
=
,
4
57
6
19 6
114 114
114
4
1
4
1
8
57
65
13
1
+
=
+
=
+
=
=
,
− × −
5
57
10
285 10
570 570
570
114
3
1
1
4
1
4
so × −
+ =− × −
+ .
4
57
6
5
57
10
5.
1
1
2
2
− x+ = x+ ,
9
8
5
3
2 1
− −
9 5
x=
2 1
− ,
3 8
13
19
x= ,
45
24
13 45
x=− × ,
24 19
195
,
x=−
152
−
Check: We will use method similar to the one in #4. First, we will reduce the fractions to lowest terms and then we will multiply by LCD.
2
1
1 13 × 15 2
−13 × 15
− ×
+ =− ×
+ ?
9
8 × 19
8
5
8 × 19
3
2 13 × 5 1
13 × 3 2
×
+ =−
+ ?
3 8 × 19 8
8 × 19 3
The least common denominator is LCD = 3 × 8 × 19 so we multiply both sides of
the equality with it.
2 13 × 5 1
×
+
3 8 × 19 8
× 3 × 8 × 19 =
13 × 3 2
−
+
8 × 19 3
3 × 8 × 19?
2 × 13 × 5 + 3 × 19 = −13 × 3 × 3 + 2 × 8 × 19?
so
130 + 57 = −117 + 304 ⇔ 187 = 187?
2 13 × 5 1
13 × 3 2
× 3 × 8 × 19 = −
3 × 8 × 19.
×
+
+
3 8 × 19 8
8 × 19 3
44
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
2.2.3 AAb3: Solve ax + b = cx + d when some of a, b, c, d are letters
We use the same method to solve equations even when we have some parameters we do
not know. We isolate the variable we want to solve for. In #1, we subtract b from both
sides and we subtract as well 1/5x.
1.
1
2
2
x+b= x+ ,
5
5
3
1
1
1
2
2
x − x + b − b = x − x + − b,
5
5
5
5
3
1
2
x = − b.
5
3
On the right side, we add 23 and (−b), for which we use LCD = 3. We look at
number b as having denominator 1 ( 1b ).
2 3b
1
x= − ,
5
3
3
1
2 − 3b
x=
.
5
3
We multiply both sides of the equation by 5:
x=
2.
1
2
x + b = x + d,
7
5
3. ax +
5(2 − 3b)
.
3
1
1
1
2
x − x + b − b = x − x + d − b,
7
5
5
5
3
x = d − b,
35
35(d − b)
x=
.
3
1
1
= x + d,
7
3
1
1 1
1
1
1
ax − x + − = x − x + d − ,
3
7 7
3
3
7
1
1
a−
x=d− ,
3
7
1
We assume a 6= . We can divide both side by (a − 1/3) because a − 1/3 6= 0.
3
d − 17
,
x=
a − 13
45
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
x=
(7d − 1)/7
7d − 1 3
=
× ,
(3a − 1)/3
3a − 1 7
x=
3(7d − 1)
.
7(3a − 1)
Let now a = 1/3. Then,
1
x+
3
1
x−
3
1
1
= x + d,
7
3
1
1
x=d− ,
3
7
1
0=d− .
7
There is no variable x in the equation and on the right side we have d − 1/7. This
identity is true only if d = 1/7. In this case, the equation has infinitely many solutions. If d 6= 1/7, then there is no solution for the equation.
4. ax +
1
1
= cx + ,
6
10
1
1
−
10 6
10
4
1
6
−
=− =− ,
(a − c)x =
60 60
60
15
ax − cx =
Assume a 6= c. We can divide by a − c because the assumption ensures a − c 6= 0.
x=−
1
.
15(a − c)
We can multiply the numerator and the denominator by (−1). This will remove the
minus sign and in the denominator we have to exchange a and c: (a − c) = −(c − a):
x=
1
.
15(c − a)
We look at the case a = c. We simplify
ax +
1
1
= ax + ,
6
10
1
1
− ,
10 6
1
0=− .
15
This is never true so the equation does not have any solutions.
ax − ax =
46
CHAPTER 2. BASIC ALGEBRA
1
2
2
5. − x + b = x + ,
9
5
3
2.2. LINEAR EQUATIONS
2 1
+
9 5
2
x=b− ,
3
10
3b 2
9
=
+
− ,
45 45
3
3
3b − 2
19
x=
,
45
3
(3b − 2)45
x=
,
19 × 3
15(3b − 2)
x=
.
19
47
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
2.2.4 AAb4: Set up and solve verbal problems using linear equations.
1. I have fifteen coins. Some are nickels, and the rest are dimes. I have $1.25 total. How
many dimes do I have?
Let d denotes the number of dimes and n denotes the number of nickels.
We have two unknowns so that we need two different equations to answer the question.
The first equation we derive using the total number of coins. We have 15 coins —
the sum of the number of nickels (n) and the number of dimes(d) is fifteen.
n + d = 15.
Second, we know how much the total value of the coin is. The total value of nickels
is 5n cents, and of dimes is 10d cents. Their sum is $1.25 or, 125 cents.
5n + 10d = 125.
We find from the first equation how many nickels there are: n = 15 − d. We put this
value in the second equation and then, we solve it for d:
5(15 − d) + 10d = 125,
75 − 5d + 10d = 125,
5d = 125 − 75,
5d = 50,
d = 10,
n = 5.
We found that I have ten dimes and five nickels.
We check the answer: five nickels and ten dimes are 15 coins and their total value is
$1.25.
NOTE! How could you check that your answer is reasonable?
First, the answer is a positive whole number. If you get fraction for the number of
coins, there is an error in your computation.
Second, if you get a NEGATIVE number for the number of coins, there is an error
too.
48
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
2. A pilot notices that when the wind speed is 20 mph he can fly 60 miles with the
wind in the same time needed to fly 40 miles against the wind. How fast can he fly
when there is no wind? (Adapted from Bardell and Spitzbart.)
distance_1 is 60 mi for time t
A)
total speed is (v+20)
60=(v+20)t
wind
plane
t=
60
v+20
60 miles
wind speed is 20 mph (w)
A) and B)
B)
40 miles
plane
distance_2 is 40 mi for time t
wind
total speed is (v-20)
40=(v-20)t
40
t=
v-20
We denote the time by t. Let the speed in still air be v. Then, the speed with the
wind is (v + 20) mph and the speed against the wind is (v − 20) mph. We know that
the time for both is one and the same. We use the identity connecting the distance,
speed and time.
distance = time × speed
or
time =
distance
.
speed
We express t for the two flights — with the wind and against:
distance 1
60
=
,
speed 1
v + 20
40
distance 2
=
.
t=
speed 2
v − 20
t=
where the distance 1 is distance flown with the wind, and the other one — against
accordingly. The left sides are equal, so we can equate the right sides of equations:
60
40
=
.
v + 20
v − 20
49
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
We solve the resulting equation:
60(v − 20) = 40(v + 20),
60v − 1200 = 40v + 800,
20v = 2000,
v = 100.
The pilot can fly with 100 mph in still air.
Check: With the wind of 20 mph and a speed of 100 mph, the pilot will fly 60 miles
for time
t=
1
60
= hours.
100 + 20
2
t=
40
1
= hours.
100 − 20
2
Against the wind the time is
The times are equal.
50
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
3. What part of a 40% solution of alcohol in water must be replaced by pure alcohol to
get a 60% solution of alcohol in water? (From Bardell and Spitzbart.)
The amount of solution is not mentioned. In such case, we can use any amount, and
simplest is 1 liter. Using this amount we find that in it, 40% is 0.4 liters and 60% is
0.6 liters.
One may guess that we could replace 20% of the solution to get 60% mix. Let’s
check:
(1 − 0.2) × .4 + 0.2 = .32 + .2 = .52 = 52%.
1 liter 40%
alcohol
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x liters 100%
alcohol
40%
x
x
100%
(1-x) 40% +x100% alcohol
=1 liter 60%
This answer is wrong so that we need other way of computing how much of the
solution we need to change. Let us denote that amount with x. From our original
solution we have remaining only (1 − x) × 0.4 liters of alcohol. We need to add to
this quantity x liters of alcohol (it is 100% pure). So the result must have 0.60 parts
of alcohol. We translate into an equation:
(1 − x) × 0.4 + x = 0.6,
0.4 − 0.4x + x = 0.6,
0.6x = 0.2,
We must replace
1
3
6x = 2,
1
x = liters.
3
part of 40% solution in order to make a 60% solution.
51
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
Check: if we replace 1/3 part of a 40% solution with pure alcohol, we will have
alcohol content
1
1
2
40
1
(1 − ) × 0.4 + = ×
+
3
3
3 100 3
5
9
3
4
+
=
= = 0.60.
=
15 15
15
5
Since the alcohol is 0.6 liters in 1 liter liquid we have 60% alcohol solution.
52
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
4. In an alloy of copper and tin weighing 25 pounds, 35% is copper. How many
pounds of copper must be added so that the alloy is 50% copper? (From Bardell
and Spitzbart.)
We can compute how many pounds is copper and how many pounds is tin. The tin
quantity is (0.65 × 25) pounds. The amount for the copper is (0.35 × 25) pounds. We
need to add some copper so its quantity becomes equal to that of tin.
Let’s denote the added amount with x. Then, we will have the quantity of tin
(0.65 × 25) equal to the amount of copper (0.35 × 25 + x). The equation is:
0.65 × 25 = (0.35 × 25) + x,
x = 16.25 − 8.75 = 7.5 pounds.
We need to add 7.5 pounds of copper to get an alloy that is 50% alloy copper and
tin.
NOTE! This problem differs from the one before because now we change the amount
of alloy. The ending amount is 25 + 7.5 = 32.5 pounds. A common error students
make is not to multiply one of the amounts with the total amount of alloy. In such
case, one can produce negative amount to be added, which makes no sense.
Check: If we add 7.5 pounds of copper to 25 pounds of 35% copper and tin, we will
have
7.5 + .35 × 25 = 7.5 + 8.75 = 16.25 pounds cooper.
The tin in the alloy is the same amount as in the original alloy and is
0.65 × 25 = 16.25 pounds tin.
The quantities are the same so the alloy is 50% tin and 50% copper.
53
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
5. What proportions of a 20% solution of alcohol in water and a 30% solution of alcohol in water must be combined to get a solution that is 28% alcohol? (Adapted from
Bardell and Spitzbart.)
This problem is similar to problem #3. We need to construct two equations because
we have two proportions of two solutions to mix. Let us denote the portion of the
first solution by x and second by y. We need parts, so we think parts of one - we will
have 1 liter of each solution, and we will produce 1 liter final solution.
20 % alcohol
solution
0.2x
28 % alcohol
solution
30 % alcohol
solution
0.2x
x
x
y
+
=
0.3y
0.3y
x + y = 1 liter
y
(1)
0.2x+0.3y=0.28 x1 (2)
The total amount we need is one liter. So, we can say that the sum of the amounts
of solution one (x), and two (y) is 1:
x + y = 1.
At the same time, we need 28% alcohol solution. First solution is 20%, so the alcohol
in it is 0.2 × x, and in the second is 0.3 × y. The resulting solution is 28%. Then the
alcohol content in it is 0.28 × 1 (since we make 1 liter). The equation is
0.2x + 0.3y = 0.28 × 1.
We now solve the system:
x + y = 1,
0.2x + 0.3y = 0.28 × 1.
We multiply the second equation in the system by 5 and subtract the first one from
it:
x + y = 1,
x + 1.5y = 1.4.
54
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
We solve this equation:
0.5y = 0.4,
y = 0.8.
Now, we use the first equation (x + y = 1) to find the amount of the other solution.
It is 0.2 liters.
In parts, it is (0.2/1) parts 20% solution and (0.8/1) 30% solution to make 28% solution. We can make the numerators of these fractions whole numbers - 1/5 part 20%
solution and 4/5 parts 30% solution to make solution which is 28%.
Check: If we combine 1/5 part 20% solution and 4/5 parts 30% solution we will have
alcohol content (in 1 liter solution):
20
4
30
1
×
+ ×
5 100 5 100
1
140
(20 + 4 × 30) =
= 0.28 liters.
500
500
This is exactly 28% solution.
=
55
2.2. LINEAR EQUATIONS
CHAPTER 2. BASIC ALGEBRA
2.2.5 AAb5: Solve ax + b = cx + d after removing grouping symbols.
1. 2(x + 9) − 2 = 5x + 12.
We multiply 2 by x and 9:
2x + 18 − 2 = 5x + 12.
We subtract 5x and 16 from both sides of the equation
2x + 16 − 5x − 16 = 5x + 12 − 5x − 16,
−3x = −4,
4
x= .
3
We check our work by substituting the solution into the original equation:
62 − 6
4
+9 −2=
,
2
3
3
4
62 − 6
5 × + 12 =
,
3
3
4
4
+ 9 − 2 = 5 × + 12.
so 2
3
3
2. 3(x − 9) − x = 5(x + 12).
3x − 27 − x = 5x + 60,
−3x = 87,
x = −29.
Check:
3(−29 − 9) − (−29) = −85,
5(−29 + 12) = 5(−17) = −85,
so
3(−29 − 9) − (−29) = 5(−29 + 12) = 5(−17).
3. 2(x − 4) − 2x = x + 12.
2x − 8 − 2x = x + 12,
x = −20.
Check:
2(−20 − 4) − 2(−20) = −48 + 40 = −8,
so
−20 + 12 = −8,
2(−20 − 4) − 2(−20) = −20 + 12.
56
CHAPTER 2. BASIC ALGEBRA
2.2. LINEAR EQUATIONS
4. 12(x − 19) + 2 = 3(x − 12).
12x − 12 × 19 + 2 = 3x − 3 × 12,
12x − 3x − 12 × 19 + 2 + 12 × 19 − 2 = 3x − 3x − 3 × 12 + 12 × 19 − 2,
9x = 12(19 − 3) − 2 = 190,
x=
Check:
12(
190
.
9
190 − 171
82
190
− 19) + 2 = 12(
+2= ,
9
9
3
190 − 108
82
190
− 12) = 3 ×
= ,
3(
9
9
3
so
12(
190
190
− 19) + 2 = 3(
− 12).
9
9
5. 2(x + 9)/3 − 2 = 5x + 12.
Check:
so
2
x + 6 − 2 = 5x + 12,
3
13
− x = 8,
3
24
x=− .
13
−24 + 117
24
/3 − 2
2 − + 9 /3 − 2 = 2 ×
13
13
93
62
62 − 26
36
=2×
−2=
−2=
= ,
13 × 3
13
13
13
24
−120 + 12 × 13
156 − 120
36
5× −
+ 12 =
=
= ,
13
13
13
13
24
24
+ 12.
2(− + 9)/3 − 2 = 5 × −
13
13
57
2.3. MULTIPLICATION AND FACTORING
2.3
CHAPTER 2. BASIC ALGEBRA
Multiplication and Factoring
2.3.1 AAc1: Find products of the form (ax + b)(cx + d) and simplify
DISTRIBUTIVE PROPERTY: We usually multiply each term in the first parenthesis with
each term in the second parenthesis. We have then, before simplifying, a total of 4 terms
(2 × 2). If you have fewer or more than 4, you probably made an error. The mnemonic
device student use a lot is called FOIL - we multiply First terms, Outer terms, Inner terms
and Last terms.
1. (2x + 1)(3x − 2)
Using FOIL: First terms are 2x and 3x, Outer terms are 2x and (−2), Inner terms are
(+1) and 3x, Last terms are (+1) and (−2).
= 2x × 3x + 2x × (−2) + 1 × 3x + 1 × (−2)
= 6x2 − 4x + 3x − 2
= 6x2 − x − 2.
2. (3x + 2)(4x − 2)
F = 3x × 4x = 12x2 .
O = 3x × (−2) = −6x.
I = 2 × 4x = 8x.
L = 2 × (−2) = −4.
= 12x2 − 6x + 8x − 4
= 12x2 + 2x − 4.
3. (−2x + 11)(3x + 2)
= −6x2 − 4x + 33x + 22
= −6x2 + 29x + 22.
4. (2x − 3)(5x + 7) The other way of multiplying binomials is using the distributive
property twice. We could think about the sum in the second parenthesis 5x + 7 as a
single number and use the distributive property.
= 2x(5x + 7) − 3(5x + 7)
Then, we have an addition of two products so we apply the distributive property
again on each product.
= 10x2 + 14x − 15x − 35
= 10x2 − x − 21.
58
CHAPTER 2. BASIC ALGEBRA
2.3. MULTIPLICATION AND FACTORING
5. (x + 4)(7x + 2)
We use the same method as in #4. The only difference is that we think of the quantity
in the first set of parentheses as a number.
= (x + 4)7x + (x + 4)2
7x2 + 28x + 2x + 8
= 7x2 + 30x + 8.
Note, that one can use the way which is most convenient. However, FOIL will not
work if one of the factors is not a binomial, while the distributive property approach
always works. See also AAa6.
59
2.3. MULTIPLICATION AND FACTORING
CHAPTER 2. BASIC ALGEBRA
2.3.2 AAc2: Find products of the form (ax + b)2 and simplify
Squares of binomials are special kind of products. For them, we can use FOIL method or
we can use formula, which you can memorize, as these appear frequently and the formula
will save you time and effort:
(a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 .
1. (3x − 2)2
We can use the FOIL device for this:
= 9x2 − 6x − 6x + 4
= 9x2 − 12x + 4.
2. (3x + 2)2
We can organize the product as a multiplication table, as this:
3x
2
3x
9x2
6x
2
6x
4
Then, the product is the sum of all the terms in the table:
= 9x2 + 6x + 6x + 4
= 9x2 + 12x + 4.
3. (−2x + 11)2
= (−2x)2 + 2(−2x)11 + 112
= 4x2 − 44x + 121.
Note, if we subtract the second term, or have minus sign before first term, the only
difference is that the double product is subtracted!
4. (2x − 3)2
= 4x2 − 12x + 9.
5. (ax + b)2
= (ax)2 + 2axb + b2
= a2 x2 + 2abx + b2 .
60
CHAPTER 2. BASIC ALGEBRA
2.3. MULTIPLICATION AND FACTORING
2.3.3 AAc3: Factor expressions containing a common monomial factor
We factor out the largest factor we can.
1. 3x3 + 9x + 12
The largest factor is (3):
= 3(x3 + 3x + 4).
Note that we divide all terms in the expression by the factor we took out. This is the
distributive property in reverse. A common mistake is to forget to divide one or
more terms by the factor.
2. 14x3 + 7x2 + 21x
The factor we see here is 7x.
= 7x(2x2 + x + 3).
Dividing by x, we subtract one from the exponent in each term. The last term has
exponent 0.
3. ax2 + a2 x
The monomial we can factor out is (ax).
= ax(a + x).
4. 15x2 − 25
= 5(3x2 − 5).
5. 2x4 + 4x3 + 8x2 + 16x
= 2x(x3 + 2x2 + 4x + 8).
We can check whether we worked it correctly by opening the parentheses. We must have
the expression we stared with. If not, there is an error - either in the factoring or in the
opening the parenthesis!
61
2.3. MULTIPLICATION AND FACTORING
CHAPTER 2. BASIC ALGEBRA
2.3.4 AAc4: Factor by grouping an expression like ax + ac + bx + bc
We usually group the terms two by two. Then, we find a common factor to factor out.
The continuation is finding another factor, this time it is a binomial, which we now factor
out to have complete factoring.
1. ax − bx + 2a − 2b
We can, on our first step, to combine first and second terms in a group, then the
third and fourth terms will form second group, or
we can combine first and third, second and fourth. We will show both possibilities.
(a)
= (ax − bx) + (2a − 2b)
First group has common factor x, and second has 2. We factor them out:
= x(a − b) + 2(a − b)
We see that we have now common factor (a − b), so we factor that out.
= (a − b)(x + 2);
(b) Now we are grouping the first and the third, second and fourth:
= (ax + 2a) − bx − 2b
We factor out in the second group (−1):
= (ax + 2a) − (bx + 2b)
We now factor out a from first group, and b from second.
= a(x + 2) − b(x + 2)
We see common factor (x + 2), so we finish the job:
= (x + 2)(a − b).
The important part here is that one needs to be careful with the pluses and minuses.
If a minus sign is factored out, all signs in the parenthesis must be changed! The
above example shows this.
We check our work by multiplying the factors:
(x + 2)(a − b) = ax − bx + 2a − 2b. TRUE!
62
CHAPTER 2. BASIC ALGEBRA
2. ac + 4a − bc − 4b
2.3. MULTIPLICATION AND FACTORING
= (ac + 4a) − (bc + 4b)
= a(c + 4) − b(c + 4)
= (c + 4)(a − b).
3. ac − 3a + bc − 3b
= (ac + bc) − 3a − 3b
= c(a + b) − 3(a + b)
= (a + b)(c − 3).
4. ac − 3a + 2bc − 6b
= (ac − 3a) + (2bc − 6b)
= a(c − 3) + 2b(c − 3)
= (c − 3)(a + 2b).
5. ac − 3a + 2c − 6
= (ac + 2c) − 3a − 6
= c(a + 2) − 3(a + 2)
= (a + 2)(c − 3).
We can check by multiplying the factors we found. If we worked correctly, we will obtain
the original expression.
63
2.3. MULTIPLICATION AND FACTORING
CHAPTER 2. BASIC ALGEBRA
2.3.5 AAc5: Factor expressions like x2 + bx + c
The most widely used method for students is called (again!) FOIL method. However, this
method does not work in all cases. We look for factors with the following properties:
(a) The sum of the numbers is the coefficient b and
(b) the product is the coefficient c.
In practice, we factor c (into only two factors) and check the sum of them. We find our
numbers when the sum matches coefficient b.
1. x2 + 5x + 6
Using this method, we factor 6: Factors could be
(1) (+1) and (+6), or
(2) (+2) and (+3).
The sum of factors (1) is (+7), which is not equal to (+5); the sum of factors (2) is
(+5), so we can factor the expression now:
= (x + 2)(x + 3).
We check by multiplying the factors:
(x + 2)(x + 3) = x2 + 3x + 2x + 6 = x2 + 5x + 6. TRUE!
2. t2 − t − 2
When coefficient c is negative, we need one negative factor and one positive factor.
The negative factor has to be bigger by its absolute value because our coefficient b
is negative.
The factors are: (−2) and (+1).
= (t − 2)(t + 1).
3. x2 + 6x + 9
Factors: (+1) and (+9) or (+3) and (+3).
We see that the proper factors are equal, and the result is
= (x + 3)(x + 3) = (x + 3)2 .
Note, that this is a formula and there are other ways of factoring it. Recognizing,
we have sum of first number or expression square (x)2 plus doubled product of first
and second term (2 × x × 3 = 6x) plus second number or expression square (32 = 9),
we find that it is the sum of the two terms square (x + 3)2 (see section AAc2).
64
CHAPTER 2. BASIC ALGEBRA
2.3. MULTIPLICATION AND FACTORING
4. a2 − 11a − 60
Factors:
(1)(−60) and (+1), sum (−59) -not equal to (−11);
(2) (−30) and (+2), sum (−28) 6= (−11);
(3) (−20) and (+3), (−17) 6= (−11);
(4) (−15) and (+4), (−11) = (−11)
Our factors are (−15) and (+4).
= (a − 15)(a + 4).
5. x2 + 11x − 60
This expression has almost the same coefficients as the one above. The only one
different is coefficient b. It is the opposite of the coefficient b in previous example.
We simply need to reverse the signs of them: (+15) and (−4).
= (x + 15)(x − 4).
Note the rule: If the number you find is the opposite of the coefficient b, reverse the
signs of the factors to obtain them.
65
2.4. POSITIVE INTEGER EXPONENTS
2.4
CHAPTER 2. BASIC ALGEBRA
Positive Integer Exponents
2.4.1 AAd1: Simplify exponents using the “add exponents” rule
We multiply exponents with common base by adding their exponents. The convention is
to list numerical values first, then symbols in alphabetical order. The convention is to list
numerical values first, then the symbols in alphabetical order. For example, 2w2 xy.
1. xy 3 z 3 × x2 y 3
= x1+2 y 3+3 z 3
= x3 y 6 z 3 .
2. x2 y 3 × 2x3 yz
= 2x2+3 y 3+1 z
= 2x5 y 4 z.
3. 3x2 y 4 × 4x3 y 5 u
= 3 × 4 × x2+3 y 4+5 u
= 12ux5 y 9 .
4. 6ab2 c3 × 2xac2
= 12a1+1 b2 c3+2 x
= 12a2 b2 c5 x.
5. xyx × x2 y 7
= x1+1+2 y 1+7
= x4 y 8 .
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CHAPTER 2. BASIC ALGEBRA
2.4. POSITIVE INTEGER EXPONENTS
2.4.2 AAd2: Simplify expressions using “subtract exponents” rule
1.
x3 y 4
x2 y 2
= x3−2 y 4−2
= xy 2 .
We check the work by multiplying the answer and the divisor:
xy 2 × x2 y 2 = x1+2 y 2+2 = x3 y 4 .
2.
x−3 y 4
x2 y 7
= x−3−2 y 4−7
= x−5 y −3 .
Check:
x−5 y −3 × x2 y 7 = x−5+2 y −3+7 = x−3 y 4 .
x3 y 4
3. −2 8
x y
= x3−(−2) y 4−8
= x5 y −4 .
Check:
x5 y −4 × x−2 y 8 = x5−2 y −4+8 = x3 y 4 .
x3 y 4
4.
zx2 y 2
= x3−2 y 4−2 z 0−1
= xy 2 z −1 .
Check:
xy 2 z −1 × zx2 y 2 = x1+2 y 2+2 z −1+1 = x3 y 4 .
5.
z 4 x3 y 4
z 3 x2 y 2
= x3−2 y 4−2 z 4−3
= xy 2 z.
Check:
xy 2 z × z 3 x2 y 2 = x1+2 y 2+2 z 1+3 = x3 y 4 z 4 .
67
2.4. POSITIVE INTEGER EXPONENTS
CHAPTER 2. BASIC ALGEBRA
2.4.3 AAd3: Find power of power using “multiply exponents” rule
When we have an exponent powered on another power, we compute it by using the same
base and multiplying the powers.
1. (x5 y 3 )2
Recall that squaring means multiplying the number twice.
= x5 y 3 × x5 y 3
= x5×2 y 3×2
= x10 y 6 .
2. (x−2 y 3 )2
= x(−2)×2 y 3×2
= x−4 y 6 .
3. (x−2 y −3 )2
= x(−2)×2 y (−3)×2
= x−4 y −6 .
4. (x2 y 3 )−2
= x2×(−2) y 3×(−2)
= x−4 y −6 .
5. (x7 y 3 )5
= x7×5 y 3×5
= x35 y 15 .
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CHAPTER 2. BASIC ALGEBRA
2.4. POSITIVE INTEGER EXPONENTS
2.4.4 AAd4: Find whole number power of a rational number
We exponentiate rational numbers with whole number exponents by exponentiating the
numerator and denominator separately, and the power of the numerator is the numerator
of the answer, and the same for the denominator.
4
3
1.
2
=
81
34
= .
4
2
16
=
15
1
= .
5
2
32
=
9
32
= .
2
7
49
5
1
2.
2
2
3
3.
7
4.
−5
2
3
(−5)3
23
125
−125
=−
.
=
8
8
Note that we use parenthesis to indicate that the base is a negative number. The
answer is negative because we multiply a negative number an odd number of times.
In the next example, we have an even power so that we multiply negative number
an even number times, and therefore, the answer is positive.
4
−3
5.
2
81
(−3)4
=
.
=
24
16
=
69
2.4. POSITIVE INTEGER EXPONENTS
CHAPTER 2. BASIC ALGEBRA
70
Appendix A
Glossary
Base amount:
1
this is the amount which is designed to represent the whole or 100%.
Common multiple: this is a natural number which may be divided evenly by each number in a given finite set of natural numbers. Example: 12 is a common multiple of 2
and 3 because 12 = 6 × 2 and 12 = 4 × 3.
Least common multiple: this is the smallest common multiple of two or more natural
numbers. Example: 6 is the least common multiple of 2 and 3.
Dividend: this is the number to be divided.
Divisor: this is the number we divide by.
Common divisor a number which divides each number in a finite set evenly (the reminder of each division is zero). Example: 2 is a common divisor for 8 and 12.
Greatest common divisor: this is the largest number which divides each number in a
finite set evenly. Example: 4 is the greatest common divisor for 8 and 12.
Magnitude: the distance from a number to zero on the real number line or to the origin
in the complex plane.
1
The definitions of base amount, common multiple, least common multiple, greatest common divisor are taken from BMCC:Math Tutorials and rewritten as necessary.
Web page:
http://www.bmcc.org/nish/MathTutorials/
71
APPENDIX A. GLOSSARY
Minuend: this is the number we subtract from another number. Example: in the subtraction 7 − 4 = 3 the minuend is 4.
Numbers, natural:
2
this is the set of numbers we count with: {1, 2, 3, 4, 5, ...}.
Numbers, whole: this is the set containing zero and the natural numbers: {0, 1, 2, 3, 4, 5, ...}.
Numbers, integers: this is the set containing zero, the natural numbers and their opposites (zero is its own opposite): {0, +1, −1, +2, −2, +3, −3, +4, −4, ...}.
Numbers, rational: this is a set of all numbers of a type a/b where a is an integer and b is
a natural number.
Numbers, irrational: this is the set of all numbers which are not rational. Their
√ decimal
representation is a decimal number which never repeats itself. Example: 2 is not a
rational number.
Numbers, real: this is the union of rational and irrational numbers.
Rounding: 3 reducing the significant digits in a number while trying to keep its value
similar. The result is less accurate, but easier
to use. Example: we often need to
√
approximate irrational numbers such as: 2 ≈ 1.4142.
Scientific notation: a number is in scientific notation if it is written in a form n = a × 10r ,
where a is number between 1 and 10, i.e. 1 ≤ a ≤ 10 and r is a whole number.
Example: 35 = 3.5 × 101 .
Subtrahend: this is the number we subtract from. Example: in the subtraction 7 − 4 = 3
the subtrahend is 7.
2
The definitions of scientific notation,
natural numbers,
whole numbers,
integers are taken from WTAMU Beginning Algebra Tutorial web page.
Address:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/int algebra/int alg tut23 exppart1.htm
3
The definition for rounding is from Wikipedia.
72