Download Student Worksheets for Important Concepts

Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #1 – The Distributive Property
An interesting property of our number system is illustrated by the following example. We get the same
result regardless of which way we perform the process. This property of our number system is called
the Distributive Property.
In the first line we add the numbers in the parenthesis using the order of operations followed by
multiplying the sum by 5.
5(3 + 4) = 5*7 = 35
In the next example we multiply the 5 by each number in the parenthesis and then add. We say that
we have distributed the 5 to each number in the sum and then added the results. Here we are adding
the numbers in the parenthesis and multiplying last. We are applying the Distributive Property.
5(3 + 4) = 5*3 + 5*4 = 15 + 20 = 35
This property of our number system is called the distributive property and it will be used extensively
as you progress in your study of algebra.
Notice we can follow the same process regardless of the number of items being added inside the
parenthesis.
5(3 + 4 + 7) = 5*14 = 70 Following the order of operations and adding in the parenthesis first
5(3 + 4 + 7) = 5*3 + 5*4 + 5*7 = 15 + 20 + 35 = 70 Distributing the 5 to each term and adding the results
We can also use the same property with subtraction by rewriting as addition.
4*(8-3) = 4*5 = 20
4(8 – 3) = 4[8 + (-3)] = 4*8 + 4*(-3) = 32 + (-12) = 20
Notice this is the same as the answer if you follow the order of operations and subtract in the
parenthesis first followed by multiplying by 4, verses distributing the 4 to each term in the parenthesis
and then adding. We can distribute multiplication over addition and subtraction.
Work the following three examples both ways showing your work as demonstrated above.
Problem: 8(3 + 7) =
Problem: 7(9 + 5 + 1) =
Problem: 6(7 - 5) =
The next example has unlike variable terms. Since we can’t combine unlike terms, all we can do is
distribute the number each front to each term in the parenthesis.
7(3x + 4y - 7z + 1) = 7*3x + 7*4y + 7*(-7z) + 7*1 = 21x + 28y - 49z + 7
Unlike terms can’t be combined.
Simplify the following expressions by distributing the 6 to each term in the parenthesis.
9(5x – 3y + 4z - 2) =
Here is one more example and problem where a negative number is distributed across the parenthesis.
-5(3x – 4y + 7z – 1) = -5[3x + (-4y) + 7z + (–1)] = -15x + 20y – 35z + 5
Problem:
-5(3x – 4y + z – 7) =
You may struggle with distributing and simplifying terms with fractional coefficients. Examples like the
following will help you to understand how to work with distributing and simplifying expressions with
fractional coefficients.
Notice that
3𝑥
4
3
𝑥
3
= 4 ∗ 1 = 4 𝑥 and
𝑥
7
1
𝑥
1
= 7∗ 1 = 7𝑥
3 𝑥 2
3 𝑥
3 2 𝑥
3 −5
3
6
15 3𝑥
𝑥 15
= 𝑥− 𝑥−
=
𝑥− −
[ + 𝑥 − 5] = ∗ + ∗ ∗ + ∗
4 2 3
4 2
4 3 1
4 1
8
12
4
8
2 4
Simplify the following problem:
5 𝑥 2
[ + 𝑥 − 4] =
7 3 9
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra - For the Student – Level 3
Concept #2 – Comparing Addition verses Multiplication
Look at the first example. In addition, notice that you can’t divide out/cancel the 3’s, or the 2 into 4. If
you try to divide out/cancel the 3 into the 3 and the 2 into the 4, and then add, you don’t get the correct
answer.
3+2 5
=
3+4 7
In multiplication, you can divide out/cancel the common factors and get the same answer. You can
divide out/cancel the 5 into the 10 and the 7 into the 21.
5 7
1 1 1
∗
= ∗ =
10 21 2 3 6
Notice that you could multiply the numerators and denominators first, you can still divide out/cancel the
common factor of 35 to get the same result.
5 7
35
35 ∗ 1
1 1
∗
=
=
=1∗ =
10 21 210 35 ∗ 6
6 6
Work each of the two following problems using the correct method showing all steps. Simplify as
needed and cancel common factors where possible.
Problem:
7+6
=
7 + 12
Problem:
3 2
∗
=
9 10
The following examples give you some slight variations to the original problems to practice your skills.
3
3
=
3+4 7
Notice we can’t divide out/cancel since the 3 and 4 are added. They are not factors. You don’t get the
same answer if you tried to divide out/cancel the 3’s.
3 ∗ 4 12
=
3+4
7
Notice we can’t divide out/ cancel since the 3 and 4 in the denominator are added, not multiplied. The 3
and 4 in the denominator are not factors. You don’t get the same answer if you try to divide out/cancel
the 3’s and 4’s.
3∗2
1 1
3∗2
6
1
= 1 ∗ = 𝐼𝑓 𝑤𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑓𝑖𝑟𝑠𝑡 𝑤𝑒 𝑔𝑒𝑡
=
=
3∗4
2 2
3 ∗ 4 12 2
In the last example, we could divide out/cancel the 3’s and the 2 into the 4 since it is multiplication and
we are dividing out/canceling factors.
Simplify the following problems showing your steps.
4
=
4+9
7∗4
=
7+4
3∗5
=
3 ∗ 20
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra– For the Student – Level 3
Concept #3 - Distributing -1 to a Quantity
Distributing a -1 throughout a quantity and can be demonstrated with a few examples.
−(5 − 8) = −1[5 + (−8)] = (−1) ∗ 5 + (−1) ∗ (−8) = −5 + 8 = 8 − 5 = 3
Notice we could perform the parenthesis first and get the same answer.
- (5 – 8) = -[5 + (-8)] = - (-3) = 3
We can introduce the same concept with variable terms.
−(𝑥 − 8) = −1[𝑥 + (−8)] = −1 ∗ 𝑥 + −1 ∗ (−8) = −𝑥 + 8 = 8 − 𝑥
-(x – y + 4z – 4) = -1[x + (–y) + 4z + (–4)] = (-1)*x + (-1)*(-y) + (-1)*4z + ( -1)*(-4) = -x + y + -4z + 4
Work the following examples both ways as shown above.
Problem: - (4 – 9) =
Problem: - (3 – 7) =
Show your steps in distributing the negative one and simplifying.
Problem: - (x – 7) =
Problem: - (x – 7y + 2z – 11) =
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra - For the Student – Level 3
Concept #4 - Multiplication – Rewriting to use the Distributive Property
An alternative way to perform multiplication is shown in the next example. You are using repeated
distribution to get the final product by multiplying each number in the first quantity times each number
in the second quantity, and adding the results to get the final answer. The 30 is distributed to the 40
and the 5. The 2 is then distributed to the 40 and the 5. The sums are added to get you final product.
(32)(45) = (30+2)(40+5) = 30*40 + 30*5 + 2*40 + 2*5 = 1200 + 150 + 80 + 10 = 1440
There are many ways to write each number as a sum or difference of two numbers as can be seen in the
following example.
(32)(45) = (40 – 8)(40 + 5) = 40*40 + 40*5 - 8*40 - 8*5 = 1600 +200 -320 – 40 = 1440
Perform the following multiplications using the process illustrated above.
(32)(25) = (30 + 2)(20 + 5) =
For two sums and follow the process illustrated above to get the answer.
(31)(24) =
Some other examples are given.
(32)(45) = [40 + (– 8)](40 + 5) = 40*40 + 40*5 (-8)*40 + (-8)*5 = 1600 +200 -320 – 40 = 1440
Here is an example involving multiplying a negative times a negative. There are many ways we could
perform the multiplication.
(47)(13) = [50+ (– 3)][20 +(– 7)] = 50*20 + 50*(-7) + (-3)*20 + (-3)*-7 = 1000 – 350 - 60 + 21 = 611
Work this problem showing your work as demonstrated above.
(27)(68) = (30 – 3)(70-2) = [30 + (-3)][70 + (-2)] =
Work this problem by forming two differences.
(27)(39) =
We could use the same concept when multiplying mixed numbers such as the following problem.
1
3
1
3
3 1
1 3
3 ∗ 2 = (3 + ) ∗ (2 + ) = 3 ∗ 2 + 3 ∗ + ∗ 2 + ∗
2
4
2
4
4 2
2 4
9
3 48 18 8 3 77
5
6+ +1+ =
+
+ + =
=9
4
8
8
8 8 8
8
8
Remember, we have taught you to rewrite the problem mixed numbers as improper fractions and then
multiply.
1
3 7 11 77
5
3 ∗2 = ∗
=
=9
2
4 2 4
8
8
Work the following problem as shown above using the distributive property and check by rewriting the
mixed numbers as improper fractions and multiplying.
3
1
1 ∗3 =
4
2
Below is an example where we divide the numbers into three sums and then distribute to multiply.
(125)(213) =
= (100 + 20 + 5)(200 + 10 + 3) = 100*200 + 100*10 + 100*3 + 20*200 + 20*10 + 20*3 + 5*200 + 5*10 + 5*3
= 20000 + 1000 +300 + 4000 +200 + 60 + 1000 + 50 + 15 = 26,625
Form sums as in the example above and use your distributive property to get the correct product.
Problem: (423)(165) =
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #5 - Using the Distributive Property in Reverse - Factoring
The next two examples show how to use the distributive property in reverse. We have found the
greatest common factor 5, and written each number as a product of 5 and the other factor that makes it
equivalent to the original number. Notice we can move the 5 to the front, (We say we have factored
out the 5 in the 1st example, 7 in the 2nd example), you can then add the numbers left in the
parenthesis and multiply to get the same answer. In each case we are finding the greatest common
factor and factoring it from each of the terms of the expression.
Example:
10 + 15 + 25 = 50
10 + 15 + 25 = 5*2 + 5*3 + 5*5 = 5(2 + 3 + 5) = 5*10 = 50
Example:
42 + 56 = 98
42 + 56 = 7*6 + 7*8 = 7(6+8) = 7(14) = 98
The next example has variable terms with a common factor in each term which we will factor out.
Example: 10x + 20y + 35z + 5 = 5*2x + 5*4y + 5*7z + 5*1 = 5(2x + 5y + 7z+ 1)
Follow the pattern of the three examples above to remove the greatest common factor and simplify.
Problem:
8 + 40 + 48 =
Problem: 27 + 45 =
Problem: 7x + 28y + 24z + 14 =
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #6 – Division by Zero
Division by zero is always an important concept we work on with students and introducing examples
such as the following is helpful. It is important to start with this concept. Notice that in normal division,
you should always be able to multiply the divisor time the quotient to get back to the dividend.
6
= 3 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 2 ∗ 3 = 6
2
0
= 0 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 2 ∗ 0 = 0
2
6
= ? 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 0 ∗? = 6
0
Since there is no number times zero that will give us 6, we say that division by zero is undefined.
Answer the following problems.
12
=
𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ____ ∗ ___ = ____
2 _____
0
= ____ 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ____ ∗ ____ = ____
7
8
= ___ 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ____ ∗ ____ = ____
0
Use the order of operations to add the numbers in the numerator and subtract the numbers in the
denominator. Once we have a division by zero, we know the answer is undefined.
12 + 8 20
=
= 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
4−4
0
In the following example, we subtract the numbers in the numerator, add the numbers in the
denominator, and once we see we have zero divided by a number we know the answer is zero.
15 − 15 0
= =0
5+5
5
Simplify each numerator and denominator to find the correct result for the following problems.
2−2
=
7+4
2+3
=
5−5
5+4
=
9−9
3−3
=
3+3
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra - For the Student – Level 3
Concept #7 - Simplifying Rational Expressions – Identifying Restricted Values
In working with rational expressions as well as when given a rational function, we emphasize that
division by zero is not possible.
𝑓(𝑥) =
𝑥−8
𝑥−3
It is important to note that the numerator can be zero so that in the example above, if x = 8, the rational
expression would be equivalent to 0/5 = 0. If x = 3, the rational expression would be equivalent to -5/0
which is undefined. When simplifying rational expressions and/or functions, and when multiplying
and dividing rational expressions it is important to note that our restrictions on the acceptable values
for the rational expression (domain) are based on the original rational expression before any dividing
out/canceling is done.
In the following example, after you factor the numerator and denominator, it is important to note
that x can’t equal -3 or 8 even though we cancel the factor (x + 3). Our restriction is based on the
original rational expression or function given. We have factored the numerator and the denominator
and divided out/canceled the common factor to completely simplify the rational equation.
𝑓(𝑥) =
𝑥 2 − 2𝑥 − 15 (𝑥 + 3)(𝑥 − 5) (𝑥 − 5) 𝑥 − 5
=
=
=
𝑥 2 − 5𝑥 − 24 (𝑥 + 3)(𝑥 − 8) (𝑥 − 8) 𝑥 − 8
For the following problem, factor the numerator and denominator, identify all restricted values of x, and
then simplify the rational function by dividing out/canceling any like terms.
Problem:
𝑓(𝑥) =
𝑥 2 + 2𝑥 − 15
=
𝑥 2 − 14𝑥 + 45
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #8 – Simplifying Rational Expressions with Numerical Values
We will use this concept to simplify a quotient as shown in the following examples.
Keep in mind the idea of reducing fractions to lowest terms.
Example:
12
15
3∗4
3
4
4
4
= 3∗5 = 3 ∗ 5 = 1 ∗ 5 = 5
We are able to divide out/cancel common factors, in this case 3.
Example: We will write each number as a product of 7 and the other factor. We then factor 7 from each
sum. Since 7*(6 + 8) is a product of two factors, and 7*(4 + 9) is a product of two factors, we can divide
out/cancel the common factor of 7 and simplify.
42 + 56
7∗6 + 7∗8
7(6 + 8) (6 + 8) 14
=
=
=
=
28 + 63
7∗4+7∗9
7(4 + 9) (4 + 9) 13
Notice we can’t divide out/cancel the 4 and the 8 or take a 3 from the 6 and 9 since these numbers are
added. We can only divide out/cancel common factors.
Notice we could solve this problem by following the order of operations. We would add the numbers in
the numerator and denominator, and reduce to lowest terms.
42 + 56 98 7 ∗ 14
14 14
=
=
=1∗
=
28 + 63 91 7 ∗ 13
13 13
Problem: Work the following problem both ways showing all steps to prove your answer.
18 + 24
=
12 + 30
In the following example, you can see that both 3 and the quantity (3+4) are common factors that can
be divided out/canceled.
9 + 12
3∗3 + 3∗4
3(3 + 4) 1(3 + 4) 1
=
=
=
=
27 + 36
9∗3+9∗4
9(3 + 4) 3(3 + 4) 3
Notice we could just simplify by adding the numerators and denominators and then reducing to lowest
terms.
9 + 12
21 7 ∗ 3
3 1
=
=
=1∗ =
27 + 36 63 7 ∗ 9
9 3
Work the following problem both ways showing all steps to prove your answer.
Problem:
8 + 20
=
16 + 40
Here is a similar example and problem using subtraction.
30 − 6
6 ∗ 5 + (6) ∗ (−1) 6[5 + (−1)] 6(4) 1 ∗ 1
=
=
=
=
=1
48 − 24 6 ∗ 8 + (6) ∗ (−4) 6[8 + ( −4)] 6(4) 1 ∗ 1
Notice we could follow the order of operations and subtract in the numerator and denominator first.
30 − 6
24
=
=1
48 − 24 24
Work the following problem both ways. Begin by using the same pattern above illustrated by factoring
out a 3 from the numerator and denominator.
15 − 5
=
45 − 35
Here is an example with a single number in the denominator.
9 + 15 3 ∗ 3 + 3 ∗ 5 3(3 + 5) 3(8) 8
=
=
=
= =2
12
3∗4
3(4)
3(4) 4
We could follow the order o f operations and add the numbers in the numerator to get the same
answer.
9 + 15 24 2
=
= =2
12
12 1
Work the following problem by factoring out the greatest common factor in the numerator as shown
above and check by adding the numbers in the numerator first and then reducing.
40 + 56
=
32
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra - For the Student – Level 3
Concept #9 - Simplifying Rational Expressions Equal to One
Another example to be presented in Math 0960 and Math 1310 would be this one dealing with dealing
with a quotient equal to -1. Begin with the examples below and progress to the fraction.
−(5 − 8) = −1(5 − 8) = −1 ∗ 5 + −1 ∗ −8 = −5 + 8 = 8 − 5
−(𝑥 − 8) = −1(𝑥 − 8) = −1 ∗ 𝑥 + −1 ∗ −8 = −𝑥 + 8 = 8 − 𝑥
The first example shows that we can make the quotient equal to -1 after we have a common factor. The
2nd example shows the same process with a variable expression.
5 − 8 −3
−1(−5 + 8) −1(8 − 5)
=
= −1 =
=
= −1
8−5
3
8−5
(8 − 5)
𝑥 − 8 −1(−𝑥 + 8) −1(8 − 𝑥)
=
=
= −1
8−𝑥
8−𝑥
(8 − 𝑥)
Simplify each of the following problems demonstrating the same process.
Problem:
9−5
=
5−9
Problem:
2−𝑥
=
𝑥−2
This example shows us to factor out the -1 to divide out/cancel the quantity (5 - 8) and simplify the
numerator and denominator since nothing more can be divided out/canceled. The 2nd example is similar
with variable factors, the factor (x – 8) can be divided out/canceled.
(8 − 5)(7 − 3)
−1(−8 + 5)(7 − 3)
−1(5 − 8)(7 − 3)
−1(7 − 3) −4
=
=
=
=
(5
(5
(5 − 8)(6 + 9)
− 8)(6 + 9)
− 8)(6 + 9)
(6 + 9)
15
(8 − 𝑥)(𝑥 − 3)
−1(−8 + x)(x − 3)
−1(x − 8)(x − 3)
−1(𝑥 − 3) −𝑥 + 3
=
=
=
=
(𝑥
(𝑥
(𝑥 − 8)(𝑥 + 9)
− 8)(𝑥 + 9)
− 8)(𝑥 + 9)
(𝑥 + 9)
𝑥+9
Work the following two problems showing your process as illustrated above.
Problem:
(11 − 3)(7 − 3)
=
(3 − 11)(6 + 9)
Problem:
(12 − 𝑥)(𝑥 − 7)
=
(𝑥 − 12)(𝑥 + 8)
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #10 – Simplifying Rational Expressions with Variable Terms
The first example requires us to factor -1 from both factors. You can also divide out/cancel the common
quantity (3+4). The second example involves variable terms and illustrates taking out a common factor
of 3 and dividing out/canceling the common quantity which is the factor (3x + 4). In both examples we
are dividing out/canceling only common factors.
−9 − 12
3(−3 − 4)
−3(3 + 4)
−1
=
=
=
27 + 36
9(3 + 4)
9(3 + 4)
3
Notice you get the same answer if you simplify the numerator and denominator first, and then reduce.
−9 − 12 −21 7 ∗ −3 −3 −1
=
=
=
=
27 + 36
63
7∗9
9
3
We now introduce variable terms into the rational expressions
−9𝑥 − 12
3(−3x − 4)
−3(3x + 4)
−1
=
=
=
27𝑥 + 36
9(3𝑥 + 4)
9(3𝑥 + 4)
3
Factor out the greatest common factor from the numerator and denominator and simplify the next two
examples. Look at the examples above.
Problem:
−14 − 21
=
28 + 42
Problem:
−10𝑥 − 5
20𝑥 + 10
This example allows us to factor out and then divide out/cancel a common factor of 2, but we can’t
divide out/cancel a common quantity.
(−4 + 3)
−8 + 6
2(−4 + 3)
−1
=
=
=
(7 − 2)
14 − 4
2(7 − 2)
5
Notice we get the same answer if we simplify the numerator and denominator first and then reduce.
−8 + 6
−2
−1
=
=
14 − 4
10
5
The second example is similar, but has variable factors and after we divide out/cancel the 2’s, we are left
with unlike linear factors that can’t be simplified any further.
(−4x + 3)
−8𝑥 + 6
2(−4x + 3)
−4𝑥 + 3
=
=
=
(7𝑥 − 2)
14𝑥 − 4
2(7𝑥 − 2)
7𝑥 − 2
Show your steps in factoring out the greatest common factor and simplify if possible.
Problem:
−12 + 15
24 − 9
Problem:
−20𝑥 + 30
=
35𝑥 − 25
The last example and problem involves having a single term in the numerator. We factor the greatest
common factor 4 from the denominator and then divide out/cancel the factor of 4.
16
4∗4
4∗4
4
4
=
=
=
=
4 + 8 4 ∗ 1 + 4 ∗ 2 4(1 + 2) (1 + 2) 3
Notice we can’t divide out/ cancel the 2 into the 4 on the next to last step since they are not factors.
We could follow the order of operations and add the numbers in the numerator to get the same answer.
16
16 4 ∗ 4 4
=
=
=
4 + 8 12 4 ∗ 3 3
Work the following problem by factoring out the greatest common factor in the denominator as shown
above and check by adding the numbers in the denominator first and then reducing.
24
=
12 + 18
We could have a similar type of problem with variable terms.
18𝑥
3 ∗ 6𝑥
6𝑥
=
=
6𝑥 + 9 3(2𝑥 + 3) 2𝑥 + 3
Notice we can’t divide out/cancel the 2x into the 6x since the 2x is added in the denominator and not a
factor. We can divide out/cancel the factor of 3 since it is a factor in both the numerator and
denominator.
Simplify the following rational expression by factoring the denominator.
Problem:
42𝑎
=
14𝑎 + 35
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra - For the Student – Level 3
Concept #11 - Location of Negative Sign in a Fraction
We have discussed that a +/- or -/+ equal a negative answer. So the following is true:
3 −3
3
− =
=
4
4
−4
You can see that moving the negative sign from the numerator to the denominator makes an equivalent
expression.
A simple explanation for this might be that
3
−4
3
1
3
−1
= −4 ∗ 1 = −4 ∗ −1 =
−3
4
You are not changing the value of the fraction, just rewriting it by multiplying by a well chosen one, in
this case, -1/-1 = +1. This shows you can move the negative sign from the numerator to the
denominator without changing the value of the fraction.
Given this and thinking about adding fractions, we look at the following examples.
Example:
2 3 5
+ =
7 7 7
Example:
5
3
5 −3 5 + (−3) 2
+
= +
=
=
7 −7 7
7
7
7
Work the following problems showing your work.
Problem:
1 4
+ =
9 9
Problem:
2
5
+
=
13 −13
This example includes finding the common denominator
Example:
2
3
2 −3 2 4 −3 3
8 −9 −1
+
= +
= ∗ +
∗ =
+
=
3 −4 3
4
3 4
4 3 12 12 12
Work the following problem using the same process and show your steps.
Problem:
1
2
+
=
7 −3
Try this problem:
4
4
+
=
9 −9
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra - For the Student – Level 3
Concept #12 - Clearing Fractions in an Equation
For the concept of clearing fractions when solving linear equations in Math 0960, presenting these
examples may help the student to understand why we are able to clear the fractions in an equation. We
begin by showing you a true equation which you can see is true when you get a common denominator
and compare the left and right sides.
−1 3 5
+ =
28 4 7
−1 21 20
20 20
+
=
𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑒𝑒 𝑡ℎ𝑎𝑡
=
28 28 28
28 28
Then put the variable x in for the -1 and solve for x:
𝑥 3 5
+ =
28 4 7
𝑥 21 20
21
+
=
𝑠𝑜 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡
𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠
28 28 28
28
[
𝑥
−1
=
]
28 28
Multiply both sides by 28 to get:
28 𝑥
−1 28
[ =
]
1 28 28 1
𝑥 = −1
This shows the justification for clearing fractions as we show in Math 0960.
28 𝑥 3 5 28
[ + = ]
1 28 4 7 1
𝑥+7∗3=5∗4
𝑥 + 21 = 20
𝑥 = −1
Solve the following problem without clearing fractions by getting a common denominator for all terms
as illustrated above and then solve it by clearing the fractions as illustrated in the 2nd part of the
example.
𝑥 5 2
− =
18 9 3
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #13 – Adding Rational Expressions with Variable Terms
A common problem you will see in intermediate algebra is the process of adding rational expressions.
We need to have a common denominator when we add rational expressions with denominators
opposite in sign and rational expressions with different denominators.
We will begin with the addition of rational expressions with denominators that differ only in the sign
of the terms in the denominator.
3
4
3
−1 (4)
3
−4
3
−4
3 + (−4) −1
1
+
=
+
=
+
=
+
=
=
=−
8 − 5 5 − 8 8 − 5 −1 (5 − 8) 8 − 5 (−5 + 8) 8 − 5 8 − 5
8−5
3
3
We could also simplify the denominator to show that our answer is the same.
3
4
3
4
3 −1 (4)
3 −4 −1
1
+
= +
= +
= +
=
=−
8 − 5 5 − 8 3 (−3) 3 −1 (−3) 3
3
3
3
Work the following problem both ways as illustrated above.
Problem:
7
3
+
=
9−4 4−9
Now we will work the same type of problem with variable denominators. This is a tough problem for
many students and needs to be emphasized.
3
4
3
−1 (4)
3
−4
3
−4
−1
+
=
+
=
+
=
+
=
𝑥 − 5 5 − 𝑥 𝑥 − 5 −1 (5 − 𝑥) 𝑥 − 5 (−5 + 𝑥) 𝑥 − 5 𝑥 − 5 𝑥 − 5
Simplify the following sum using the method illustrated above.
Problem:
9
5
+
=
𝑥−7 7−𝑥
Here is a similar type problem where we are subtracting two rational expressions.
9
5
9
−5
9
−1
(−5)
−
=
+
=
+
∗
=
𝑥 − 11 11 − 𝑥 𝑥 − 11 11 − 𝑥 𝑥 − 11 −1 (11 − 𝑥)
9
5
9
5
14
+
=
+
=
𝑥 − 11 (−11 + 𝑥) 𝑥 − 11 𝑥 − 11 𝑥 − 11
Simplify the following subtraction of two rational expressions.
Problem:
11
5
−
=
𝑥−1 1−𝑥
We will now present numerical examples with different denominators and show how they are similar to
sums of rational expressions with variable denominators. We must identify the least common
denominator (LCD). In the numerical example we must multiply each fraction by a well chosen one that
will make the denominator equal to the LCD as in the following examples.
2 3 2+3
5
+ =
=∗
7 7
7
7
𝐻𝑒𝑟𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑎 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑜𝑓 7.
2 4 2 5 4 3 10 12 22
+ = ∗ + ∗ =
+
=
3 5 3 5 5 3 15 15 15
𝐻𝑒𝑟𝑒 𝑤𝑒 𝑟𝑒𝑤𝑟𝑜𝑡𝑒 𝑒𝑎𝑐ℎ 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑡𝑜 ℎ𝑎𝑣𝑒 𝑎𝑛 𝐿𝐶𝐷 𝑜𝑓 15.
The next example replaces the numerical denominators with the variables x and y. It will work exactly
the same. It is important to see that the same process is required to get the common denominator. The
LCD is the term xy so we must multiply each term by the well chosen one that will make the
denominator equal to the LCD as shown in the following example.
2 4
2 𝑦
4 𝑥
2𝑦
4𝑥
2𝑦 + 4𝑥
+ = ∗ + ∗ =
+
=
𝑥 𝑦
𝑥 𝑦
𝑦 𝑥
𝑥𝑦
𝑥𝑦
𝑥𝑦
Find the Least Common Denominator where needed for each of the following problems showing each
step and give you final simplified answer.
Problem:
2
5
+
=
11 11
Problem:
2 4
+ =
5 7
Problem:
3 5
+ =
𝑤 𝑣
We will look at the problem of finding the LCD in the next examples.
7
9
+
15 20
At times it can be difficult to find the LCD of two or more factors. Sometimes just having a common
factor between the two denominators can make it difficult to find the LCD. For 15 and 20, the LCD is 60.
It can help to look at the prime factors of:
15 = 3*5 and 20 = 2*2*5
Note that in the LCD we need each factor the most number of times it appears in either denominator.
So in this case we need 2*2*3*5 = 60 for the LCD. We multiply each term by the well chosen one that
will make the denominator equal to 60 for each term. Notice we can reduce the final fraction.
7
9
7 4 9 3 28 27 55 5 ∗ 11
11
+
=
∗ +
∗ =
+
=
=
=
15 20 15 4 20 3 60 60 60 5 ∗ 12
12
Let’s look at a problem with variable denominators.
2
4
+
=
3𝑥 5𝑦
Since each denominator has unique prime factors, the LCD needs to contain 3*5*x*y. The LCD is 15xy.
Multiply each term by the well chosen one that will make the denominator equal to 15xy. Notice we
can’t cancel at the end of the problem because of the addition. We can only cancel factors.
2 5𝑦 4 3𝑥
10𝑥
12𝑦
10𝑥 + 12𝑦
∗
+
∗
=
+
=
3𝑥 5𝑦 5𝑦 3𝑥 15𝑥𝑦 15𝑥𝑦
15𝑥𝑦
Work each of the following problems showing how you find the LCD, and the entire set of steps to
rewrite the rational expressions with a common denominator as shown in the above examples.
Problem:
7
9
+
=
30 45
Problem:
5
2
+
=
7𝑣 3𝑤
The following example shows why it is a good idea to find the prime factors of each denominator to help
identify the LCD.
15xz = 3*5*x*z and 20yz = 2*2*5*y*z
so the LCD must have 2*2*3*5*x*y*z = 60xyz.
Multiply each term by the well chosen one that will make the denominator equal to 60xyz. Notice that
5z is common to both denominators.
2𝑤
3𝑣
2𝑤 4𝑦
3𝑣 3𝑥
8𝑦𝑤
9𝑣𝑥
8𝑤𝑦 + 9𝑥𝑣
+
=
∗
+
∗
=
+
=
15𝑥𝑧 20𝑦𝑧 15𝑥𝑧 4𝑦 20𝑦𝑧 3𝑥 60𝑥𝑦𝑧 60𝑥𝑦𝑧
60𝑥𝑦𝑧
Demonstrate the same procedures to simplify and rewrite the following rational expression with the
least common denominator.
Problem:
5𝑣
7𝑤
+
=
12𝑦𝑧 10𝑥𝑧
Dynamic Solution Exercise for Intermediate Algebra
Arithmetic to Algebra – For the Student – Level 3
Concept #14 – Solving Rational Equations
Rational expressions can become part of a rational equation once an equal sign is introduced. We begin
by presenting a fractional equation that is shown to be a true equation.
2 2 16
+ =
5 3 15
Get a common denominator to show that the equation is true.
6 10 16
+
=
15 15 15
Multiply both sides by 15 to clear the fraction.
15 6 10 16 15
= ]
[ +
1 15 15 15 1
We get a true statement showing that our original equation is a true statement.
6 + 10 = 16
We could contrast this problem with the following rational equation. As we begin the process of
solving this rational equation, we will note that division by zero is undefined and letting x = -3 would
cause the equation to contain a denominator that is zero. We would say that x = -3 is excluded from
the possible solutions (domain) for x since it would cause the equation to be undefined.
2
2 16
+ =
𝑥 + 3 3 15
We talk about clearing the fractions, or multiplying by the Least Common Denominator of all the
denominators in the equation. In this case, the LCD is 15(x + 3).
15(𝑥 + 3) 2
2 16 15(𝑥 + 3)
+ = ]
[
1
𝑥 + 3 3 15
1
We multiply each term in the equation by the LCD which allows us to cancel the common factors which
will remove the denominators from our equation and give us an equivalent equation without any
denominators.
15(2) + 5(𝑥 + 3)(2) = 16(𝑥 + 3)
Simplify the left and right sides and solve the resulting equation. We will multiply the 2*5 in the 2nd
term before we distribute.
30 + 10(𝑥 + 3) = 16(𝑥 + 3)
30 + 10𝑥 + 30 = 16𝑥 + 48
10𝑥 + 60 = 16𝑥 + 48
Subtract 10x from both sides and subtract 48 from both sides to obtain.
12 = 6𝑥
Divide by 6 on both sides to obtain the solution for x.
2 = 𝑥 𝑜𝑟 𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒 𝑥 = 2
This confirms our original statement and we see that by substituting into a fractional statement, we
created a rational equation which could be solved by identifying the least common denominator and
multiplying both sides of the equation by that LCD to clear the denominators and give us a linear and
sometimes a quadratic equation to solve for x.
Solve the following problem by finding the LCD and multiplying both sides by the LCD to clear the
denominators. Show your work in getting to the final solution. Identify the restricted values for x
that would cause a division by zero error. Check your solution once you have found x and show that
you have a true fractional equation.
Problem:
2
2 20
+ =
𝑥 − 5 3 21