Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 5.1
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Exercise 12 Page 214

Practice makes perfect
a

To solve the equation, we have to isolate x on one side. Using the Multiplication Property of Equality, we will multiply both sides of the equation by 3. This will eliminate the denominator of the fraction and isolate x.

x/3=4/5
x/3(3) = 4/5(3)
x = 4/5(3)
x = 12/5

b

Like in Part A, to solve the equation, we will use the Multiplication Property of Equality. Hence, we will start by multiplying both sides of the equation by the denominators one at a time. Then we will simplify the result and isolate x.

x/x+1=5/7
7x/x+1=5
7x=5(x+1)
7x=5x+5
2x=5
x=5/2

c

This time, we will start with rewriting 2 as a fraction to combine terms on the right-hand side of the equation.

6/15=2-x/5
6/15=10/5-x/5
6/15=10-x/5

Next, like in previous parts, we will use the Multiplication Property of Equality to isolate x.

6/15=10-x/5
30/15=10-x
2=10-x
2+x=10
x=8

d

For the last part, we will start with adding the fractions on the left-hand side of the equation. To do that we need to rewrite both fractions so they have a common denominator.

2/3+x/5=6
10/15+x/5=6
10/15+3x/15=6
10+3x/15=6

Now, like in Parts A through C, we will use the the Properties of Equality to isolate the variable. Let's start with Multiplication Property of Equality to get rid of the denominator.

10+3x/15=6
10+3x=15(6)
10+3x=90
3x=80
x=80/3