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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Chapter Review

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Exercise 8 Page 153

Start by using the Division Property of Equality.

14

Practice makes perfect

To solve the equation, we have to isolate b on one side. We will begin by dividing both sides by 3. This is allowed by the Division Property of Equality.

3b=42

.LHS /3.=.RHS /3.

b=42/3

Calculate quotient

b=14

Therefore, b=14 is the solution to the equation. To check that we have the correct solution, we can substitute b=14 into the original equation and simplify.

3b=42

b= 14

3( 14)? =42

Multiply

42=42

Since the left-hand side and the right-hand side are equal, b=14 is the correct solution.

Subchapter links

Exercises p.152-156