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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 16 Page 153

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

11

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the left-hand side of the equation.

7(s-5)=42

Distribute 7

7* s+7*(- 5)=42

a(- b)=- a * b

7s-35=42

LHS+35=RHS+35

7s-35+35=42+35

Add terms

7s=77

.LHS /7.=.RHS /7.

s=11

The solution to the equation is s=11. We can check our solution by substituting it into the original equation.

7(s-5)=42

s= 11

7( 11-5)? =42

Subtract term

7* 6? =42

Multiply

42=42

Since the left-hand side is equal to the right-hand side, our solution is correct.

Subchapter links

Exercises p.152-156