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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 12 Page 153

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

- 5

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 start by subtracting 8 from both sides of the equation.

8+3m=- 7

LHS-8=RHS-8

8+3m-8=- 7-8

CommutativePropAdd

Commutative Property of Addition

3m+8-8=- 7-8

Subtract term

3m=- 15

By dividing both sides of the equation by 3, we can isolate m.

3m=- 15

.LHS /3.=.RHS /3.

3m/3=- 15/3

MoveNegNumToFrac

Put minus sign in front of fraction

3m/3=- 15/3

Calculate quotient

m=- 5

The solution to the equation is m=-5. We can check our solution by substituting it into the original equation.

8+3m=- 7

m= - 5

8+3( - 5)? =- 7

a(- b)=- a * b

8- 3* 5? =- 7

Multiply

8-15? = - 7

Subtract term

- 7=- 7

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

Subchapter links

Exercises p.152-156