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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

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Exercise 12 Page 231

Apply inverse operations to gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

- 3

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.

5x+3=- 12

LHS-3=RHS-3

5x=- 12 - 3

Subtract term

5x=- 15

.LHS /5.=.RHS /5.

x = - 15/5

Calculate quotient

x=- 3

The solution to the equation is x=- 3. We can check our solution by substituting it into the original equation.

5x+3=- 12

x= - 3

5( - 3) +3 ? = - 12

a(- b)=- a * b

- 15 + 3 ? = - 12

Add terms

- 12 = - 12 ✓

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