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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

2. Measuring and Constructing Segments

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Exercise 44 Page 394

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

$x = - 13$

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. In this case, we will start by using the Multiplication Property of Equality to remove the fraction on the left-hand side of the equation.

\frac{- 5 + x}{2} = - 9

$LHS \cdot 2 = RHS \cdot 2$

2 \cdot \frac{- 5 + x}{2} = 2 \cdot (- 9)

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

- 5 + x = 2 \cdot (- 9)

$a (- b) = - a \cdot b$

- 5 + x = - 18

$LHS + 5 = RHS + 5$

- 5 + x + 5 = - 18 + 5

Add terms

x = - 13

The solution to the equation is

x = - 13 .

Subchapter links

Communicate Your Answer p.387

Monitoring Progress p.388-391

Exercises p.392-394