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McGraw Hill Glencoe Geometry, 2012

MH

McGraw Hill Glencoe Geometry, 2012 View details

2. The Pythagorean Theorem and Its Converse

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Exercise 73 Page 555

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 need to start by using the Division Property of Equality to isolate the variable term.

18 = 3 x 3

$LHS / 3 = RHS / 3$

6 = x 3

$LHS / 3 = RHS / 3$

\frac{6}{3} = x

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

\frac{6 \cdot 3}{3 \cdot 3} = x

Multiply

\frac{6 3}{( 3 ) ^{2}} = x

$(a)^{2} = a$

\frac{6 3}{3} = x

$\frac{a}{b} = \frac{a / 3}{b / 3}$

23 = x

Rearrange equation

x = 23

The solution to the equation is

x = 23 .

Subchapter links

Check Your Understanding p.551-552

Practice and Problem Solving p.552-554

H.O.T. Problems p.554

Standardized Test Practice p.555

Spiral Review p.555

Skills Review p.555