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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

4. Congruence and Transformations

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Exercise 41 Page 206

To solve an equation, we should 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 Distributive Property to simplify the left-hand side of the equation.

7 (2 n + 11) = 4 n

Distribute $7$

14 n + 77 = 4 n

Now we can continue to solve using the Properties of Equality.

14 n + 77 = 4 n

$LHS - 4 n = RHS - 4 n$

14 n + 77 - 4 n = 4 n - 4 n

Subtract terms

10 n + 77 = 0

$LHS - 77 = RHS - 77$

10 n + 77 - 77 = 0 - 77

Subtract terms

10 n = - 77

$LHS / 10 = RHS / 10$

\frac{1 0 n}{1 0} = \frac{- 7 7}{1 0}

Calculate quotient

n = - 7.7

The solution to the equation is

n = - 7.7 .

We can check our solution by substituting it into the original equation.

7 (2 n + 11) = 4 n

$n = - 7.7$

7 (2 (- 7.7) + 11) = ? 4 (- 7.7)

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Simplify

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

7 (- 15.4 + 11) = ? - 30.8

Add terms

7 (- 4.4) = ? - 30.8

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

- 30.8 = - 30.8

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

Subchapter links

Communicate Your Answer p.199

Monitoring Progress p.200-203