Exercise 22 Page 32

Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

- 3 ∣ ∣ ∣ ∣ ∣ 1 - \frac{2}{3} v ∣ ∣ ∣ ∣ ∣ = - 9

$LHS / (- 3) = RHS / (- 3)$

∣ ∣ ∣ ∣ ∣ 1 - \frac{2}{3} v ∣ ∣ ∣ ∣ ∣ = 3

An absolute value measures an expression's distance from a midpoint on a number line.

∣ ∣ ∣ ∣ ∣ 1 - \frac{2}{3} v ∣ ∣ ∣ ∣ ∣ = 3

This equation means that the distance from

1

3,

either in the

positive

direction or the

negative

direction.

1 - \frac{2}{3} v = 3 and 1 - \frac{2}{3} v = - 3

To find the solutions to the absolute value equation, we need to solve both of these cases for

v .

Let's start with the positive case.

1 - \frac{2}{3} v = 3

▼

Solve for

v

$LHS - 1 = RHS - 1$

- \frac{2}{3} v = 2

$LHS \cdot (- \frac{3}{2}) = RHS \cdot (- \frac{3}{2})$

- \frac{2}{3} v (- \frac{3}{2}) = 2 (- \frac{3}{2})

$\frac{a}{b} \cdot \frac{b}{a} = 1$

v = 2 (- \frac{3}{2})

Put minus sign in numerator

v = 2 (\frac{- 3}{2})

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

v = - 3

Similarly, we can solve the negative case.

1 - \frac{2}{3} v = - 3

▼

Solve for

v

$LHS - 1 = RHS - 1$

- \frac{2}{3} v = - 4

$LHS \cdot (- \frac{3}{2}) = RHS \cdot (- \frac{3}{2})$

- \frac{2}{3} v (- \frac{3}{2}) = - 4 (- \frac{3}{2})

$\frac{a}{b} \cdot \frac{b}{a} = 1$

v = - 4 (- \frac{3}{2})

Put minus sign in numerator

v = - 4 (\frac{- 3}{2})

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

v = \frac{- 4 ( - 3 )}{2}

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

v = \frac{1 2}{2}

Calculate quotient

v = 6

Both

- 3

and

6

are solutions to the absolute value equation. Let's graph these solutions on a number line.

Hints