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

∣ 5 f - 3 ∣ = 12

This equation means that the distance is

12,

either in the

positive

direction or the

negative

direction.

∣ 5 f - 3 ∣ = 12 \Rightarrow 5 f - 3 = - 12 5 f - 3 = - 12

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

f .

∣ 5 f - 3 ∣ = 12

$5 f - 3 \geq 0 : 5 f - 3 = 12 5 f - 3 < 0 : 5 f - 3 = - 12 (I) (II)$

5 f - 3 = 12 5 f - 3 = - 12 (I) (II)

$(I), (II):$ $LHS + 3 = RHS + 3$

5 f = 15 5 f = - 9

$(I), (II):$ $LHS / 5 = RHS / 5$

f_{1} = 3 f_{2} = - \frac{9}{5}

Both

f = - \frac{9}{5}

and

f = 3

are solutions to the absolute value equation. We want to graph these solutions. In the fraction

f = - \frac{9}{5},

5 .

To graph this, we divide the segments between each whole number on a number line into

5

equal spaces as shown below.

To graph $f = - \frac{9}{5},$ we move $9$ spaces on the number line to the left of $0$ as shown below.

Let's now mark the other solution $f = 3$ on the number line.

Exercise 39 Page 106

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