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Big Ideas Math Integrated I, 2016 View details

4. Solving Absolute Value Equations

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Exercise 59 Page 34

Interpret $∣ x + 2 ∣$ as a single quantity, then isolate it.

$x = 1$ and $x = - 5,$ see solution.

Practice makes perfect

Take a look at the given absolute value equation.

8 ∣ x + 2 ∣ - 6 = 5 ∣ x + 2 ∣ + 3

We have the absolute value expression

∣ x + 2 ∣

on both sides, but with different coefficients. As a first step, we can interpret this as a single quantity and subtract

5 ∣ x + 2 ∣

from both sides. This will help us isolate it. Let's begin!

8 ∣ x + 2 ∣ - 6 = 5 ∣ x + 2 ∣ + 3

SubEqn

$LHS - 5 ∣ x + 2 ∣ = RHS - 5 ∣ x + 2 ∣$

3 ∣ x + 2 ∣ - 6 = 3

AddEqn

$LHS + 6 = RHS + 6$

3 ∣ x + 2 ∣ = 9

DivEqn

$LHS / 3 = RHS / 3$

∣ x + 2 ∣ = 3

Now we have a much simpler absolute value equation that we can solve by splitting it into two cases. When doing so, we need to keep in mind that we are considering a positive and a negative case.

∣ x + 2 ∣ = 3

$x + 2 \geq 0 : x + 2 = 3 x + 2 < 0 : x + 2 = - 3 (I) (II)$

x + 2 = 3 x + 2 = - 3 (I) (II)

$(I), (II):$ $LHS - 2 = RHS - 2$

x_{1} = 1 x_{2} = - 5

Subchapter links

Communicate Your Answer p.27

Monitoring Progress p.28-31

Exercises p.32-34

Exercise 59 Page 34

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