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Big Ideas Math Integrated I, 2016

BI

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|+3We 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

LHS-5|x+2|=RHS-5|x+2|

3|x+2|-6=3

LHS+6=RHS+6

3|x+2|=9

.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

lc x+2 ≥ 0:x+2 = 3 & (I) x+2 < 0:x+2 = - 3 & (II)

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

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

lx_1=1 x_2=-5

Subchapter links

Communicate Your Answer p.27

Monitoring Progress p.28-31

Exercises p.32-34