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Solution - Absolute value equations

Exact form: z=32,-1
z=\frac{3}{2} , -1
Mixed number form: z=112,-1
z=1\frac{1}{2} , -1
Decimal form: z=1.5,1
z=1.5 , -1

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|z+6|=5|z|
without the absolute value bars:

|x|=|y||z+6|=5|z|
x=+y(z+6)=5(z)
x=y(z+6)=5((z))
+x=y(z+6)=5(z)
x=y(z+6)=5(z)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||z+6|=5|z|
x=+y , +x=y(z+6)=5(z)
x=y , x=y(z+6)=5((z))

2. Solve the two equations for z

12 additional steps

(z+6)=5z

Subtract from both sides:

(z+6)-5z=(5z)-5z

Group like terms:

(z-5z)+6=(5z)-5z

Simplify the arithmetic:

-4z+6=(5z)-5z

Simplify the arithmetic:

4z+6=0

Subtract from both sides:

(-4z+6)-6=0-6

Simplify the arithmetic:

4z=06

Simplify the arithmetic:

4z=6

Divide both sides by :

(-4z)-4=-6-4

Cancel out the negatives:

4z4=-6-4

Simplify the fraction:

z=-6-4

Cancel out the negatives:

z=64

Find the greatest common factor of the numerator and denominator:

z=(3·2)(2·2)

Factor out and cancel the greatest common factor:

z=32

11 additional steps

(z+6)=5·-z

Group like terms:

(z+6)=(5·-1)z

Multiply the coefficients:

(z+6)=-5z

Add to both sides:

(z+6)+5z=(-5z)+5z

Group like terms:

(z+5z)+6=(-5z)+5z

Simplify the arithmetic:

6z+6=(-5z)+5z

Simplify the arithmetic:

6z+6=0

Subtract from both sides:

(6z+6)-6=0-6

Simplify the arithmetic:

6z=06

Simplify the arithmetic:

6z=6

Divide both sides by :

(6z)6=-66

Simplify the fraction:

z=-66

Simplify the fraction:

z=1

3. List the solutions

z=32,-1
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|z+6|
y=5|z|
The equation is true where the two lines cross.

Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.