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Solution - Absolute value equations

Exact form: z=1,-12
z=1 , -\frac{1}{2}
Decimal form: z=1,0.5
z=1 , -0.5

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+2|=3|z|
without the absolute value bars:

|x|=|y||z+2|=3|z|
x=+y(z+2)=3(z)
x=y(z+2)=3((z))
+x=y(z+2)=3(z)
x=y(z+2)=3(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+2|=3|z|
x=+y , +x=y(z+2)=3(z)
x=y , x=y(z+2)=3((z))

2. Solve the two equations for z

11 additional steps

(z+2)=3z

Subtract from both sides:

(z+2)-3z=(3z)-3z

Group like terms:

(z-3z)+2=(3z)-3z

Simplify the arithmetic:

-2z+2=(3z)-3z

Simplify the arithmetic:

2z+2=0

Subtract from both sides:

(-2z+2)-2=0-2

Simplify the arithmetic:

2z=02

Simplify the arithmetic:

2z=2

Divide both sides by :

(-2z)-2=-2-2

Cancel out the negatives:

2z2=-2-2

Simplify the fraction:

z=-2-2

Cancel out the negatives:

z=22

Simplify the fraction:

z=1

12 additional steps

(z+2)=3·-z

Group like terms:

(z+2)=(3·-1)z

Multiply the coefficients:

(z+2)=-3z

Add to both sides:

(z+2)+3z=(-3z)+3z

Group like terms:

(z+3z)+2=(-3z)+3z

Simplify the arithmetic:

4z+2=(-3z)+3z

Simplify the arithmetic:

4z+2=0

Subtract from both sides:

(4z+2)-2=0-2

Simplify the arithmetic:

4z=02

Simplify the arithmetic:

4z=2

Divide both sides by :

(4z)4=-24

Simplify the fraction:

z=-24

Find the greatest common factor of the numerator and denominator:

z=(-1·2)(2·2)

Factor out and cancel the greatest common factor:

z=-12

3. List the solutions

z=1,-12
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|z+2|
y=3|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.