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Solution - Absolute value equations

Exact form: a=-1,12
a=-1 , \frac{1}{2}
Decimal form: a=1,0.5
a=-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
|a2|=|3a|
without the absolute value bars:

|x|=|y||a2|=|3a|
x=+y(a2)=(3a)
x=y(a2)=(3a)
+x=y(a2)=(3a)
x=y(a2)=(3a)

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||a2|=|3a|
x=+y , +x=y(a2)=(3a)
x=y , x=y(a2)=(3a)

2. Solve the two equations for a

11 additional steps

(a-2)=3a

Subtract from both sides:

(a-2)-3a=(3a)-3a

Group like terms:

(a-3a)-2=(3a)-3a

Simplify the arithmetic:

-2a-2=(3a)-3a

Simplify the arithmetic:

2a2=0

Add to both sides:

(-2a-2)+2=0+2

Simplify the arithmetic:

2a=0+2

Simplify the arithmetic:

2a=2

Divide both sides by :

(-2a)-2=2-2

Cancel out the negatives:

2a2=2-2

Simplify the fraction:

a=2-2

Move the negative sign from the denominator to the numerator:

a=-22

Simplify the fraction:

a=1

9 additional steps

(a-2)=-3a

Add to both sides:

(a-2)+2=(-3a)+2

Simplify the arithmetic:

a=(-3a)+2

Add to both sides:

a+3a=((-3a)+2)+3a

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

4a=2

Divide both sides by :

(4a)4=24

Simplify the fraction:

a=24

Find the greatest common factor of the numerator and denominator:

a=(1·2)(2·2)

Factor out and cancel the greatest common factor:

a=12

3. List the solutions

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

4. Graph

Each line represents the function of one side of the equation:
y=|a2|
y=|3a|
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.