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Solution - Absolute value equations

Exact form: x=85,2
x=\frac{8}{5} , 2
Mixed number form: x=135,2
x=1\frac{3}{5} , 2
Decimal form: x=1.6,2
x=1.6 , 2

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
|7x12|=|3x+4|
without the absolute value bars:

|x|=|y||7x12|=|3x+4|
x=+y(7x12)=(3x+4)
x=y(7x12)=(3x+4)
+x=y(7x12)=(3x+4)
x=y(7x12)=(3x+4)

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||7x12|=|3x+4|
x=+y , +x=y(7x12)=(3x+4)
x=y , x=y(7x12)=(3x+4)

2. Solve the two equations for x

11 additional steps

(7x-12)=(-3x+4)

Add to both sides:

(7x-12)+3x=(-3x+4)+3x

Group like terms:

(7x+3x)-12=(-3x+4)+3x

Simplify the arithmetic:

10x-12=(-3x+4)+3x

Group like terms:

10x-12=(-3x+3x)+4

Simplify the arithmetic:

10x12=4

Add to both sides:

(10x-12)+12=4+12

Simplify the arithmetic:

10x=4+12

Simplify the arithmetic:

10x=16

Divide both sides by :

(10x)10=1610

Simplify the fraction:

x=1610

Find the greatest common factor of the numerator and denominator:

x=(8·2)(5·2)

Factor out and cancel the greatest common factor:

x=85

12 additional steps

(7x-12)=-(-3x+4)

Expand the parentheses:

(7x-12)=3x-4

Subtract from both sides:

(7x-12)-3x=(3x-4)-3x

Group like terms:

(7x-3x)-12=(3x-4)-3x

Simplify the arithmetic:

4x-12=(3x-4)-3x

Group like terms:

4x-12=(3x-3x)-4

Simplify the arithmetic:

4x12=4

Add to both sides:

(4x-12)+12=-4+12

Simplify the arithmetic:

4x=4+12

Simplify the arithmetic:

4x=8

Divide both sides by :

(4x)4=84

Simplify the fraction:

x=84

Find the greatest common factor of the numerator and denominator:

x=(2·4)(1·4)

Factor out and cancel the greatest common factor:

x=2

3. List the solutions

x=85,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|7x12|
y=|3x+4|
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.