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Solution - Absolute value equations

Exact form: x=52,12
x=\frac{5}{2} , \frac{1}{2}
Mixed number form: x=212,12
x=2\frac{1}{2} , \frac{1}{2}
Decimal form: x=2.5,0.5
x=2.5 , 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
|6x7|=|2x+3|
without the absolute value bars:

|x|=|y||6x7|=|2x+3|
x=+y(6x7)=(2x+3)
x=y(6x7)=(2x+3)
+x=y(6x7)=(2x+3)
x=y(6x7)=(2x+3)

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||6x7|=|2x+3|
x=+y , +x=y(6x7)=(2x+3)
x=y , x=y(6x7)=(2x+3)

2. Solve the two equations for x

11 additional steps

(6x-7)=(2x+3)

Subtract from both sides:

(6x-7)-2x=(2x+3)-2x

Group like terms:

(6x-2x)-7=(2x+3)-2x

Simplify the arithmetic:

4x-7=(2x+3)-2x

Group like terms:

4x-7=(2x-2x)+3

Simplify the arithmetic:

4x7=3

Add to both sides:

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

Simplify the arithmetic:

4x=3+7

Simplify the arithmetic:

4x=10

Divide both sides by :

(4x)4=104

Simplify the fraction:

x=104

Find the greatest common factor of the numerator and denominator:

x=(5·2)(2·2)

Factor out and cancel the greatest common factor:

x=52

12 additional steps

(6x-7)=-(2x+3)

Expand the parentheses:

(6x-7)=-2x-3

Add to both sides:

(6x-7)+2x=(-2x-3)+2x

Group like terms:

(6x+2x)-7=(-2x-3)+2x

Simplify the arithmetic:

8x-7=(-2x-3)+2x

Group like terms:

8x-7=(-2x+2x)-3

Simplify the arithmetic:

8x7=3

Add to both sides:

(8x-7)+7=-3+7

Simplify the arithmetic:

8x=3+7

Simplify the arithmetic:

8x=4

Divide both sides by :

(8x)8=48

Simplify the fraction:

x=48

Find the greatest common factor of the numerator and denominator:

x=(1·4)(2·4)

Factor out and cancel the greatest common factor:

x=12

3. List the solutions

x=52,12
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|6x7|
y=|2x+3|
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.