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Solution - Absolute value equations

Exact form: x=45,-32
x=\frac{4}{5} , -\frac{3}{2}
Mixed number form: x=45,-112
x=\frac{4}{5} , -1\frac{1}{2}
Decimal form: x=0.8,1.5
x=0.8 , -1.5

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
|7x1|=|3x+7|
without the absolute value bars:

|x|=|y||7x1|=|3x+7|
x=+y(7x1)=(3x+7)
x=y(7x1)=(3x+7)
+x=y(7x1)=(3x+7)
x=y(7x1)=(3x+7)

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||7x1|=|3x+7|
x=+y , +x=y(7x1)=(3x+7)
x=y , x=y(7x1)=(3x+7)

2. Solve the two equations for x

11 additional steps

(7x-1)=(-3x+7)

Add to both sides:

(7x-1)+3x=(-3x+7)+3x

Group like terms:

(7x+3x)-1=(-3x+7)+3x

Simplify the arithmetic:

10x-1=(-3x+7)+3x

Group like terms:

10x-1=(-3x+3x)+7

Simplify the arithmetic:

10x1=7

Add to both sides:

(10x-1)+1=7+1

Simplify the arithmetic:

10x=7+1

Simplify the arithmetic:

10x=8

Divide both sides by :

(10x)10=810

Simplify the fraction:

x=810

Find the greatest common factor of the numerator and denominator:

x=(4·2)(5·2)

Factor out and cancel the greatest common factor:

x=45

12 additional steps

(7x-1)=-(-3x+7)

Expand the parentheses:

(7x-1)=3x-7

Subtract from both sides:

(7x-1)-3x=(3x-7)-3x

Group like terms:

(7x-3x)-1=(3x-7)-3x

Simplify the arithmetic:

4x-1=(3x-7)-3x

Group like terms:

4x-1=(3x-3x)-7

Simplify the arithmetic:

4x1=7

Add to both sides:

(4x-1)+1=-7+1

Simplify the arithmetic:

4x=7+1

Simplify the arithmetic:

4x=6

Divide both sides by :

(4x)4=-64

Simplify the fraction:

x=-64

Find the greatest common factor of the numerator and denominator:

x=(-3·2)(2·2)

Factor out and cancel the greatest common factor:

x=-32

3. List the solutions

x=45,-32
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|7x1|
y=|3x+7|
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.