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Solution - Absolute value equations

Exact form: x=25,-32
x=\frac{2}{5} , -\frac{3}{2}
Mixed number form: x=25,-112
x=\frac{2}{5} , -1\frac{1}{2}
Decimal form: x=0.4,1.5
x=0.4 , -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
|7x+1|=|3x+5|
without the absolute value bars:

|x|=|y||7x+1|=|3x+5|
x=+y(7x+1)=(3x+5)
x=y(7x+1)=(3x+5)
+x=y(7x+1)=(3x+5)
x=y(7x+1)=(3x+5)

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||7x+1|=|3x+5|
x=+y , +x=y(7x+1)=(3x+5)
x=y , x=y(7x+1)=(3x+5)

2. Solve the two equations for x

11 additional steps

(7x+1)=(-3x+5)

Add to both sides:

(7x+1)+3x=(-3x+5)+3x

Group like terms:

(7x+3x)+1=(-3x+5)+3x

Simplify the arithmetic:

10x+1=(-3x+5)+3x

Group like terms:

10x+1=(-3x+3x)+5

Simplify the arithmetic:

10x+1=5

Subtract from both sides:

(10x+1)-1=5-1

Simplify the arithmetic:

10x=51

Simplify the arithmetic:

10x=4

Divide both sides by :

(10x)10=410

Simplify the fraction:

x=410

Find the greatest common factor of the numerator and denominator:

x=(2·2)(5·2)

Factor out and cancel the greatest common factor:

x=25

12 additional steps

(7x+1)=-(-3x+5)

Expand the parentheses:

(7x+1)=3x-5

Subtract from both sides:

(7x+1)-3x=(3x-5)-3x

Group like terms:

(7x-3x)+1=(3x-5)-3x

Simplify the arithmetic:

4x+1=(3x-5)-3x

Group like terms:

4x+1=(3x-3x)-5

Simplify the arithmetic:

4x+1=5

Subtract from both sides:

(4x+1)-1=-5-1

Simplify the arithmetic:

4x=51

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=25,-32
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|7x+1|
y=|3x+5|
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.