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Solution - Absolute value equations

Exact form: x=2,-15
x=2 , -\frac{1}{5}
Decimal form: x=2,0.2
x=2 , -0.2

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
|4x+3|=|6x1|
without the absolute value bars:

|x|=|y||4x+3|=|6x1|
x=+y(4x+3)=(6x1)
x=y(4x+3)=(6x1)
+x=y(4x+3)=(6x1)
x=y(4x+3)=(6x1)

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||4x+3|=|6x1|
x=+y , +x=y(4x+3)=(6x1)
x=y , x=y(4x+3)=(6x1)

2. Solve the two equations for x

13 additional steps

(4x+3)=(6x-1)

Subtract from both sides:

(4x+3)-6x=(6x-1)-6x

Group like terms:

(4x-6x)+3=(6x-1)-6x

Simplify the arithmetic:

-2x+3=(6x-1)-6x

Group like terms:

-2x+3=(6x-6x)-1

Simplify the arithmetic:

2x+3=1

Subtract from both sides:

(-2x+3)-3=-1-3

Simplify the arithmetic:

2x=13

Simplify the arithmetic:

2x=4

Divide both sides by :

(-2x)-2=-4-2

Cancel out the negatives:

2x2=-4-2

Simplify the fraction:

x=-4-2

Cancel out the negatives:

x=42

Find the greatest common factor of the numerator and denominator:

x=(2·2)(1·2)

Factor out and cancel the greatest common factor:

x=2

12 additional steps

(4x+3)=-(6x-1)

Expand the parentheses:

(4x+3)=-6x+1

Add to both sides:

(4x+3)+6x=(-6x+1)+6x

Group like terms:

(4x+6x)+3=(-6x+1)+6x

Simplify the arithmetic:

10x+3=(-6x+1)+6x

Group like terms:

10x+3=(-6x+6x)+1

Simplify the arithmetic:

10x+3=1

Subtract from both sides:

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

Simplify the arithmetic:

10x=13

Simplify the arithmetic:

10x=2

Divide both sides by :

(10x)10=-210

Simplify the fraction:

x=-210

Find the greatest common factor of the numerator and denominator:

x=(-1·2)(5·2)

Factor out and cancel the greatest common factor:

x=-15

3. List the solutions

x=2,-15
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x+3|
y=|6x1|
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.