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Solution - Absolute value equations

Exact form: x=-116
x=-\frac{1}{16}
Decimal form: x=0.062
x=-0.062

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
|8x3|=4|2x+1|
without the absolute value bars:

|x|=|y||8x3|=4|2x+1|
x=+y(8x3)=4(2x+1)
x=y(8x3)=4((2x+1))
+x=y(8x3)=4(2x+1)
x=y(8x3)=4(2x+1)

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||8x3|=4|2x+1|
x=+y , +x=y(8x3)=4(2x+1)
x=y , x=y(8x3)=4((2x+1))

2. Solve the two equations for x

8 additional steps

(8x-3)=4·(2x+1)

Expand the parentheses:

(8x-3)=4·2x+4·1

Multiply the coefficients:

(8x-3)=8x+4·1

Simplify the arithmetic:

(8x-3)=8x+4

Subtract from both sides:

(8x-3)-8x=(8x+4)-8x

Group like terms:

(8x-8x)-3=(8x+4)-8x

Simplify the arithmetic:

-3=(8x+4)-8x

Group like terms:

-3=(8x-8x)+4

Simplify the arithmetic:

3=4

The statement is false:

3=4

The equation is false so it has no solution.

13 additional steps

(8x-3)=4·(-(2x+1))

Expand the parentheses:

(8x-3)=4·(-2x-1)

Expand the parentheses:

(8x-3)=4·-2x+4·-1

Multiply the coefficients:

(8x-3)=-8x+4·-1

Simplify the arithmetic:

(8x-3)=-8x-4

Add to both sides:

(8x-3)+8x=(-8x-4)+8x

Group like terms:

(8x+8x)-3=(-8x-4)+8x

Simplify the arithmetic:

16x-3=(-8x-4)+8x

Group like terms:

16x-3=(-8x+8x)-4

Simplify the arithmetic:

16x3=4

Add to both sides:

(16x-3)+3=-4+3

Simplify the arithmetic:

16x=4+3

Simplify the arithmetic:

16x=1

Divide both sides by :

(16x)16=-116

Simplify the fraction:

x=-116

3. Graph

Each line represents the function of one side of the equation:
y=|8x3|
y=4|2x+1|
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.