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Solution - Absolute value equations

Exact form: x=132,-32
x=\frac{13}{2} , -\frac{3}{2}
Mixed number form: x=612,-112
x=6\frac{1}{2} , -1\frac{1}{2}
Decimal form: x=6.5,1.5
x=6.5 , -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
|4x2|=|2x+11|
without the absolute value bars:

|x|=|y||4x2|=|2x+11|
x=+y(4x2)=(2x+11)
x=y(4x2)=(2x+11)
+x=y(4x2)=(2x+11)
x=y(4x2)=(2x+11)

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||4x2|=|2x+11|
x=+y , +x=y(4x2)=(2x+11)
x=y , x=y(4x2)=(2x+11)

2. Solve the two equations for x

9 additional steps

(4x-2)=(2x+11)

Subtract from both sides:

(4x-2)-2x=(2x+11)-2x

Group like terms:

(4x-2x)-2=(2x+11)-2x

Simplify the arithmetic:

2x-2=(2x+11)-2x

Group like terms:

2x-2=(2x-2x)+11

Simplify the arithmetic:

2x2=11

Add to both sides:

(2x-2)+2=11+2

Simplify the arithmetic:

2x=11+2

Simplify the arithmetic:

2x=13

Divide both sides by :

(2x)2=132

Simplify the fraction:

x=132

12 additional steps

(4x-2)=-(2x+11)

Expand the parentheses:

(4x-2)=-2x-11

Add to both sides:

(4x-2)+2x=(-2x-11)+2x

Group like terms:

(4x+2x)-2=(-2x-11)+2x

Simplify the arithmetic:

6x-2=(-2x-11)+2x

Group like terms:

6x-2=(-2x+2x)-11

Simplify the arithmetic:

6x2=11

Add to both sides:

(6x-2)+2=-11+2

Simplify the arithmetic:

6x=11+2

Simplify the arithmetic:

6x=9

Divide both sides by :

(6x)6=-96

Simplify the fraction:

x=-96

Find the greatest common factor of the numerator and denominator:

x=(-3·3)(2·3)

Factor out and cancel the greatest common factor:

x=-32

3. List the solutions

x=132,-32
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x2|
y=|2x+11|
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.