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Solution - Absolute value equations

Exact form: x=1,16
x=1 , \frac{1}{6}
Decimal form: x=1,0.167
x=1 , 0.167

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

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

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

2. Solve the two equations for x

12 additional steps

(4x+1)=(8x-3)

Subtract from both sides:

(4x+1)-8x=(8x-3)-8x

Group like terms:

(4x-8x)+1=(8x-3)-8x

Simplify the arithmetic:

-4x+1=(8x-3)-8x

Group like terms:

-4x+1=(8x-8x)-3

Simplify the arithmetic:

4x+1=3

Subtract from both sides:

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

Simplify the arithmetic:

4x=31

Simplify the arithmetic:

4x=4

Divide both sides by :

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

Cancel out the negatives:

4x4=-4-4

Simplify the fraction:

x=-4-4

Cancel out the negatives:

x=44

Simplify the fraction:

x=1

12 additional steps

(4x+1)=-(8x-3)

Expand the parentheses:

(4x+1)=-8x+3

Add to both sides:

(4x+1)+8x=(-8x+3)+8x

Group like terms:

(4x+8x)+1=(-8x+3)+8x

Simplify the arithmetic:

12x+1=(-8x+3)+8x

Group like terms:

12x+1=(-8x+8x)+3

Simplify the arithmetic:

12x+1=3

Subtract from both sides:

(12x+1)-1=3-1

Simplify the arithmetic:

12x=31

Simplify the arithmetic:

12x=2

Divide both sides by :

(12x)12=212

Simplify the fraction:

x=212

Find the greatest common factor of the numerator and denominator:

x=(1·2)(6·2)

Factor out and cancel the greatest common factor:

x=16

3. List the solutions

x=1,16
(2 solution(s))

4. Graph

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