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Solution - Absolute value equations

Exact form: x=1,45
x=1 , \frac{4}{5}
Decimal form: x=1,0.8
x=1 , 0.8

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|=|6x+5|
without the absolute value bars:

|x|=|y||4x+3|=|6x+5|
x=+y(4x+3)=(6x+5)
x=y(4x+3)=(6x+5)
+x=y(4x+3)=(6x+5)
x=y(4x+3)=(6x+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||4x+3|=|6x+5|
x=+y , +x=y(4x+3)=(6x+5)
x=y , x=y(4x+3)=(6x+5)

2. Solve the two equations for x

10 additional steps

(-4x+3)=(-6x+5)

Add to both sides:

(-4x+3)+6x=(-6x+5)+6x

Group like terms:

(-4x+6x)+3=(-6x+5)+6x

Simplify the arithmetic:

2x+3=(-6x+5)+6x

Group like terms:

2x+3=(-6x+6x)+5

Simplify the arithmetic:

2x+3=5

Subtract from both sides:

(2x+3)-3=5-3

Simplify the arithmetic:

2x=53

Simplify the arithmetic:

2x=2

Divide both sides by :

(2x)2=22

Simplify the fraction:

x=22

Simplify the fraction:

x=1

14 additional steps

(-4x+3)=-(-6x+5)

Expand the parentheses:

(-4x+3)=6x-5

Subtract from both sides:

(-4x+3)-6x=(6x-5)-6x

Group like terms:

(-4x-6x)+3=(6x-5)-6x

Simplify the arithmetic:

-10x+3=(6x-5)-6x

Group like terms:

-10x+3=(6x-6x)-5

Simplify the arithmetic:

10x+3=5

Subtract from both sides:

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

Simplify the arithmetic:

10x=53

Simplify the arithmetic:

10x=8

Divide both sides by :

(-10x)-10=-8-10

Cancel out the negatives:

10x10=-8-10

Simplify the fraction:

x=-8-10

Cancel out the negatives:

x=810

Find the greatest common factor of the numerator and denominator:

x=(4·2)(5·2)

Factor out and cancel the greatest common factor:

x=45

3. List the solutions

x=1,45
(2 solution(s))

4. Graph

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