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Solution - Absolute value equations

Exact form: x=16,2
x=16 , -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
|4x1|=3|x+5|
without the absolute value bars:

|x|=|y||4x1|=3|x+5|
x=+y(4x1)=3(x+5)
x=y(4x1)=3((x+5))
+x=y(4x1)=3(x+5)
x=y(4x1)=3(x+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||4x1|=3|x+5|
x=+y , +x=y(4x1)=3(x+5)
x=y , x=y(4x1)=3((x+5))

2. Solve the two equations for x

9 additional steps

(4x-1)=3·(x+5)

Expand the parentheses:

(4x-1)=3x+3·5

Simplify the arithmetic:

(4x-1)=3x+15

Subtract from both sides:

(4x-1)-3x=(3x+15)-3x

Group like terms:

(4x-3x)-1=(3x+15)-3x

Simplify the arithmetic:

x-1=(3x+15)-3x

Group like terms:

x-1=(3x-3x)+15

Simplify the arithmetic:

x1=15

Add to both sides:

(x-1)+1=15+1

Simplify the arithmetic:

x=15+1

Simplify the arithmetic:

x=16

16 additional steps

(4x-1)=3·(-(x+5))

Expand the parentheses:

(4x-1)=3·(-x-5)

(4x-1)=3·-x+3·-5

Group like terms:

(4x-1)=(3·-1)x+3·-5

Multiply the coefficients:

(4x-1)=-3x+3·-5

Simplify the arithmetic:

(4x-1)=-3x-15

Add to both sides:

(4x-1)+3x=(-3x-15)+3x

Group like terms:

(4x+3x)-1=(-3x-15)+3x

Simplify the arithmetic:

7x-1=(-3x-15)+3x

Group like terms:

7x-1=(-3x+3x)-15

Simplify the arithmetic:

7x1=15

Add to both sides:

(7x-1)+1=-15+1

Simplify the arithmetic:

7x=15+1

Simplify the arithmetic:

7x=14

Divide both sides by :

(7x)7=-147

Simplify the fraction:

x=-147

Find the greatest common factor of the numerator and denominator:

x=(-2·7)(1·7)

Factor out and cancel the greatest common factor:

x=2

3. List the solutions

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

4. Graph

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