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Solution - Absolute value equations

Exact form: x=-5,13
x=-5 , \frac{1}{3}
Decimal form: x=5,0.333
x=-5 , 0.333

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
|5x+1|=4|x1|
without the absolute value bars:

|x|=|y||5x+1|=4|x1|
x=+y(5x+1)=4(x1)
x=y(5x+1)=4((x1))
+x=y(5x+1)=4(x1)
x=y(5x+1)=4(x1)

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||5x+1|=4|x1|
x=+y , +x=y(5x+1)=4(x1)
x=y , x=y(5x+1)=4((x1))

2. Solve the two equations for x

9 additional steps

(5x+1)=4·(x-1)

Expand the parentheses:

(5x+1)=4x+4·-1

Simplify the arithmetic:

(5x+1)=4x-4

Subtract from both sides:

(5x+1)-4x=(4x-4)-4x

Group like terms:

(5x-4x)+1=(4x-4)-4x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+1=4

Subtract from both sides:

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

Simplify the arithmetic:

x=41

Simplify the arithmetic:

x=5

16 additional steps

(5x+1)=4·(-(x-1))

Expand the parentheses:

(5x+1)=4·(-x+1)

(5x+1)=4·-x+4·1

Group like terms:

(5x+1)=(4·-1)x+4·1

Multiply the coefficients:

(5x+1)=-4x+4·1

Simplify the arithmetic:

(5x+1)=-4x+4

Add to both sides:

(5x+1)+4x=(-4x+4)+4x

Group like terms:

(5x+4x)+1=(-4x+4)+4x

Simplify the arithmetic:

9x+1=(-4x+4)+4x

Group like terms:

9x+1=(-4x+4x)+4

Simplify the arithmetic:

9x+1=4

Subtract from both sides:

(9x+1)-1=4-1

Simplify the arithmetic:

9x=41

Simplify the arithmetic:

9x=3

Divide both sides by :

(9x)9=39

Simplify the fraction:

x=39

Find the greatest common factor of the numerator and denominator:

x=(1·3)(3·3)

Factor out and cancel the greatest common factor:

x=13

3. List the solutions

x=-5,13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x+1|
y=4|x1|
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.