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Solution - Absolute value equations

Exact form: x=-6,-43
x=-6 , -\frac{4}{3}
Mixed number form: x=-6,-113
x=-6 , -1\frac{1}{3}
Decimal form: x=6,1.333
x=-6 , -1.333

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
|2x+5|=|x1|
without the absolute value bars:

|x|=|y||2x+5|=|x1|
x=+y(2x+5)=(x1)
x=y(2x+5)=(x1)
+x=y(2x+5)=(x1)
x=y(2x+5)=(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||2x+5|=|x1|
x=+y , +x=y(2x+5)=(x1)
x=y , x=y(2x+5)=(x1)

2. Solve the two equations for x

7 additional steps

(2x+5)=(x-1)

Subtract from both sides:

(2x+5)-x=(x-1)-x

Group like terms:

(2x-x)+5=(x-1)-x

Simplify the arithmetic:

x+5=(x-1)-x

Group like terms:

x+5=(x-x)-1

Simplify the arithmetic:

x+5=1

Subtract from both sides:

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

Simplify the arithmetic:

x=15

Simplify the arithmetic:

x=6

10 additional steps

(2x+5)=-(x-1)

Expand the parentheses:

(2x+5)=-x+1

Add to both sides:

(2x+5)+x=(-x+1)+x

Group like terms:

(2x+x)+5=(-x+1)+x

Simplify the arithmetic:

3x+5=(-x+1)+x

Group like terms:

3x+5=(-x+x)+1

Simplify the arithmetic:

3x+5=1

Subtract from both sides:

(3x+5)-5=1-5

Simplify the arithmetic:

3x=15

Simplify the arithmetic:

3x=4

Divide both sides by :

(3x)3=-43

Simplify the fraction:

x=-43

3. List the solutions

x=-6,-43
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x+5|
y=|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.