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Solution - Absolute value equations

Exact form: x=2.25
x=2.25
See steps

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|x−3.5|+|−x+1|=0

Add −|−x+1| to both sides of the equation:

|x−3.5|+|−x+1|−|−x+1|=−|−x+1|

Simplify the arithmetic

|x−3.5|=−|−x+1|

2. 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
|x−3.5|=−|−x+1|
without the absolute value bars:

|x|=|y||x−3.5|=−|−x+1|
x=+y(x−3.5)=−(−x+1)
x=−y(x−3.5)=−−(−x+1)
+x=y(x−3.5)=−(−x+1)
−x=y−(x−3.5)=−(−x+1)

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||x−3.5|=−|−x+1|
x=+y , +x=y(x−3.5)=−(−x+1)
x=−y , −x=y(x−3.5)=−−(−x+1)

3. Solve the two equations for x

6 additional steps

(x-3.5)=-(-x+1)

Expand the parentheses:

(x-3.5)=x-1

Subtract from both sides:

(x-3.5)-x=(x-1)-x

Group like terms:

(x-x)-3.5=(x-1)-x

Simplify the arithmetic:

-3.5=(x-1)-x

Group like terms:

-3.5=(x-x)-1

Simplify the arithmetic:

−3.5=−1

The statement is false:

−3.5=−1

The equation is false so it has no solution.

11 additional steps

(x-3.5)=-(-(-x+1))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(x-3.5)=-x+1

Add to both sides:

(x-3.5)+x=(-x+1)+x

Group like terms:

(x+x)-3.5=(-x+1)+x

Simplify the arithmetic:

2x-3.5=(-x+1)+x

Group like terms:

2x-3.5=(-x+x)+1

Simplify the arithmetic:

2x−3.5=1

Add to both sides:

(2x-3.5)+3.5=1+3.5

Simplify the arithmetic:

2x=1+3.5

Simplify the arithmetic:

2x=4.5

Divide both sides by :

(2x)2=4.52

Simplify the fraction:

x=4.52

Simplify the arithmetic:

x=2.25

4. Graph

Each line represents the function of one side of the equation:
y=|x−3.5|
y=−|−x+1|
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.

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