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Solution - Absolute value equations

Exact form: p=14,2
p=14 , 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
4|p5|=|2p+8|
without the absolute value bars:

|x|=|y|4|p5|=|2p+8|
x=+y4(p5)=(2p+8)
x=y4(p5)=(2p+8)
+x=y4(p5)=(2p+8)
x=y4((p5))=(2p+8)

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|4|p5|=|2p+8|
x=+y , +x=y4(p5)=(2p+8)
x=y , x=y4(p5)=(2p+8)

2. Solve the two equations for p

13 additional steps

4·(p-5)=(2p+8)

Expand the parentheses:

4p+4·-5=(2p+8)

Simplify the arithmetic:

4p-20=(2p+8)

Subtract from both sides:

(4p-20)-2p=(2p+8)-2p

Group like terms:

(4p-2p)-20=(2p+8)-2p

Simplify the arithmetic:

2p-20=(2p+8)-2p

Group like terms:

2p-20=(2p-2p)+8

Simplify the arithmetic:

2p20=8

Add to both sides:

(2p-20)+20=8+20

Simplify the arithmetic:

2p=8+20

Simplify the arithmetic:

2p=28

Divide both sides by :

(2p)2=282

Simplify the fraction:

p=282

Find the greatest common factor of the numerator and denominator:

p=(14·2)(1·2)

Factor out and cancel the greatest common factor:

p=14

14 additional steps

4·(p-5)=-(2p+8)

Expand the parentheses:

4p+4·-5=-(2p+8)

Simplify the arithmetic:

4p-20=-(2p+8)

Expand the parentheses:

4p20=2p8

Add to both sides:

(4p-20)+2p=(-2p-8)+2p

Group like terms:

(4p+2p)-20=(-2p-8)+2p

Simplify the arithmetic:

6p-20=(-2p-8)+2p

Group like terms:

6p-20=(-2p+2p)-8

Simplify the arithmetic:

6p20=8

Add to both sides:

(6p-20)+20=-8+20

Simplify the arithmetic:

6p=8+20

Simplify the arithmetic:

6p=12

Divide both sides by :

(6p)6=126

Simplify the fraction:

p=126

Find the greatest common factor of the numerator and denominator:

p=(2·6)(1·6)

Factor out and cancel the greatest common factor:

p=2

3. List the solutions

p=14,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=4|p5|
y=|2p+8|
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.