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Solution - Absolute value equations

Exact form: p=112,-56
p=\frac{11}{2} , -\frac{5}{6}
Mixed number form: p=512,-56
p=5\frac{1}{2} , -\frac{5}{6}
Decimal form: p=5.5,0.833
p=5.5 , -0.833

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
|4p3|=|2p+8|
without the absolute value bars:

|x|=|y||4p3|=|2p+8|
x=+y(4p3)=(2p+8)
x=y(4p3)=(2p+8)
+x=y(4p3)=(2p+8)
x=y(4p3)=(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||4p3|=|2p+8|
x=+y , +x=y(4p3)=(2p+8)
x=y , x=y(4p3)=(2p+8)

2. Solve the two equations for p

9 additional steps

(4p-3)=(2p+8)

Subtract from both sides:

(4p-3)-2p=(2p+8)-2p

Group like terms:

(4p-2p)-3=(2p+8)-2p

Simplify the arithmetic:

2p-3=(2p+8)-2p

Group like terms:

2p-3=(2p-2p)+8

Simplify the arithmetic:

2p3=8

Add to both sides:

(2p-3)+3=8+3

Simplify the arithmetic:

2p=8+3

Simplify the arithmetic:

2p=11

Divide both sides by :

(2p)2=112

Simplify the fraction:

p=112

10 additional steps

(4p-3)=-(2p+8)

Expand the parentheses:

(4p-3)=-2p-8

Add to both sides:

(4p-3)+2p=(-2p-8)+2p

Group like terms:

(4p+2p)-3=(-2p-8)+2p

Simplify the arithmetic:

6p-3=(-2p-8)+2p

Group like terms:

6p-3=(-2p+2p)-8

Simplify the arithmetic:

6p3=8

Add to both sides:

(6p-3)+3=-8+3

Simplify the arithmetic:

6p=8+3

Simplify the arithmetic:

6p=5

Divide both sides by :

(6p)6=-56

Simplify the fraction:

p=-56

3. List the solutions

p=112,-56
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4p3|
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.