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Solution - Absolute value equations

Exact form: p=-6,-12
p=-6 , -\frac{1}{2}
Decimal form: p=6,0.5
p=-6 , -0.5

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
|p5|=|3p+7|
without the absolute value bars:

|x|=|y||p5|=|3p+7|
x=+y(p5)=(3p+7)
x=y(p5)=(3p+7)
+x=y(p5)=(3p+7)
x=y(p5)=(3p+7)

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||p5|=|3p+7|
x=+y , +x=y(p5)=(3p+7)
x=y , x=y(p5)=(3p+7)

2. Solve the two equations for p

13 additional steps

(p-5)=(3p+7)

Subtract from both sides:

(p-5)-3p=(3p+7)-3p

Group like terms:

(p-3p)-5=(3p+7)-3p

Simplify the arithmetic:

-2p-5=(3p+7)-3p

Group like terms:

-2p-5=(3p-3p)+7

Simplify the arithmetic:

2p5=7

Add to both sides:

(-2p-5)+5=7+5

Simplify the arithmetic:

2p=7+5

Simplify the arithmetic:

2p=12

Divide both sides by :

(-2p)-2=12-2

Cancel out the negatives:

2p2=12-2

Simplify the fraction:

p=12-2

Move the negative sign from the denominator to the numerator:

p=-122

Find the greatest common factor of the numerator and denominator:

p=(-6·2)(1·2)

Factor out and cancel the greatest common factor:

p=6

12 additional steps

(p-5)=-(3p+7)

Expand the parentheses:

(p-5)=-3p-7

Add to both sides:

(p-5)+3p=(-3p-7)+3p

Group like terms:

(p+3p)-5=(-3p-7)+3p

Simplify the arithmetic:

4p-5=(-3p-7)+3p

Group like terms:

4p-5=(-3p+3p)-7

Simplify the arithmetic:

4p5=7

Add to both sides:

(4p-5)+5=-7+5

Simplify the arithmetic:

4p=7+5

Simplify the arithmetic:

4p=2

Divide both sides by :

(4p)4=-24

Simplify the fraction:

p=-24

Find the greatest common factor of the numerator and denominator:

p=(-1·2)(2·2)

Factor out and cancel the greatest common factor:

p=-12

3. List the solutions

p=-6,-12
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|p5|
y=|3p+7|
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.