Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: y=-20,-1007
y=-20 , -\frac{100}{7}
Mixed number form: y=-20,-1427
y=-20 , -14\frac{2}{7}
Decimal form: y=20,14.286
y=-20 , -14.286

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
|12y+8|=|15y+2|
without the absolute value bars:

|x|=|y||12y+8|=|15y+2|
x=+y(12y+8)=(15y+2)
x=-y(12y+8)=-(15y+2)
+x=y(12y+8)=(15y+2)
-x=y-(12y+8)=(15y+2)

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||12y+8|=|15y+2|
x=+y , +x=y(12y+8)=(15y+2)
x=-y , -x=y(12y+8)=-(15y+2)

2. Solve the two equations for y

21 additional steps

(12·y+8)=(15y+2)

Subtract from both sides:

(12y+8)-15·y=(15y+2)-15y

Group like terms:

(12·y+-15·y)+8=(15·y+2)-15y

Group the coefficients:

(12+-15)y+8=(15·y+2)-15y

Find the lowest common denominator:

((1·5)(2·5)+(-1·2)(5·2))y+8=(15·y+2)-15y

Multiply the denominators:

((1·5)10+(-1·2)10)y+8=(15·y+2)-15y

Multiply the numerators:

(510+-210)y+8=(15·y+2)-15y

Combine the fractions:

(5-2)10·y+8=(15·y+2)-15y

Combine the numerators:

310·y+8=(15·y+2)-15y

Group like terms:

310·y+8=(15·y+-15y)+2

Combine the fractions:

310·y+8=(1-1)5y+2

Combine the numerators:

310·y+8=05y+2

Reduce the zero numerator:

310y+8=0y+2

Simplify the arithmetic:

310y+8=2

Subtract from both sides:

(310y+8)-8=2-8

Simplify the arithmetic:

310y=2-8

Simplify the arithmetic:

310y=-6

Multiply both sides by inverse fraction :

(310y)·103=-6·103

Group like terms:

(310·103)y=-6·103

Multiply the coefficients:

(3·10)(10·3)y=-6·103

Simplify the fraction:

y=-6·103

Multiply the fraction(s):

y=(-6·10)3

Simplify the arithmetic:

y=20

22 additional steps

(12y+8)=-(15y+2)

Expand the parentheses:

(12·y+8)=-15y-2

Add to both sides:

(12y+8)+15·y=(-15y-2)+15y

Group like terms:

(12·y+15·y)+8=(-15·y-2)+15y

Group the coefficients:

(12+15)y+8=(-15·y-2)+15y

Find the lowest common denominator:

((1·5)(2·5)+(1·2)(5·2))y+8=(-15·y-2)+15y

Multiply the denominators:

((1·5)10+(1·2)10)y+8=(-15·y-2)+15y

Multiply the numerators:

(510+210)y+8=(-15·y-2)+15y

Combine the fractions:

(5+2)10·y+8=(-15·y-2)+15y

Combine the numerators:

710·y+8=(-15·y-2)+15y

Group like terms:

710·y+8=(-15·y+15y)-2

Combine the fractions:

710·y+8=(-1+1)5y-2

Combine the numerators:

710·y+8=05y-2

Reduce the zero numerator:

710y+8=0y-2

Simplify the arithmetic:

710y+8=-2

Subtract from both sides:

(710y+8)-8=-2-8

Simplify the arithmetic:

710y=-2-8

Simplify the arithmetic:

710y=-10

Multiply both sides by inverse fraction :

(710y)·107=-10·107

Group like terms:

(710·107)y=-10·107

Multiply the coefficients:

(7·10)(10·7)y=-10·107

Simplify the fraction:

y=-10·107

Multiply the fraction(s):

y=(-10·10)7

Simplify the arithmetic:

y=-1007

3. List the solutions

y=-20,-1007
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|12y+8|
y=|15y+2|
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.