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Solution - Absolute value equations

Exact form: y=-5,-43
y=-5 , -\frac{4}{3}
Mixed number form: y=-5,-113
y=-5 , -1\frac{1}{3}
Decimal form: y=5,1.333
y=-5 , -1.333

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
|4y+9|=|2y1|
without the absolute value bars:

|x|=|y||4y+9|=|2y1|
x=+y(4y+9)=(2y1)
x=y(4y+9)=(2y1)
+x=y(4y+9)=(2y1)
x=y(4y+9)=(2y1)

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||4y+9|=|2y1|
x=+y , +x=y(4y+9)=(2y1)
x=y , x=y(4y+9)=(2y1)

2. Solve the two equations for y

11 additional steps

(4y+9)=(2y-1)

Subtract from both sides:

(4y+9)-2y=(2y-1)-2y

Group like terms:

(4y-2y)+9=(2y-1)-2y

Simplify the arithmetic:

2y+9=(2y-1)-2y

Group like terms:

2y+9=(2y-2y)-1

Simplify the arithmetic:

2y+9=1

Subtract from both sides:

(2y+9)-9=-1-9

Simplify the arithmetic:

2y=19

Simplify the arithmetic:

2y=10

Divide both sides by :

(2y)2=-102

Simplify the fraction:

y=-102

Find the greatest common factor of the numerator and denominator:

y=(-5·2)(1·2)

Factor out and cancel the greatest common factor:

y=5

12 additional steps

(4y+9)=-(2y-1)

Expand the parentheses:

(4y+9)=-2y+1

Add to both sides:

(4y+9)+2y=(-2y+1)+2y

Group like terms:

(4y+2y)+9=(-2y+1)+2y

Simplify the arithmetic:

6y+9=(-2y+1)+2y

Group like terms:

6y+9=(-2y+2y)+1

Simplify the arithmetic:

6y+9=1

Subtract from both sides:

(6y+9)-9=1-9

Simplify the arithmetic:

6y=19

Simplify the arithmetic:

6y=8

Divide both sides by :

(6y)6=-86

Simplify the fraction:

y=-86

Find the greatest common factor of the numerator and denominator:

y=(-4·2)(3·2)

Factor out and cancel the greatest common factor:

y=-43

3. List the solutions

y=-5,-43
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4y+9|
y=|2y1|
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.