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Solution - Absolute value equations

Exact form: y=0,0
y=0 , 0

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
|25y|=|45y|
without the absolute value bars:

|x|=|y||25y|=|45y|
x=+y(25y)=(45y)
x=-y(25y)=-(45y)
+x=y(25y)=(45y)
-x=y-(25y)=(45y)

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||25y|=|45y|
x=+y , +x=y(25y)=(45y)
x=-y , -x=y(25y)=-(45y)

2. Solve the two equations for y

7 additional steps

25·y=45y

Subtract from both sides:

(25y)-45·y=(45y)-45y

Combine the fractions:

(2-4)5·y=(45·y)-45y

Combine the numerators:

-25·y=(45·y)-45y

Combine the fractions:

-25·y=(4-4)5y

Combine the numerators:

-25·y=05y

Reduce the zero numerator:

-25y=0y

Simplify the arithmetic:

-25y=0

Divide both sides by the coefficient:

y=0

10 additional steps

25·y=-45y

Multiply both sides by inverse fraction :

(25y)·52=(-45y)·52

Group like terms:

(25·52)y=(-45y)·52

Multiply the coefficients:

(2·5)(5·2)·y=(-45y)·52

Simplify the fraction:

y=(-45y)·52

Group like terms:

y=(-45·52)y

Multiply the coefficients:

y=(-4·5)(5·2)y

Simplify the arithmetic:

y=2y

Add to both sides:

y+2y=(-2y)+2y

Simplify the arithmetic:

3y=(-2y)+2y

Simplify the arithmetic:

3y=0

Divide both sides by the coefficient:

y=0

3. List the solutions

y=0,0
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|25y|
y=|45y|
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.