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Solution - Absolute value equations

Exact form: y=34,-3
y=\frac{3}{4} , -3
Decimal form: y=0.75,3
y=0.75 , -3

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
|4y3|=|4y+3|
without the absolute value bars:

|x|=|y||4y3|=|4y+3|
x=+y(4y3)=(4y+3)
x=y(4y3)=(4y+3)
+x=y(4y3)=(4y+3)
x=y(4y3)=(4y+3)

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||4y3|=|4y+3|
x=+y , +x=y(4y3)=(4y+3)
x=y , x=y(4y3)=(4y+3)

2. Solve the two equations for y

11 additional steps

(4y-3)=(-4y+3)

Add to both sides:

(4y-3)+4y=(-4y+3)+4y

Group like terms:

(4y+4y)-3=(-4y+3)+4y

Simplify the arithmetic:

8y-3=(-4y+3)+4y

Group like terms:

8y-3=(-4y+4y)+3

Simplify the arithmetic:

8y3=3

Add to both sides:

(8y-3)+3=3+3

Simplify the arithmetic:

8y=3+3

Simplify the arithmetic:

8y=6

Divide both sides by :

(8y)8=68

Simplify the fraction:

y=68

Find the greatest common factor of the numerator and denominator:

y=(3·2)(4·2)

Factor out and cancel the greatest common factor:

y=34

5 additional steps

(4y-3)=-(-4y+3)

Expand the parentheses:

(4y-3)=4y-3

Subtract from both sides:

(4y-3)-4y=(4y-3)-4y

Group like terms:

(4y-4y)-3=(4y-3)-4y

Simplify the arithmetic:

-3=(4y-3)-4y

Group like terms:

-3=(4y-4y)-3

Simplify the arithmetic:

3=3

3. List the solutions

y=34,-3
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4y3|
y=|4y+3|
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.