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Solution - Absolute value equations

Exact form: y=23,-1
y=\frac{2}{3} , -1
Decimal form: y=0.667,1
y=0.667 , -1

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|4y1|+|2y3|=0

Add |2y3| to both sides of the equation:

|4y1|+|2y3||2y3|=|2y3|

Simplify the arithmetic

|4y1|=|2y3|

2. 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
|4y1|=|2y3|
without the absolute value bars:

|x|=|y||4y1|=|2y3|
x=+y(4y1)=(2y3)
x=y(4y1)=(2y3)
+x=y(4y1)=(2y3)
x=y(4y1)=(2y3)

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||4y1|=|2y3|
x=+y , +x=y(4y1)=(2y3)
x=y , x=y(4y1)=(2y3)

3. Solve the two equations for y

12 additional steps

(4y-1)=-(2y-3)

Expand the parentheses:

(4y-1)=-2y+3

Add to both sides:

(4y-1)+2y=(-2y+3)+2y

Group like terms:

(4y+2y)-1=(-2y+3)+2y

Simplify the arithmetic:

6y-1=(-2y+3)+2y

Group like terms:

6y-1=(-2y+2y)+3

Simplify the arithmetic:

6y1=3

Add to both sides:

(6y-1)+1=3+1

Simplify the arithmetic:

6y=3+1

Simplify the arithmetic:

6y=4

Divide both sides by :

(6y)6=46

Simplify the fraction:

y=46

Find the greatest common factor of the numerator and denominator:

y=(2·2)(3·2)

Factor out and cancel the greatest common factor:

y=23

11 additional steps

(4y-1)=-(-(2y-3))

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(4y-1)=2y-3

Subtract from both sides:

(4y-1)-2y=(2y-3)-2y

Group like terms:

(4y-2y)-1=(2y-3)-2y

Simplify the arithmetic:

2y-1=(2y-3)-2y

Group like terms:

2y-1=(2y-2y)-3

Simplify the arithmetic:

2y1=3

Add to both sides:

(2y-1)+1=-3+1

Simplify the arithmetic:

2y=3+1

Simplify the arithmetic:

2y=2

Divide both sides by :

(2y)2=-22

Simplify the fraction:

y=-22

Simplify the fraction:

y=1

4. List the solutions

y=23,-1
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
y=|4y1|
y=|2y3|
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.