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Solution - Absolute value equations

Exact form: y=12,413
y=12 , \frac{4}{13}
Decimal form: y=12,0.308
y=12 , 0.308

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
|7y8|=|6y+4|
without the absolute value bars:

|x|=|y||7y8|=|6y+4|
x=+y(7y8)=(6y+4)
x=y(7y8)=(6y+4)
+x=y(7y8)=(6y+4)
x=y(7y8)=(6y+4)

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||7y8|=|6y+4|
x=+y , +x=y(7y8)=(6y+4)
x=y , x=y(7y8)=(6y+4)

2. Solve the two equations for y

7 additional steps

(7y-8)=(6y+4)

Subtract from both sides:

(7y-8)-6y=(6y+4)-6y

Group like terms:

(7y-6y)-8=(6y+4)-6y

Simplify the arithmetic:

y-8=(6y+4)-6y

Group like terms:

y-8=(6y-6y)+4

Simplify the arithmetic:

y8=4

Add to both sides:

(y-8)+8=4+8

Simplify the arithmetic:

y=4+8

Simplify the arithmetic:

y=12

10 additional steps

(7y-8)=-(6y+4)

Expand the parentheses:

(7y-8)=-6y-4

Add to both sides:

(7y-8)+6y=(-6y-4)+6y

Group like terms:

(7y+6y)-8=(-6y-4)+6y

Simplify the arithmetic:

13y-8=(-6y-4)+6y

Group like terms:

13y-8=(-6y+6y)-4

Simplify the arithmetic:

13y8=4

Add to both sides:

(13y-8)+8=-4+8

Simplify the arithmetic:

13y=4+8

Simplify the arithmetic:

13y=4

Divide both sides by :

(13y)13=413

Simplify the fraction:

y=413

3. List the solutions

y=12,413
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|7y8|
y=|6y+4|
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.