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Solution - Absolute value equations

Exact form: k=1611,0
k=\frac{16}{11} , 0
Mixed number form: k=1511,0
k=1\frac{5}{11} , 0
Decimal form: k=1.455,0
k=1.455 , 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
2|6k4|=|k+8|
without the absolute value bars:

|x|=|y|2|6k4|=|k+8|
x=+y2(6k4)=(k+8)
x=y2(6k4)=(k+8)
+x=y2(6k4)=(k+8)
x=y2((6k4))=(k+8)

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|2|6k4|=|k+8|
x=+y , +x=y2(6k4)=(k+8)
x=y , x=y2(6k4)=(k+8)

2. Solve the two equations for k

12 additional steps

2·(6k-4)=(k+8)

Expand the parentheses:

2·6k+2·-4=(k+8)

Multiply the coefficients:

12k+2·-4=(k+8)

Simplify the arithmetic:

12k-8=(k+8)

Subtract from both sides:

(12k-8)-k=(k+8)-k

Group like terms:

(12k-k)-8=(k+8)-k

Simplify the arithmetic:

11k-8=(k+8)-k

Group like terms:

11k-8=(k-k)+8

Simplify the arithmetic:

11k8=8

Add to both sides:

(11k-8)+8=8+8

Simplify the arithmetic:

11k=8+8

Simplify the arithmetic:

11k=16

Divide both sides by :

(11k)11=1611

Simplify the fraction:

k=1611

12 additional steps

2·(6k-4)=-(k+8)

Expand the parentheses:

2·6k+2·-4=-(k+8)

Multiply the coefficients:

12k+2·-4=-(k+8)

Simplify the arithmetic:

12k-8=-(k+8)

Expand the parentheses:

12k8=k8

Add to both sides:

(12k-8)+k=(-k-8)+k

Group like terms:

(12k+k)-8=(-k-8)+k

Simplify the arithmetic:

13k-8=(-k-8)+k

Group like terms:

13k-8=(-k+k)-8

Simplify the arithmetic:

13k8=8

Add to both sides:

(13k-8)+8=-8+8

Simplify the arithmetic:

13k=8+8

Simplify the arithmetic:

13k=0

Divide both sides by the coefficient:

k=0

3. List the solutions

k=1611,0
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=2|6k4|
y=|k+8|
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.