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Solution - Absolute value equations

Exact form: k=143,-1013
k=\frac{14}{3} , -\frac{10}{13}
Mixed number form: k=423,-1013
k=4\frac{2}{3} , -\frac{10}{13}
Decimal form: k=4.667,0.769
k=4.667 , -0.769

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
|5k+12|=2|4k1|
without the absolute value bars:

|x|=|y||5k+12|=2|4k1|
x=+y(5k+12)=2(4k1)
x=y(5k+12)=2((4k1))
+x=y(5k+12)=2(4k1)
x=y(5k+12)=2(4k1)

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||5k+12|=2|4k1|
x=+y , +x=y(5k+12)=2(4k1)
x=y , x=y(5k+12)=2((4k1))

2. Solve the two equations for k

14 additional steps

(5k+12)=2·(4k-1)

Expand the parentheses:

(5k+12)=2·4k+2·-1

Multiply the coefficients:

(5k+12)=8k+2·-1

Simplify the arithmetic:

(5k+12)=8k-2

Subtract from both sides:

(5k+12)-8k=(8k-2)-8k

Group like terms:

(5k-8k)+12=(8k-2)-8k

Simplify the arithmetic:

-3k+12=(8k-2)-8k

Group like terms:

-3k+12=(8k-8k)-2

Simplify the arithmetic:

3k+12=2

Subtract from both sides:

(-3k+12)-12=-2-12

Simplify the arithmetic:

3k=212

Simplify the arithmetic:

3k=14

Divide both sides by :

(-3k)-3=-14-3

Cancel out the negatives:

3k3=-14-3

Simplify the fraction:

k=-14-3

Cancel out the negatives:

k=143

13 additional steps

(5k+12)=2·(-(4k-1))

Expand the parentheses:

(5k+12)=2·(-4k+1)

Expand the parentheses:

(5k+12)=2·-4k+2·1

Multiply the coefficients:

(5k+12)=-8k+2·1

Simplify the arithmetic:

(5k+12)=-8k+2

Add to both sides:

(5k+12)+8k=(-8k+2)+8k

Group like terms:

(5k+8k)+12=(-8k+2)+8k

Simplify the arithmetic:

13k+12=(-8k+2)+8k

Group like terms:

13k+12=(-8k+8k)+2

Simplify the arithmetic:

13k+12=2

Subtract from both sides:

(13k+12)-12=2-12

Simplify the arithmetic:

13k=212

Simplify the arithmetic:

13k=10

Divide both sides by :

(13k)13=-1013

Simplify the fraction:

k=-1013

3. List the solutions

k=143,-1013
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5k+12|
y=2|4k1|
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.