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Solution - Absolute value equations

Exact form: k=1,-513
k=1 , -\frac{5}{13}
Decimal form: k=1,0.385
k=1 , -0.385

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
|6k+3|=|7k+2|
without the absolute value bars:

|x|=|y||6k+3|=|7k+2|
x=+y(6k+3)=(7k+2)
x=y(6k+3)=(7k+2)
+x=y(6k+3)=(7k+2)
x=y(6k+3)=(7k+2)

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||6k+3|=|7k+2|
x=+y , +x=y(6k+3)=(7k+2)
x=y , x=y(6k+3)=(7k+2)

2. Solve the two equations for k

10 additional steps

(6k+3)=(7k+2)

Subtract from both sides:

(6k+3)-7k=(7k+2)-7k

Group like terms:

(6k-7k)+3=(7k+2)-7k

Simplify the arithmetic:

-k+3=(7k+2)-7k

Group like terms:

-k+3=(7k-7k)+2

Simplify the arithmetic:

k+3=2

Subtract from both sides:

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

Simplify the arithmetic:

k=23

Simplify the arithmetic:

k=1

Multiply both sides by :

-k·-1=-1·-1

Remove the one(s):

k=-1·-1

Simplify the arithmetic:

k=1

10 additional steps

(6k+3)=-(7k+2)

Expand the parentheses:

(6k+3)=-7k-2

Add to both sides:

(6k+3)+7k=(-7k-2)+7k

Group like terms:

(6k+7k)+3=(-7k-2)+7k

Simplify the arithmetic:

13k+3=(-7k-2)+7k

Group like terms:

13k+3=(-7k+7k)-2

Simplify the arithmetic:

13k+3=2

Subtract from both sides:

(13k+3)-3=-2-3

Simplify the arithmetic:

13k=23

Simplify the arithmetic:

13k=5

Divide both sides by :

(13k)13=-513

Simplify the fraction:

k=-513

3. List the solutions

k=1,-513
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|6k+3|
y=|7k+2|
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.