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Solution - Absolute value equations

Exact form: k=6,-25
k=6 , -\frac{2}{5}
Decimal form: k=6,0.4
k=6 , -0.4

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
|3k2|=2|k+2|
without the absolute value bars:

|x|=|y||3k2|=2|k+2|
x=+y(3k2)=2(k+2)
x=y(3k2)=2((k+2))
+x=y(3k2)=2(k+2)
x=y(3k2)=2(k+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||3k2|=2|k+2|
x=+y , +x=y(3k2)=2(k+2)
x=y , x=y(3k2)=2((k+2))

2. Solve the two equations for k

9 additional steps

(3k-2)=2·(k+2)

Expand the parentheses:

(3k-2)=2k+2·2

Simplify the arithmetic:

(3k-2)=2k+4

Subtract from both sides:

(3k-2)-2k=(2k+4)-2k

Group like terms:

(3k-2k)-2=(2k+4)-2k

Simplify the arithmetic:

k-2=(2k+4)-2k

Group like terms:

k-2=(2k-2k)+4

Simplify the arithmetic:

k2=4

Add to both sides:

(k-2)+2=4+2

Simplify the arithmetic:

k=4+2

Simplify the arithmetic:

k=6

14 additional steps

(3k-2)=2·(-(k+2))

Expand the parentheses:

(3k-2)=2·(-k-2)

(3k-2)=2·-k+2·-2

Group like terms:

(3k-2)=(2·-1)k+2·-2

Multiply the coefficients:

(3k-2)=-2k+2·-2

Simplify the arithmetic:

(3k-2)=-2k-4

Add to both sides:

(3k-2)+2k=(-2k-4)+2k

Group like terms:

(3k+2k)-2=(-2k-4)+2k

Simplify the arithmetic:

5k-2=(-2k-4)+2k

Group like terms:

5k-2=(-2k+2k)-4

Simplify the arithmetic:

5k2=4

Add to both sides:

(5k-2)+2=-4+2

Simplify the arithmetic:

5k=4+2

Simplify the arithmetic:

5k=2

Divide both sides by :

(5k)5=-25

Simplify the fraction:

k=-25

3. List the solutions

k=6,-25
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3k2|
y=2|k+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.