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Solution - Absolute value equations

Exact form: k=-23
k=-\frac{2}{3}
Decimal form: k=0.667
k=-0.667

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

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

2. Solve the two equations for k

7 additional steps

(3k-2)=3·(k+2)

Expand the parentheses:

(3k-2)=3k+3·2

Simplify the arithmetic:

(3k-2)=3k+6

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2=6

The statement is false:

2=6

The equation is false so it has no solution.

16 additional steps

(3k-2)=3·(-(k+2))

Expand the parentheses:

(3k-2)=3·(-k-2)

(3k-2)=3·-k+3·-2

Group like terms:

(3k-2)=(3·-1)k+3·-2

Multiply the coefficients:

(3k-2)=-3k+3·-2

Simplify the arithmetic:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

6k2=6

Add to both sides:

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

Simplify the arithmetic:

6k=6+2

Simplify the arithmetic:

6k=4

Divide both sides by :

(6k)6=-46

Simplify the fraction:

k=-46

Find the greatest common factor of the numerator and denominator:

k=(-2·2)(3·2)

Factor out and cancel the greatest common factor:

k=-23

3. Graph

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