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Solution - Absolute value equations

Exact form:

x = - 3, 1

x=-3 , 1

See steps

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
$| x | = | y |$ → $x = \pm y$ and $| x | = | y |$ → $\pm x = y$
to write all four options of the equation
$| 6 x | = | 3 x - 9 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 6 x \| = \| 3 x - 9 \|$
$x = + y$	$(6 x) = (3 x - 9)$
$x = - y$	$(6 x) = - (3 x - 9)$
$+ x = y$	$(6 x) = (3 x - 9)$
$- x = y$	$- (6 x) = (3 x - 9)$

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:

$\| x \| = \| y \|$	$\| 6 x \| = \| 3 x - 9 \|$
$x = + y$ , $+ x = y$	$(6 x) = (3 x - 9)$
$x = - y$ , $- x = y$	$(6 x) = - (3 x - 9)$

2. Solve the two equations for x

7 additional steps

$6 x = (3 x - 9)$

Subtract from both sides:

$(6 x) - 3 x = (3 x - 9) - 3 x$

Simplify the arithmetic:

$3 x = (3 x - 9) - 3 x$

Group like terms:

$3 x = (3 x - 3 x) - 9$

Simplify the arithmetic:

$3 x = - 9$

Divide both sides by :

$\frac{(3 x)}{3} = \frac{- 9}{3}$

Simplify the fraction:

$x = \frac{- 9}{3}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(- 3 \cdot 3)}{(1 \cdot 3)}$

Factor out and cancel the greatest common factor:

$x = - 3$

7 additional steps

$6 x = - (3 x - 9)$

Expand the parentheses:

$6 x = - 3 x + 9$

Add to both sides:

$(6 x) + 3 x = (- 3 x + 9) + 3 x$

Simplify the arithmetic:

$9 x = (- 3 x + 9) + 3 x$

Group like terms:

$9 x = (- 3 x + 3 x) + 9$

Simplify the arithmetic:

$9 x = 9$

Divide both sides by :

$\frac{(9 x)}{9} = \frac{9}{9}$

Simplify the fraction:

$x = \frac{9}{9}$

Simplify the fraction:

$x = 1$

3. List the solutions

$x = - 3, 1$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 6 x |$
$y = | 3 x - 9 |$
The equation is true where the two lines cross.

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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.

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

2. Solve the two equations for x

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

2. Solve the two equations for x

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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