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Solution - Absolute value equations

Exact form:

x = \frac{12}{5}, \frac{15}{4}

x=\frac{12}{5} , \frac{15}{4}

Mixed number form:

x = 2 \frac{2}{5}, 3 \frac{3}{4}

x=2\frac{2}{5} , 3\frac{3}{4}

Decimal form:

x = 2.4, 3.75

x=2.4 , 3.75

See steps

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

$9 | x - 3 | + | x + 3 | = 0$

Add $- | x + 3 |$ to both sides of the equation:

$9 | x - 3 | + | x + 3 | - | x + 3 | = - | x + 3 |$

Simplify the arithmetic

$9 | x - 3 | = - | x + 3 |$

2. 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
$9 | x - 3 | = - | x + 3 |$
without the absolute value bars:

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

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:

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

3. Solve the two equations for x

14 additional steps

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

Expand the parentheses:

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

Simplify the arithmetic:

$9 x - 27 = - (x + 3)$

Expand the parentheses:

$9 x - 27 = - x - 3$

Add to both sides:

$(9 x - 27) + x = (- x - 3) + x$

Group like terms:

$(9 x + x) - 27 = (- x - 3) + x$

Simplify the arithmetic:

$10 x - 27 = (- x - 3) + x$

Group like terms:

$10 x - 27 = (- x + x) - 3$

Simplify the arithmetic:

$10 x - 27 = - 3$

Add to both sides:

$(10 x - 27) + 27 = - 3 + 27$

Simplify the arithmetic:

$10 x = - 3 + 27$

Simplify the arithmetic:

$10 x = 24$

Divide both sides by :

$\frac{(10 x)}{10} = \frac{24}{10}$

Simplify the fraction:

$x = \frac{24}{10}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(12 \cdot 2)}{(5 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = \frac{12}{5}$

14 additional steps

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

Expand the parentheses:

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

Simplify the arithmetic:

$9 x - 27 = - (- (x + 3))$

Resolve the double minus:

$9 x - 27 = x + 3$

Subtract from both sides:

$(9 x - 27) - x = (x + 3) - x$

Group like terms:

$(9 x - x) - 27 = (x + 3) - x$

Simplify the arithmetic:

$8 x - 27 = (x + 3) - x$

Group like terms:

$8 x - 27 = (x - x) + 3$

Simplify the arithmetic:

$8 x - 27 = 3$

Add to both sides:

$(8 x - 27) + 27 = 3 + 27$

Simplify the arithmetic:

$8 x = 3 + 27$

Simplify the arithmetic:

$8 x = 30$

Divide both sides by :

$\frac{(8 x)}{8} = \frac{30}{8}$

Simplify the fraction:

$x = \frac{30}{8}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(15 \cdot 2)}{(4 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = \frac{15}{4}$

4. List the solutions

$x = \frac{12}{5}, \frac{15}{4}$
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
$y = 9 | x - 3 |$
$y = - | x + 3 |$
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 with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

Why learn this

Terms and topics

Related links

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved