Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

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

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

Mixed number form:

x = - 1 \frac{1}{2}, 5

x=-1\frac{1}{2} , 5

Decimal form:

x = - 1.5, 5

x=-1.5 , 5

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
$| - 6 x + 4 | = | 2 x + 16 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| - 6 x + 4 \| = \| 2 x + 16 \|$
$x = + y$	$(- 6 x + 4) = (2 x + 16)$
$x = - y$	$(- 6 x + 4) = - (2 x + 16)$
$+ x = y$	$(- 6 x + 4) = (2 x + 16)$
$- x = y$	$- (- 6 x + 4) = (2 x + 16)$

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:

$\| x \| = \| y \|$	$\| - 6 x + 4 \| = \| 2 x + 16 \|$
$x = + y$ , $+ x = y$	$(- 6 x + 4) = (2 x + 16)$
$x = - y$ , $- x = y$	$(- 6 x + 4) = - (2 x + 16)$

2. Solve the two equations for x

13 additional steps

$(- 6 x + 4) = (2 x + 16)$

Subtract from both sides:

$(- 6 x + 4) - 2 x = (2 x + 16) - 2 x$

Group like terms:

$(- 6 x - 2 x) + 4 = (2 x + 16) - 2 x$

Simplify the arithmetic:

$- 8 x + 4 = (2 x + 16) - 2 x$

Group like terms:

$- 8 x + 4 = (2 x - 2 x) + 16$

Simplify the arithmetic:

$- 8 x + 4 = 16$

Subtract from both sides:

$(- 8 x + 4) - 4 = 16 - 4$

Simplify the arithmetic:

$- 8 x = 16 - 4$

Simplify the arithmetic:

$- 8 x = 12$

Divide both sides by :

$\frac{(- 8 x)}{- 8} = \frac{12}{- 8}$

Cancel out the negatives:

$\frac{8 x}{8} = \frac{12}{- 8}$

Simplify the fraction:

$x = \frac{12}{- 8}$

Move the negative sign from the denominator to the numerator:

$x = \frac{- 12}{8}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(- 3 \cdot 4)}{(2 \cdot 4)}$

Factor out and cancel the greatest common factor:

$x = \frac{- 3}{2}$

14 additional steps

$(- 6 x + 4) = - (2 x + 16)$

Expand the parentheses:

$(- 6 x + 4) = - 2 x - 16$

Add to both sides:

$(- 6 x + 4) + 2 x = (- 2 x - 16) + 2 x$

Group like terms:

$(- 6 x + 2 x) + 4 = (- 2 x - 16) + 2 x$

Simplify the arithmetic:

$- 4 x + 4 = (- 2 x - 16) + 2 x$

Group like terms:

$- 4 x + 4 = (- 2 x + 2 x) - 16$

Simplify the arithmetic:

$- 4 x + 4 = - 16$

Subtract from both sides:

$(- 4 x + 4) - 4 = - 16 - 4$

Simplify the arithmetic:

$- 4 x = - 16 - 4$

Simplify the arithmetic:

$- 4 x = - 20$

Divide both sides by :

$\frac{(- 4 x)}{- 4} = \frac{- 20}{- 4}$

Cancel out the negatives:

$\frac{4 x}{4} = \frac{- 20}{- 4}$

Simplify the fraction:

$x = \frac{- 20}{- 4}$

Cancel out the negatives:

$x = \frac{20}{4}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(5 \cdot 4)}{(1 \cdot 4)}$

Factor out and cancel the greatest common factor:

$x = 5$

3. List the solutions

$x = - \frac{3}{2}, 5$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | - 6 x + 4 |$
$y = | 2 x + 16 |$
The equation is true where the two lines cross.

How did we do?

Please leave us feedback.

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

Latest Related Drills Solved