Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

x = - \frac{1}{4}, \frac{5}{8}

x=-\frac{1}{4} , \frac{5}{8}

Decimal form:

x = - 0.25, 0.625

x=-0.25 , 0.625

See steps

Step-by-step explanation

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

$| - 6 x + 2 | + | 2 x - 3 | = 0$

Add $- | 2 x - 3 |$ to both sides of the equation:

$| - 6 x + 2 | + | 2 x - 3 | - | 2 x - 3 | = - | 2 x - 3 |$

Simplify the arithmetic

$| - 6 x + 2 | = - | 2 x - 3 |$

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
$| - 6 x + 2 | = - | 2 x - 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| - 6 x + 2 \| = - \| 2 x - 3 \|$
$x = + y$	$(- 6 x + 2) = - (2 x - 3)$
$x = - y$	$(- 6 x + 2) = - - (2 x - 3)$
$+ x = y$	$(- 6 x + 2) = - (2 x - 3)$
$- x = y$	$- (- 6 x + 2) = - (2 x - 3)$

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:

$\| x \| = \| y \|$	$\| - 6 x + 2 \| = - \| 2 x - 3 \|$
$x = + y$ , $+ x = y$	$(- 6 x + 2) = - (2 x - 3)$
$x = - y$ , $- x = y$	$(- 6 x + 2) = - - (2 x - 3)$

3. Solve the two equations for x

12 additional steps

$(- 6 x + 2) = - (2 x - 3)$

Expand the parentheses:

$(- 6 x + 2) = - 2 x + 3$

Add to both sides:

$(- 6 x + 2) + 2 x = (- 2 x + 3) + 2 x$

Group like terms:

$(- 6 x + 2 x) + 2 = (- 2 x + 3) + 2 x$

Simplify the arithmetic:

$- 4 x + 2 = (- 2 x + 3) + 2 x$

Group like terms:

$- 4 x + 2 = (- 2 x + 2 x) + 3$

Simplify the arithmetic:

$- 4 x + 2 = 3$

Subtract from both sides:

$(- 4 x + 2) - 2 = 3 - 2$

Simplify the arithmetic:

$- 4 x = 3 - 2$

Simplify the arithmetic:

$- 4 x = 1$

Divide both sides by :

$\frac{(- 4 x)}{- 4} = \frac{1}{- 4}$

Cancel out the negatives:

$\frac{4 x}{4} = \frac{1}{- 4}$

Simplify the fraction:

$x = \frac{1}{- 4}$

Move the negative sign from the denominator to the numerator:

$x = \frac{- 1}{4}$

12 additional steps

$(- 6 x + 2) = - (- (2 x - 3))$

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

$(- 6 x + 2) = 2 x - 3$

Subtract from both sides:

$(- 6 x + 2) - 2 x = (2 x - 3) - 2 x$

Group like terms:

$(- 6 x - 2 x) + 2 = (2 x - 3) - 2 x$

Simplify the arithmetic:

$- 8 x + 2 = (2 x - 3) - 2 x$

Group like terms:

$- 8 x + 2 = (2 x - 2 x) - 3$

Simplify the arithmetic:

$- 8 x + 2 = - 3$

Subtract from both sides:

$(- 8 x + 2) - 2 = - 3 - 2$

Simplify the arithmetic:

$- 8 x = - 3 - 2$

Simplify the arithmetic:

$- 8 x = - 5$

Divide both sides by :

$\frac{(- 8 x)}{- 8} = \frac{- 5}{- 8}$

Cancel out the negatives:

$\frac{8 x}{8} = \frac{- 5}{- 8}$

Simplify the fraction:

$x = \frac{- 5}{- 8}$

Cancel out the negatives:

$x = \frac{5}{8}$

4. List the solutions

$x = - \frac{1}{4}, \frac{5}{8}$
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
$y = | - 6 x + 2 |$
$y = - | 2 x - 3 |$
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 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