Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

x = - \frac{16}{69}, \frac{32}{123}

x=-\frac{16}{69} , \frac{32}{123}

Decimal form:

x = - 0.232, 0.260

x=-0.232 , 0.260

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
$| \frac{9}{4} x - 2 | = | 8 x - \frac{2}{3} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{9}{4} x - 2 \| = \| 8 x - \frac{2}{3} \|$
$x = + y$	$(\frac{9}{4} x - 2) = (8 x - \frac{2}{3})$
$x = - y$	$(\frac{9}{4} x - 2) = - (8 x - \frac{2}{3})$
$+ x = y$	$(\frac{9}{4} x - 2) = (8 x - \frac{2}{3})$
$- x = y$	$- (\frac{9}{4} x - 2) = (8 x - \frac{2}{3})$

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:

$\| x \| = \| y \|$	$\| \frac{9}{4} x - 2 \| = \| 8 x - \frac{2}{3} \|$
$x = + y$ , $+ x = y$	$(\frac{9}{4} x - 2) = (8 x - \frac{2}{3})$
$x = - y$ , $- x = y$	$(\frac{9}{4} x - 2) = - (8 x - \frac{2}{3})$

2. Solve the two equations for x

22 additional steps

$(\frac{9}{4} x - 2) = (8 x + \frac{- 2}{3})$

Subtract from both sides:

$(\frac{9}{4} x - 2) - 8 x = (8 x + \frac{- 2}{3}) - 8 x$

Group like terms:

$(\frac{9}{4} x - 8 x) - 2 = (8 x + \frac{- 2}{3}) - 8 x$

Group the coefficients:

$(\frac{9}{4} - 8) x - 2 = (8 x + \frac{- 2}{3}) - 8 x$

Convert the integer into a fraction:

$(\frac{9}{4} + \frac{- 32}{4}) x - 2 = (8 x + \frac{- 2}{3}) - 8 x$

Combine the fractions:

$\frac{(9 - 32)}{4} x - 2 = (8 x + \frac{- 2}{3}) - 8 x$

Combine the numerators:

$\frac{- 23}{4} x - 2 = (8 x + \frac{- 2}{3}) - 8 x$

Group like terms:

$\frac{- 23}{4} x - 2 = (8 x - 8 x) + \frac{- 2}{3}$

Simplify the arithmetic:

$\frac{- 23}{4} x - 2 = \frac{- 2}{3}$

Add to both sides:

$(\frac{- 23}{4} x - 2) + 2 = (\frac{- 2}{3}) + 2$

Simplify the arithmetic:

$\frac{- 23}{4} x = (\frac{- 2}{3}) + 2$

Convert the integer into a fraction:

$\frac{- 23}{4} x = \frac{- 2}{3} + \frac{6}{3}$

Combine the fractions:

$\frac{- 23}{4} x = \frac{(- 2 + 6)}{3}$

Combine the numerators:

$\frac{- 23}{4} x = \frac{4}{3}$

Multiply both sides by inverse fraction :

$(\frac{- 23}{4} x) \cdot \frac{4}{- 23} = (\frac{4}{3}) \cdot \frac{4}{- 23}$

Move the negative sign from the denominator to the numerator:

$\frac{- 23}{4} x \cdot \frac{- 4}{23} = (\frac{4}{3}) \cdot \frac{4}{- 23}$

Group like terms:

$(\frac{- 23}{4} \cdot \frac{- 4}{23}) x = (\frac{4}{3}) \cdot \frac{4}{- 23}$

Multiply the coefficients:

$\frac{(- 23 \cdot - 4)}{(4 \cdot 23)} x = (\frac{4}{3}) \cdot \frac{4}{- 23}$

Simplify the arithmetic:

$1 x = (\frac{4}{3}) \cdot \frac{4}{- 23}$

$x = (\frac{4}{3}) \cdot \frac{4}{- 23}$

Move the negative sign from the denominator to the numerator:

$x = \frac{4}{3} \cdot \frac{- 4}{23}$

Multiply the fraction(s):

$x = \frac{(4 \cdot - 4)}{(3 \cdot 23)}$

Simplify the arithmetic:

$x = \frac{- 16}{(3 \cdot 23)}$

$x = \frac{- 16}{69}$

20 additional steps

$(\frac{9}{4} x - 2) = - (8 x + \frac{- 2}{3})$

Expand the parentheses:

$(\frac{9}{4} x - 2) = - 8 x + \frac{2}{3}$

Add to both sides:

$(\frac{9}{4} x - 2) + 8 x = (- 8 x + \frac{2}{3}) + 8 x$

Group like terms:

$(\frac{9}{4} x + 8 x) - 2 = (- 8 x + \frac{2}{3}) + 8 x$

Group the coefficients:

$(\frac{9}{4} + 8) x - 2 = (- 8 x + \frac{2}{3}) + 8 x$

Convert the integer into a fraction:

$(\frac{9}{4} + \frac{32}{4}) x - 2 = (- 8 x + \frac{2}{3}) + 8 x$

Combine the fractions:

$\frac{(9 + 32)}{4} x - 2 = (- 8 x + \frac{2}{3}) + 8 x$

Combine the numerators:

$\frac{41}{4} x - 2 = (- 8 x + \frac{2}{3}) + 8 x$

Group like terms:

$\frac{41}{4} x - 2 = (- 8 x + 8 x) + \frac{2}{3}$

Simplify the arithmetic:

$\frac{41}{4} x - 2 = \frac{2}{3}$

Add to both sides:

$(\frac{41}{4} x - 2) + 2 = (\frac{2}{3}) + 2$

Simplify the arithmetic:

$\frac{41}{4} x = (\frac{2}{3}) + 2$

Convert the integer into a fraction:

$\frac{41}{4} x = \frac{2}{3} + \frac{6}{3}$

Combine the fractions:

$\frac{41}{4} x = \frac{(2 + 6)}{3}$

Combine the numerators:

$\frac{41}{4} x = \frac{8}{3}$

Multiply both sides by inverse fraction :

$(\frac{41}{4} x) \cdot \frac{4}{41} = (\frac{8}{3}) \cdot \frac{4}{41}$

Group like terms:

$(\frac{41}{4} \cdot \frac{4}{41}) x = (\frac{8}{3}) \cdot \frac{4}{41}$

Multiply the coefficients:

$\frac{(41 \cdot 4)}{(4 \cdot 41)} x = (\frac{8}{3}) \cdot \frac{4}{41}$

Simplify the fraction:

$x = (\frac{8}{3}) \cdot \frac{4}{41}$

Multiply the fraction(s):

$x = \frac{(8 \cdot 4)}{(3 \cdot 41)}$

Simplify the arithmetic:

$x = \frac{32}{(3 \cdot 41)}$

$x = \frac{32}{123}$

3. List the solutions

$x = - \frac{16}{69}, \frac{32}{123}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{9}{4} x - 2 |$
$y = | 8 x - \frac{2}{3} |$
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 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