Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

x = - \frac{5}{12}, \frac{15}{8}

x=-\frac{5}{12} , \frac{15}{8}

Mixed number form:

x = - \frac{5}{12}, 1 \frac{7}{8}

x=-\frac{5}{12} , 1\frac{7}{8}

Decimal form:

x = - 0.417, 1.875

x=-0.417 , 1.875

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
$| \frac{1}{5} x + 1 | = | - x + \frac{1}{2} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{5} x + 1 \| = \| - x + \frac{1}{2} \|$
$x = + y$	$(\frac{1}{5} x + 1) = (- x + \frac{1}{2})$
$x = - y$	$(\frac{1}{5} x + 1) = - (- x + \frac{1}{2})$
$+ x = y$	$(\frac{1}{5} x + 1) = (- x + \frac{1}{2})$
$- x = y$	$- (\frac{1}{5} x + 1) = (- x + \frac{1}{2})$

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:

$\| x \| = \| y \|$	$\| \frac{1}{5} x + 1 \| = \| - x + \frac{1}{2} \|$
$x = + y$ , $+ x = y$	$(\frac{1}{5} x + 1) = (- x + \frac{1}{2})$
$x = - y$ , $- x = y$	$(\frac{1}{5} x + 1) = - (- x + \frac{1}{2})$

2. Solve the two equations for x

19 additional steps

$(\frac{1}{5} x + 1) = (- x + \frac{1}{2})$

Add to both sides:

$(\frac{1}{5} x + 1) + x = (- x + \frac{1}{2}) + x$

Group like terms:

$(\frac{1}{5} x + x) + 1 = (- x + \frac{1}{2}) + x$

Group the coefficients:

$(\frac{1}{5} + 1) x + 1 = (- x + \frac{1}{2}) + x$

Convert the integer into a fraction:

$(\frac{1}{5} + \frac{5}{5}) x + 1 = (- x + \frac{1}{2}) + x$

Combine the fractions:

$\frac{(1 + 5)}{5} x + 1 = (- x + \frac{1}{2}) + x$

Combine the numerators:

$\frac{6}{5} x + 1 = (- x + \frac{1}{2}) + x$

Group like terms:

$\frac{6}{5} x + 1 = (- x + x) + \frac{1}{2}$

Simplify the arithmetic:

$\frac{6}{5} x + 1 = \frac{1}{2}$

Subtract from both sides:

$(\frac{6}{5} x + 1) - 1 = (\frac{1}{2}) - 1$

Simplify the arithmetic:

$\frac{6}{5} x = (\frac{1}{2}) - 1$

Convert the integer into a fraction:

$\frac{6}{5} x = \frac{1}{2} + \frac{- 2}{2}$

Combine the fractions:

$\frac{6}{5} x = \frac{(1 - 2)}{2}$

Combine the numerators:

$\frac{6}{5} x = \frac{- 1}{2}$

Multiply both sides by inverse fraction :

$(\frac{6}{5} x) \cdot \frac{5}{6} = (\frac{- 1}{2}) \cdot \frac{5}{6}$

Group like terms:

$(\frac{6}{5} \cdot \frac{5}{6}) x = (\frac{- 1}{2}) \cdot \frac{5}{6}$

Multiply the coefficients:

$\frac{(6 \cdot 5)}{(5 \cdot 6)} x = (\frac{- 1}{2}) \cdot \frac{5}{6}$

Simplify the fraction:

$x = (\frac{- 1}{2}) \cdot \frac{5}{6}$

Multiply the fraction(s):

$x = \frac{(- 1 \cdot 5)}{(2 \cdot 6)}$

Simplify the arithmetic:

$x = \frac{- 5}{(2 \cdot 6)}$

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

23 additional steps

$(\frac{1}{5} x + 1) = - (- x + \frac{1}{2})$

Expand the parentheses:

$(\frac{1}{5} x + 1) = x + \frac{- 1}{2}$

Subtract from both sides:

$(\frac{1}{5} x + 1) - x = (x + \frac{- 1}{2}) - x$

Group like terms:

$(\frac{1}{5} x - x) + 1 = (x + \frac{- 1}{2}) - x$

Group the coefficients:

$(\frac{1}{5} - 1) x + 1 = (x + \frac{- 1}{2}) - x$

Convert the integer into a fraction:

$(\frac{1}{5} + \frac{- 5}{5}) x + 1 = (x + \frac{- 1}{2}) - x$

Combine the fractions:

$\frac{(1 - 5)}{5} x + 1 = (x + \frac{- 1}{2}) - x$

Combine the numerators:

$\frac{- 4}{5} x + 1 = (x + \frac{- 1}{2}) - x$

Group like terms:

$\frac{- 4}{5} x + 1 = (x - x) + \frac{- 1}{2}$

Simplify the arithmetic:

$\frac{- 4}{5} x + 1 = \frac{- 1}{2}$

Subtract from both sides:

$(\frac{- 4}{5} x + 1) - 1 = (\frac{- 1}{2}) - 1$

Simplify the arithmetic:

$\frac{- 4}{5} x = (\frac{- 1}{2}) - 1$

Convert the integer into a fraction:

$\frac{- 4}{5} x = \frac{- 1}{2} + \frac{- 2}{2}$

Combine the fractions:

$\frac{- 4}{5} x = \frac{(- 1 - 2)}{2}$

Combine the numerators:

$\frac{- 4}{5} x = \frac{- 3}{2}$

Multiply both sides by inverse fraction :

$(\frac{- 4}{5} x) \cdot \frac{5}{- 4} = (\frac{- 3}{2}) \cdot \frac{5}{- 4}$

Move the negative sign from the denominator to the numerator:

$\frac{- 4}{5} x \cdot \frac{- 5}{4} = (\frac{- 3}{2}) \cdot \frac{5}{- 4}$

Group like terms:

$(\frac{- 4}{5} \cdot \frac{- 5}{4}) x = (\frac{- 3}{2}) \cdot \frac{5}{- 4}$

Multiply the coefficients:

$\frac{(- 4 \cdot - 5)}{(5 \cdot 4)} x = (\frac{- 3}{2}) \cdot \frac{5}{- 4}$

Simplify the arithmetic:

$1 x = (\frac{- 3}{2}) \cdot \frac{5}{- 4}$

$x = (\frac{- 3}{2}) \cdot \frac{5}{- 4}$

Move the negative sign from the denominator to the numerator:

$x = \frac{- 3}{2} \cdot \frac{- 5}{4}$

Multiply the fraction(s):

$x = \frac{(- 3 \cdot - 5)}{(2 \cdot 4)}$

Simplify the arithmetic:

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

$x = \frac{15}{8}$

3. List the solutions

$x = - \frac{5}{12}, \frac{15}{8}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{5} x + 1 |$
$y = | - x + \frac{1}{2} |$
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