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Solution - Absolute value equations

Exact form:

x = \frac{24}{5}, \frac{6}{7}

x=\frac{24}{5} , \frac{6}{7}

Mixed number form:

x = 4 \frac{4}{5}, \frac{6}{7}

x=4\frac{4}{5} , \frac{6}{7}

Decimal form:

x = 4.8, 0.857

x=4.8 , 0.857

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
$| \frac{1}{3} x + 3 | = | 2 x - 5 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{3} x + 3 \| = \| 2 x - 5 \|$
$x = + y$	$(\frac{1}{3} x + 3) = (2 x - 5)$
$x = - y$	$(\frac{1}{3} x + 3) = - (2 x - 5)$
$+ x = y$	$(\frac{1}{3} x + 3) = (2 x - 5)$
$- x = y$	$- (\frac{1}{3} x + 3) = (2 x - 5)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{3} x + 3 \| = \| 2 x - 5 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{3} x + 3) = (2 x - 5)$
$x = - y$ , $- x = y$	$(\frac{1}{3} x + 3) = - (2 x - 5)$

2. Solve the two equations for x

19 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Group the coefficients:

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

Convert the integer into a fraction:

$(\frac{1}{3} + \frac{- 6}{3}) x + 3 = (2 x - 5) - 2 x$

Combine the fractions:

$\frac{(1 - 6)}{3} x + 3 = (2 x - 5) - 2 x$

Combine the numerators:

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

Group like terms:

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

Simplify the arithmetic:

$\frac{- 5}{3} x + 3 = - 5$

Subtract from both sides:

$(\frac{- 5}{3} x + 3) - 3 = - 5 - 3$

Simplify the arithmetic:

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

Simplify the arithmetic:

$\frac{- 5}{3} x = - 8$

Multiply both sides by inverse fraction :

$(\frac{- 5}{3} x) \cdot \frac{3}{- 5} = - 8 \cdot \frac{3}{- 5}$

Move the negative sign from the denominator to the numerator:

$\frac{- 5}{3} x \cdot \frac{- 3}{5} = - 8 \cdot \frac{3}{- 5}$

Group like terms:

$(\frac{- 5}{3} \cdot \frac{- 3}{5}) x = - 8 \cdot \frac{3}{- 5}$

Multiply the coefficients:

$\frac{(- 5 \cdot - 3)}{(3 \cdot 5)} x = - 8 \cdot \frac{3}{- 5}$

Simplify the arithmetic:

$1 x = - 8 \cdot \frac{3}{- 5}$

$x = - 8 \cdot \frac{3}{- 5}$

Move the negative sign from the denominator to the numerator:

$x = - 8 \cdot \frac{- 3}{5}$

Multiply the fraction(s):

$x = \frac{(- 8 \cdot - 3)}{5}$

Simplify the arithmetic:

$x = \frac{24}{5}$

17 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Group the coefficients:

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

Convert the integer into a fraction:

$(\frac{1}{3} + \frac{6}{3}) x + 3 = (- 2 x + 5) + 2 x$

Combine the fractions:

$\frac{(1 + 6)}{3} x + 3 = (- 2 x + 5) + 2 x$

Combine the numerators:

$\frac{7}{3} x + 3 = (- 2 x + 5) + 2 x$

Group like terms:

$\frac{7}{3} x + 3 = (- 2 x + 2 x) + 5$

Simplify the arithmetic:

$\frac{7}{3} x + 3 = 5$

Subtract from both sides:

$(\frac{7}{3} x + 3) - 3 = 5 - 3$

Simplify the arithmetic:

$\frac{7}{3} x = 5 - 3$

Simplify the arithmetic:

$\frac{7}{3} x = 2$

Multiply both sides by inverse fraction :

$(\frac{7}{3} x) \cdot \frac{3}{7} = 2 \cdot \frac{3}{7}$

Group like terms:

$(\frac{7}{3} \cdot \frac{3}{7}) x = 2 \cdot \frac{3}{7}$

Multiply the coefficients:

$\frac{(7 \cdot 3)}{(3 \cdot 7)} x = 2 \cdot \frac{3}{7}$

Simplify the fraction:

$x = 2 \cdot \frac{3}{7}$

Multiply the fraction(s):

$x = \frac{(2 \cdot 3)}{7}$

Simplify the arithmetic:

$x = \frac{6}{7}$

3. List the solutions

$x = \frac{24}{5}, \frac{6}{7}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{3} x + 3 |$
$y = | 2 x - 5 |$
The equation is true where the two lines cross.

How did we do?

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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 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

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