Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

z = - \frac{12}{7}, - \frac{96}{13}

z=-\frac{12}{7} , -\frac{96}{13}

Mixed number form:

z = - 1 \frac{5}{7}, - 7 \frac{5}{13}

z=-1\frac{5}{7} , -7\frac{5}{13}

Decimal form:

z = - 1.714, - 7.385

z=-1.714 , -7.385

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
$| \frac{1}{2} z + 7 | = | \frac{5}{3} z + 9 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} z + 7 \| = \| \frac{5}{3} z + 9 \|$
$x = + y$	$(\frac{1}{2} z + 7) = (\frac{5}{3} z + 9)$
$x = - y$	$(\frac{1}{2} z + 7) = - (\frac{5}{3} z + 9)$
$+ x = y$	$(\frac{1}{2} z + 7) = (\frac{5}{3} z + 9)$
$- x = y$	$- (\frac{1}{2} z + 7) = (\frac{5}{3} z + 9)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} z + 7 \| = \| \frac{5}{3} z + 9 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} z + 7) = (\frac{5}{3} z + 9)$
$x = - y$ , $- x = y$	$(\frac{1}{2} z + 7) = - (\frac{5}{3} z + 9)$

2. Solve the two equations for z

24 additional steps

$(\frac{1}{2} \cdot z + 7) = (\frac{5}{3} z + 9)$

Subtract from both sides:

$(\frac{1}{2} z + 7) - \frac{5}{3} \cdot z = (\frac{5}{3} z + 9) - \frac{5}{3} z$

Group like terms:

$(\frac{1}{2} \cdot z + \frac{- 5}{3} \cdot z) + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Group the coefficients:

$(\frac{1}{2} + \frac{- 5}{3}) z + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Find the lowest common denominator:

$(\frac{(1 \cdot 3)}{(2 \cdot 3)} + \frac{(- 5 \cdot 2)}{(3 \cdot 2)}) z + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Multiply the denominators:

$(\frac{(1 \cdot 3)}{6} + \frac{(- 5 \cdot 2)}{6}) z + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Multiply the numerators:

$(\frac{3}{6} + \frac{- 10}{6}) z + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Combine the fractions:

$\frac{(3 - 10)}{6} \cdot z + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Combine the numerators:

$\frac{- 7}{6} \cdot z + 7 = (\frac{5}{3} \cdot z + 9) - \frac{5}{3} z$

Group like terms:

$\frac{- 7}{6} \cdot z + 7 = (\frac{5}{3} \cdot z + \frac{- 5}{3} z) + 9$

Combine the fractions:

$\frac{- 7}{6} \cdot z + 7 = \frac{(5 - 5)}{3} z + 9$

Combine the numerators:

$\frac{- 7}{6} \cdot z + 7 = \frac{0}{3} z + 9$

Reduce the zero numerator:

$\frac{- 7}{6} z + 7 = 0 z + 9$

Simplify the arithmetic:

$\frac{- 7}{6} z + 7 = 9$

Subtract from both sides:

$(\frac{- 7}{6} z + 7) - 7 = 9 - 7$

Simplify the arithmetic:

$\frac{- 7}{6} z = 9 - 7$

Simplify the arithmetic:

$\frac{- 7}{6} z = 2$

Multiply both sides by inverse fraction :

$(\frac{- 7}{6} z) \cdot \frac{6}{- 7} = 2 \cdot \frac{6}{- 7}$

Move the negative sign from the denominator to the numerator:

$\frac{- 7}{6} z \cdot \frac{- 6}{7} = 2 \cdot \frac{6}{- 7}$

Group like terms:

$(\frac{- 7}{6} \cdot \frac{- 6}{7}) z = 2 \cdot \frac{6}{- 7}$

Multiply the coefficients:

$\frac{(- 7 \cdot - 6)}{(6 \cdot 7)} z = 2 \cdot \frac{6}{- 7}$

Simplify the arithmetic:

$1 z = 2 \cdot \frac{6}{- 7}$

$z = 2 \cdot \frac{6}{- 7}$

Move the negative sign from the denominator to the numerator:

$z = 2 \cdot \frac{- 6}{7}$

Multiply the fraction(s):

$z = \frac{(2 \cdot - 6)}{7}$

Simplify the arithmetic:

$z = \frac{- 12}{7}$

22 additional steps

$(\frac{1}{2} z + 7) = - (\frac{5}{3} z + 9)$

Expand the parentheses:

$(\frac{1}{2} \cdot z + 7) = \frac{- 5}{3} z - 9$

Add to both sides:

$(\frac{1}{2} z + 7) + \frac{5}{3} \cdot z = (\frac{- 5}{3} z - 9) + \frac{5}{3} z$

Group like terms:

$(\frac{1}{2} \cdot z + \frac{5}{3} \cdot z) + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Group the coefficients:

$(\frac{1}{2} + \frac{5}{3}) z + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Find the lowest common denominator:

$(\frac{(1 \cdot 3)}{(2 \cdot 3)} + \frac{(5 \cdot 2)}{(3 \cdot 2)}) z + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Multiply the denominators:

$(\frac{(1 \cdot 3)}{6} + \frac{(5 \cdot 2)}{6}) z + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Multiply the numerators:

$(\frac{3}{6} + \frac{10}{6}) z + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Combine the fractions:

$\frac{(3 + 10)}{6} \cdot z + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Combine the numerators:

$\frac{13}{6} \cdot z + 7 = (\frac{- 5}{3} \cdot z - 9) + \frac{5}{3} z$

Group like terms:

$\frac{13}{6} \cdot z + 7 = (\frac{- 5}{3} \cdot z + \frac{5}{3} z) - 9$

Combine the fractions:

$\frac{13}{6} \cdot z + 7 = \frac{(- 5 + 5)}{3} z - 9$

Combine the numerators:

$\frac{13}{6} \cdot z + 7 = \frac{0}{3} z - 9$

Reduce the zero numerator:

$\frac{13}{6} z + 7 = 0 z - 9$

Simplify the arithmetic:

$\frac{13}{6} z + 7 = - 9$

Subtract from both sides:

$(\frac{13}{6} z + 7) - 7 = - 9 - 7$

Simplify the arithmetic:

$\frac{13}{6} z = - 9 - 7$

Simplify the arithmetic:

$\frac{13}{6} z = - 16$

Multiply both sides by inverse fraction :

$(\frac{13}{6} z) \cdot \frac{6}{13} = - 16 \cdot \frac{6}{13}$

Group like terms:

$(\frac{13}{6} \cdot \frac{6}{13}) z = - 16 \cdot \frac{6}{13}$

Multiply the coefficients:

$\frac{(13 \cdot 6)}{(6 \cdot 13)} z = - 16 \cdot \frac{6}{13}$

Simplify the fraction:

$z = - 16 \cdot \frac{6}{13}$

Multiply the fraction(s):

$z = \frac{(- 16 \cdot 6)}{13}$

Simplify the arithmetic:

$z = \frac{- 96}{13}$

3. List the solutions

$z = - \frac{12}{7}, - \frac{96}{13}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} z + 7 |$
$y = | \frac{5}{3} z + 9 |$
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 z

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 z

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved