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Solution - Absolute value equations

Exact form:

y = 150, - \frac{10}{9}

y=150 , -\frac{10}{9}

Mixed number form:

y = 150, - 1 \frac{1}{9}

y=150 , -1\frac{1}{9}

Decimal form:

y = 150, - 1.111

y=150 , -1.111

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
$| \frac{1}{2} y - 7 | = | \frac{2}{5} y + 8 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} y - 7 \| = \| \frac{2}{5} y + 8 \|$
$x = + y$	$(\frac{1}{2} y - 7) = (\frac{2}{5} y + 8)$
$x = - y$	$(\frac{1}{2} y - 7) = - (\frac{2}{5} y + 8)$
$+ x = y$	$(\frac{1}{2} y - 7) = (\frac{2}{5} y + 8)$
$- x = y$	$- (\frac{1}{2} y - 7) = (\frac{2}{5} y + 8)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} y - 7 \| = \| \frac{2}{5} y + 8 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} y - 7) = (\frac{2}{5} y + 8)$
$x = - y$ , $- x = y$	$(\frac{1}{2} y - 7) = - (\frac{2}{5} y + 8)$

2. Solve the two equations for y

20 additional steps

$(\frac{1}{2} \cdot y - 7) = (\frac{2}{5} y + 8)$

Subtract from both sides:

$(\frac{1}{2} y - 7) - \frac{2}{5} \cdot y = (\frac{2}{5} y + 8) - \frac{2}{5} y$

Group like terms:

$(\frac{1}{2} \cdot y + \frac{- 2}{5} \cdot y) - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Group the coefficients:

$(\frac{1}{2} + \frac{- 2}{5}) y - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Find the lowest common denominator:

$(\frac{(1 \cdot 5)}{(2 \cdot 5)} + \frac{(- 2 \cdot 2)}{(5 \cdot 2)}) y - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Multiply the denominators:

$(\frac{(1 \cdot 5)}{10} + \frac{(- 2 \cdot 2)}{10}) y - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Multiply the numerators:

$(\frac{5}{10} + \frac{- 4}{10}) y - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Combine the fractions:

$\frac{(5 - 4)}{10} \cdot y - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Combine the numerators:

$\frac{1}{10} \cdot y - 7 = (\frac{2}{5} \cdot y + 8) - \frac{2}{5} y$

Group like terms:

$\frac{1}{10} \cdot y - 7 = (\frac{2}{5} \cdot y + \frac{- 2}{5} y) + 8$

Combine the fractions:

$\frac{1}{10} \cdot y - 7 = \frac{(2 - 2)}{5} y + 8$

Combine the numerators:

$\frac{1}{10} \cdot y - 7 = \frac{0}{5} y + 8$

Reduce the zero numerator:

$\frac{1}{10} y - 7 = 0 y + 8$

Simplify the arithmetic:

$\frac{1}{10} y - 7 = 8$

Add to both sides:

$(\frac{1}{10} y - 7) + 7 = 8 + 7$

Simplify the arithmetic:

$\frac{1}{10} y = 8 + 7$

Simplify the arithmetic:

$\frac{1}{10} y = 15$

Multiply both sides by inverse fraction :

$(\frac{1}{10} y) \cdot \frac{10}{1} = 15 \cdot \frac{10}{1}$

Group like terms:

$(\frac{1}{10} \cdot 10) y = 15 \cdot \frac{10}{1}$

Multiply the coefficients:

$\frac{(1 \cdot 10)}{10} y = 15 \cdot \frac{10}{1}$

Simplify the fraction:

$y = 15 \cdot \frac{10}{1}$

Simplify the arithmetic:

$y = 150$

21 additional steps

$(\frac{1}{2} y - 7) = - (\frac{2}{5} y + 8)$

Expand the parentheses:

$(\frac{1}{2} \cdot y - 7) = \frac{- 2}{5} y - 8$

Add to both sides:

$(\frac{1}{2} y - 7) + \frac{2}{5} \cdot y = (\frac{- 2}{5} y - 8) + \frac{2}{5} y$

Group like terms:

$(\frac{1}{2} \cdot y + \frac{2}{5} \cdot y) - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Group the coefficients:

$(\frac{1}{2} + \frac{2}{5}) y - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Find the lowest common denominator:

$(\frac{(1 \cdot 5)}{(2 \cdot 5)} + \frac{(2 \cdot 2)}{(5 \cdot 2)}) y - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Multiply the denominators:

$(\frac{(1 \cdot 5)}{10} + \frac{(2 \cdot 2)}{10}) y - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Multiply the numerators:

$(\frac{5}{10} + \frac{4}{10}) y - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Combine the fractions:

$\frac{(5 + 4)}{10} \cdot y - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Combine the numerators:

$\frac{9}{10} \cdot y - 7 = (\frac{- 2}{5} \cdot y - 8) + \frac{2}{5} y$

Group like terms:

$\frac{9}{10} \cdot y - 7 = (\frac{- 2}{5} \cdot y + \frac{2}{5} y) - 8$

Combine the fractions:

$\frac{9}{10} \cdot y - 7 = \frac{(- 2 + 2)}{5} y - 8$

Combine the numerators:

$\frac{9}{10} \cdot y - 7 = \frac{0}{5} y - 8$

Reduce the zero numerator:

$\frac{9}{10} y - 7 = 0 y - 8$

Simplify the arithmetic:

$\frac{9}{10} y - 7 = - 8$

Add to both sides:

$(\frac{9}{10} y - 7) + 7 = - 8 + 7$

Simplify the arithmetic:

$\frac{9}{10} y = - 8 + 7$

Simplify the arithmetic:

$\frac{9}{10} y = - 1$

Multiply both sides by inverse fraction :

$(\frac{9}{10} y) \cdot \frac{10}{9} = - 1 \cdot \frac{10}{9}$

Group like terms:

$(\frac{9}{10} \cdot \frac{10}{9}) y = - 1 \cdot \frac{10}{9}$

Multiply the coefficients:

$\frac{(9 \cdot 10)}{(10 \cdot 9)} y = - 1 \cdot \frac{10}{9}$

Simplify the fraction:

$y = - 1 \cdot \frac{10}{9}$

Remove the one(s):

$y = \frac{- 10}{9}$

3. List the solutions

$y = 150, - \frac{10}{9}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} y - 7 |$
$y = | \frac{2}{5} y + 8 |$
The equation is true where the two lines cross.

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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 y

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 y

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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