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Solution - Absolute value equations

Exact form:

y = \frac{140}{3}, \frac{140}{27}

y=\frac{140}{3} , \frac{140}{27}

Mixed number form:

y = 46 \frac{2}{3}, 5 \frac{5}{27}

y=46\frac{2}{3} , 5\frac{5}{27}

Decimal form:

y = 46.667, 5.185

y=46.667 , 5.185

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
$| \frac{3}{5} y | = | \frac{3}{4} y - 7 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{3}{5} y \| = \| \frac{3}{4} y - 7 \|$
$x = + y$	$(\frac{3}{5} y) = (\frac{3}{4} y - 7)$
$x = - y$	$(\frac{3}{5} y) = - (\frac{3}{4} y - 7)$
$+ x = y$	$(\frac{3}{5} y) = (\frac{3}{4} y - 7)$
$- x = y$	$- (\frac{3}{5} y) = (\frac{3}{4} y - 7)$

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:

$\| x \| = \| y \|$	$\| \frac{3}{5} y \| = \| \frac{3}{4} y - 7 \|$
$x = + y$ , $+ x = y$	$(\frac{3}{5} y) = (\frac{3}{4} y - 7)$
$x = - y$ , $- x = y$	$(\frac{3}{5} y) = - (\frac{3}{4} y - 7)$

2. Solve the two equations for y

20 additional steps

$\frac{3}{5} \cdot y = (\frac{3}{4} y - 7)$

Subtract from both sides:

$(\frac{3}{5} y) - \frac{3}{4} \cdot y = (\frac{3}{4} y - 7) - \frac{3}{4} y$

Group the coefficients:

$(\frac{3}{5} + \frac{- 3}{4}) y = (\frac{3}{4} \cdot y - 7) - \frac{3}{4} y$

Find the lowest common denominator:

$(\frac{(3 \cdot 4)}{(5 \cdot 4)} + \frac{(- 3 \cdot 5)}{(4 \cdot 5)}) y = (\frac{3}{4} \cdot y - 7) - \frac{3}{4} y$

Multiply the denominators:

$(\frac{(3 \cdot 4)}{20} + \frac{(- 3 \cdot 5)}{20}) y = (\frac{3}{4} \cdot y - 7) - \frac{3}{4} y$

Multiply the numerators:

$(\frac{12}{20} + \frac{- 15}{20}) y = (\frac{3}{4} \cdot y - 7) - \frac{3}{4} y$

Combine the fractions:

$\frac{(12 - 15)}{20} \cdot y = (\frac{3}{4} \cdot y - 7) - \frac{3}{4} y$

Combine the numerators:

$\frac{- 3}{20} \cdot y = (\frac{3}{4} \cdot y - 7) - \frac{3}{4} y$

Group like terms:

$\frac{- 3}{20} \cdot y = (\frac{3}{4} \cdot y + \frac{- 3}{4} y) - 7$

Combine the fractions:

$\frac{- 3}{20} \cdot y = \frac{(3 - 3)}{4} y - 7$

Combine the numerators:

$\frac{- 3}{20} \cdot y = \frac{0}{4} y - 7$

Reduce the zero numerator:

$\frac{- 3}{20} y = 0 y - 7$

Simplify the arithmetic:

$\frac{- 3}{20} y = - 7$

Multiply both sides by inverse fraction :

$(\frac{- 3}{20} y) \cdot \frac{20}{- 3} = - 7 \cdot \frac{20}{- 3}$

Move the negative sign from the denominator to the numerator:

$\frac{- 3}{20} y \cdot \frac{- 20}{3} = - 7 \cdot \frac{20}{- 3}$

Group like terms:

$(\frac{- 3}{20} \cdot \frac{- 20}{3}) y = - 7 \cdot \frac{20}{- 3}$

Multiply the coefficients:

$\frac{(- 3 \cdot - 20)}{(20 \cdot 3)} y = - 7 \cdot \frac{20}{- 3}$

Simplify the arithmetic:

$1 y = - 7 \cdot \frac{20}{- 3}$

$y = - 7 \cdot \frac{20}{- 3}$

Move the negative sign from the denominator to the numerator:

$y = - 7 \cdot \frac{- 20}{3}$

Multiply the fraction(s):

$y = \frac{(- 7 \cdot - 20)}{3}$

Simplify the arithmetic:

$y = \frac{140}{3}$

18 additional steps

$\frac{3}{5} y = - (\frac{3}{4} y - 7)$

Expand the parentheses:

$\frac{3}{5} \cdot y = \frac{- 3}{4} y + 7$

Add to both sides:

$(\frac{3}{5} y) + \frac{3}{4} \cdot y = (\frac{- 3}{4} y + 7) + \frac{3}{4} y$

Group the coefficients:

$(\frac{3}{5} + \frac{3}{4}) y = (\frac{- 3}{4} \cdot y + 7) + \frac{3}{4} y$

Find the lowest common denominator:

$(\frac{(3 \cdot 4)}{(5 \cdot 4)} + \frac{(3 \cdot 5)}{(4 \cdot 5)}) y = (\frac{- 3}{4} \cdot y + 7) + \frac{3}{4} y$

Multiply the denominators:

$(\frac{(3 \cdot 4)}{20} + \frac{(3 \cdot 5)}{20}) y = (\frac{- 3}{4} \cdot y + 7) + \frac{3}{4} y$

Multiply the numerators:

$(\frac{12}{20} + \frac{15}{20}) y = (\frac{- 3}{4} \cdot y + 7) + \frac{3}{4} y$

Combine the fractions:

$\frac{(12 + 15)}{20} \cdot y = (\frac{- 3}{4} \cdot y + 7) + \frac{3}{4} y$

Combine the numerators:

$\frac{27}{20} \cdot y = (\frac{- 3}{4} \cdot y + 7) + \frac{3}{4} y$

Group like terms:

$\frac{27}{20} \cdot y = (\frac{- 3}{4} \cdot y + \frac{3}{4} y) + 7$

Combine the fractions:

$\frac{27}{20} \cdot y = \frac{(- 3 + 3)}{4} y + 7$

Combine the numerators:

$\frac{27}{20} \cdot y = \frac{0}{4} y + 7$

Reduce the zero numerator:

$\frac{27}{20} y = 0 y + 7$

Simplify the arithmetic:

$\frac{27}{20} y = 7$

Multiply both sides by inverse fraction :

$(\frac{27}{20} y) \cdot \frac{20}{27} = 7 \cdot \frac{20}{27}$

Group like terms:

$(\frac{27}{20} \cdot \frac{20}{27}) y = 7 \cdot \frac{20}{27}$

Multiply the coefficients:

$\frac{(27 \cdot 20)}{(20 \cdot 27)} y = 7 \cdot \frac{20}{27}$

Simplify the fraction:

$y = 7 \cdot \frac{20}{27}$

Multiply the fraction(s):

$y = \frac{(7 \cdot 20)}{27}$

Simplify the arithmetic:

$y = \frac{140}{27}$

3. List the solutions

$y = \frac{140}{3}, \frac{140}{27}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{3}{5} y |$
$y = | \frac{3}{4} y - 7 |$
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 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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