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Solution - Absolute value equations

Exact form:

y = \frac{11}{10}, - \frac{5}{2}

y=\frac{11}{10} , -\frac{5}{2}

Mixed number form:

y = 1 \frac{1}{10}, - 2 \frac{1}{2}

y=1\frac{1}{10} , -2\frac{1}{2}

Decimal form:

y = 1.1, - 2.5

y=1.1 , -2.5

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
$| - 6 y + 3 | = 4 | y - 2 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| - 6 y + 3 \| = 4 \| y - 2 \|$
$x = + y$	$(- 6 y + 3) = 4 (y - 2)$
$x = - y$	$(- 6 y + 3) = 4 (- (y - 2))$
$+ x = y$	$(- 6 y + 3) = 4 (y - 2)$
$- x = y$	$- (- 6 y + 3) = 4 (y - 2)$

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:

$\| x \| = \| y \|$	$\| - 6 y + 3 \| = 4 \| y - 2 \|$
$x = + y$ , $+ x = y$	$(- 6 y + 3) = 4 (y - 2)$
$x = - y$ , $- x = y$	$(- 6 y + 3) = 4 (- (y - 2))$

2. Solve the two equations for y

13 additional steps

$(- 6 y + 3) = 4 \cdot (y - 2)$

Expand the parentheses:

$(- 6 y + 3) = 4 y + 4 \cdot - 2$

Simplify the arithmetic:

$(- 6 y + 3) = 4 y - 8$

Subtract from both sides:

$(- 6 y + 3) - 4 y = (4 y - 8) - 4 y$

Group like terms:

$(- 6 y - 4 y) + 3 = (4 y - 8) - 4 y$

Simplify the arithmetic:

$- 10 y + 3 = (4 y - 8) - 4 y$

Group like terms:

$- 10 y + 3 = (4 y - 4 y) - 8$

Simplify the arithmetic:

$- 10 y + 3 = - 8$

Subtract from both sides:

$(- 10 y + 3) - 3 = - 8 - 3$

Simplify the arithmetic:

$- 10 y = - 8 - 3$

Simplify the arithmetic:

$- 10 y = - 11$

Divide both sides by :

$\frac{(- 10 y)}{- 10} = \frac{- 11}{- 10}$

Cancel out the negatives:

$\frac{10 y}{10} = \frac{- 11}{- 10}$

Simplify the fraction:

$y = \frac{- 11}{- 10}$

Cancel out the negatives:

$y = \frac{11}{10}$

16 additional steps

$(- 6 y + 3) = 4 \cdot (- (y - 2))$

Expand the parentheses:

$(- 6 y + 3) = 4 \cdot (- y + 2)$

$(- 6 y + 3) = 4 \cdot - y + 4 \cdot 2$

Group like terms:

$(- 6 y + 3) = (4 \cdot - 1) y + 4 \cdot 2$

Multiply the coefficients:

$(- 6 y + 3) = - 4 y + 4 \cdot 2$

Simplify the arithmetic:

$(- 6 y + 3) = - 4 y + 8$

Add to both sides:

$(- 6 y + 3) + 4 y = (- 4 y + 8) + 4 y$

Group like terms:

$(- 6 y + 4 y) + 3 = (- 4 y + 8) + 4 y$

Simplify the arithmetic:

$- 2 y + 3 = (- 4 y + 8) + 4 y$

Group like terms:

$- 2 y + 3 = (- 4 y + 4 y) + 8$

Simplify the arithmetic:

$- 2 y + 3 = 8$

Subtract from both sides:

$(- 2 y + 3) - 3 = 8 - 3$

Simplify the arithmetic:

$- 2 y = 8 - 3$

Simplify the arithmetic:

$- 2 y = 5$

Divide both sides by :

$\frac{(- 2 y)}{- 2} = \frac{5}{- 2}$

Cancel out the negatives:

$\frac{2 y}{2} = \frac{5}{- 2}$

Simplify the fraction:

$y = \frac{5}{- 2}$

Move the negative sign from the denominator to the numerator:

$y = \frac{- 5}{2}$

3. List the solutions

$y = \frac{11}{10}, - \frac{5}{2}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | - 6 y + 3 |$
$y = 4 | y - 2 |$
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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