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Solution - Absolute value equations

Exact form:

y = 1, - \frac{1}{3}

y=1 , -\frac{1}{3}

Decimal form:

y = 1, - 0.333

y=1 , -0.333

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
$| 9 y + 1 | = | 6 y + 4 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 9 y + 1 \| = \| 6 y + 4 \|$
$x = + y$	$(9 y + 1) = (6 y + 4)$
$x = - y$	$(9 y + 1) = - (6 y + 4)$
$+ x = y$	$(9 y + 1) = (6 y + 4)$
$- x = y$	$- (9 y + 1) = (6 y + 4)$

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:

$\| x \| = \| y \|$	$\| 9 y + 1 \| = \| 6 y + 4 \|$
$x = + y$ , $+ x = y$	$(9 y + 1) = (6 y + 4)$
$x = - y$ , $- x = y$	$(9 y + 1) = - (6 y + 4)$

2. Solve the two equations for y

10 additional steps

$(9 y + 1) = (6 y + 4)$

Subtract from both sides:

$(9 y + 1) - 6 y = (6 y + 4) - 6 y$

Group like terms:

$(9 y - 6 y) + 1 = (6 y + 4) - 6 y$

Simplify the arithmetic:

$3 y + 1 = (6 y + 4) - 6 y$

Group like terms:

$3 y + 1 = (6 y - 6 y) + 4$

Simplify the arithmetic:

$3 y + 1 = 4$

Subtract from both sides:

$(3 y + 1) - 1 = 4 - 1$

Simplify the arithmetic:

$3 y = 4 - 1$

Simplify the arithmetic:

$3 y = 3$

Divide both sides by :

$\frac{(3 y)}{3} = \frac{3}{3}$

Simplify the fraction:

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

Simplify the fraction:

$y = 1$

12 additional steps

$(9 y + 1) = - (6 y + 4)$

Expand the parentheses:

$(9 y + 1) = - 6 y - 4$

Add to both sides:

$(9 y + 1) + 6 y = (- 6 y - 4) + 6 y$

Group like terms:

$(9 y + 6 y) + 1 = (- 6 y - 4) + 6 y$

Simplify the arithmetic:

$15 y + 1 = (- 6 y - 4) + 6 y$

Group like terms:

$15 y + 1 = (- 6 y + 6 y) - 4$

Simplify the arithmetic:

$15 y + 1 = - 4$

Subtract from both sides:

$(15 y + 1) - 1 = - 4 - 1$

Simplify the arithmetic:

$15 y = - 4 - 1$

Simplify the arithmetic:

$15 y = - 5$

Divide both sides by :

$\frac{(15 y)}{15} = \frac{- 5}{15}$

Simplify the fraction:

$y = \frac{- 5}{15}$

Find the greatest common factor of the numerator and denominator:

$y = \frac{(- 1 \cdot 5)}{(3 \cdot 5)}$

Factor out and cancel the greatest common factor:

$y = \frac{- 1}{3}$

3. List the solutions

$y = 1, - \frac{1}{3}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 9 y + 1 |$
$y = | 6 y + 4 |$
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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