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Solution - Absolute value equations

Exact form:

a = - \frac{9}{4}, \frac{9}{8}

a=-\frac{9}{4} , \frac{9}{8}

Mixed number form:

a = - 2 \frac{1}{4}, 1 \frac{1}{8}

a=-2\frac{1}{4} , 1\frac{1}{8}

Decimal form:

a = - 2.25, 1.125

a=-2.25 , 1.125

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
$2 | 3 a | = | 2 a - 9 |$
without the absolute value bars:

$\| x \| = \| y \|$	$2 \| 3 a \| = \| 2 a - 9 \|$
$x = + y$	$2 (3 a) = (2 a - 9)$
$x = - y$	$2 (3 a) = - (2 a - 9)$
$+ x = y$	$2 (3 a) = (2 a - 9)$
$- x = y$	$2 (- (3 a)) = (2 a - 9)$

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:

$\| x \| = \| y \|$	$2 \| 3 a \| = \| 2 a - 9 \|$
$x = + y$ , $+ x = y$	$2 (3 a) = (2 a - 9)$
$x = - y$ , $- x = y$	$2 (3 a) = - (2 a - 9)$

2. Solve the two equations for a

6 additional steps

$2 \cdot 3 a = (2 a - 9)$

Multiply the coefficients:

$6 a = (2 a - 9)$

Subtract from both sides:

$(6 a) - 2 a = (2 a - 9) - 2 a$

Simplify the arithmetic:

$4 a = (2 a - 9) - 2 a$

Group like terms:

$4 a = (2 a - 2 a) - 9$

Simplify the arithmetic:

$4 a = - 9$

Divide both sides by :

$\frac{(4 a)}{4} = \frac{- 9}{4}$

Simplify the fraction:

$a = \frac{- 9}{4}$

7 additional steps

$2 \cdot 3 a = - (2 a - 9)$

Multiply the coefficients:

$6 a = - (2 a - 9)$

Expand the parentheses:

$6 a = - 2 a + 9$

Add to both sides:

$(6 a) + 2 a = (- 2 a + 9) + 2 a$

Simplify the arithmetic:

$8 a = (- 2 a + 9) + 2 a$

Group like terms:

$8 a = (- 2 a + 2 a) + 9$

Simplify the arithmetic:

$8 a = 9$

Divide both sides by :

$\frac{(8 a)}{8} = \frac{9}{8}$

Simplify the fraction:

$a = \frac{9}{8}$

3. List the solutions

$a = - \frac{9}{4}, \frac{9}{8}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = 2 | 3 a |$
$y = | 2 a - 9 |$
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 a

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 a

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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