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Solution - Absolute value equations

Exact form:

x = - \frac{26}{3}, \frac{28}{9}

x=-\frac{26}{3} , \frac{28}{9}

Mixed number form:

x = - 8 \frac{2}{3}, 3 \frac{1}{9}

x=-8\frac{2}{3} , 3\frac{1}{9}

Decimal form:

x = - 8.667, 3.111

x=-8.667 , 3.111

See steps

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

$| - 2 x + \frac{1}{3} | + | x - 9 | = 0$

Add $- | x - 9 |$ to both sides of the equation:

$| - 2 x + \frac{1}{3} | + | x - 9 | - | x - 9 | = - | x - 9 |$

Simplify the arithmetic

$| - 2 x + \frac{1}{3} | = - | x - 9 |$

2. 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
$| - 2 x + \frac{1}{3} | = - | x - 9 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| - 2 x + \frac{1}{3} \| = - \| x - 9 \|$
$x = + y$	$(- 2 x + \frac{1}{3}) = - (x - 9)$
$x = - y$	$(- 2 x + \frac{1}{3}) = - - (x - 9)$
$+ x = y$	$(- 2 x + \frac{1}{3}) = - (x - 9)$
$- x = y$	$- (- 2 x + \frac{1}{3}) = - (x - 9)$

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:

$\| x \| = \| y \|$	$\| - 2 x + \frac{1}{3} \| = - \| x - 9 \|$
$x = + y$ , $+ x = y$	$(- 2 x + \frac{1}{3}) = - (x - 9)$
$x = - y$ , $- x = y$	$(- 2 x + \frac{1}{3}) = - - (x - 9)$

3. Solve the two equations for x

16 additional steps

$(- 2 x + \frac{1}{3}) = - (x - 9)$

Expand the parentheses:

$(- 2 x + \frac{1}{3}) = - x + 9$

Add to both sides:

$(- 2 x + \frac{1}{3}) + x = (- x + 9) + x$

Group like terms:

$(- 2 x + x) + \frac{1}{3} = (- x + 9) + x$

Simplify the arithmetic:

$- x + \frac{1}{3} = (- x + 9) + x$

Group like terms:

$- x + \frac{1}{3} = (- x + x) + 9$

Simplify the arithmetic:

$- x + \frac{1}{3} = 9$

Subtract from both sides:

$(- x + \frac{1}{3}) - \frac{1}{3} = 9 - \frac{1}{3}$

Combine the fractions:

$- x + \frac{(1 - 1)}{3} = 9 - \frac{1}{3}$

Combine the numerators:

$- x + \frac{0}{3} = 9 - \frac{1}{3}$

Reduce the zero numerator:

$- x + 0 = 9 - \frac{1}{3}$

Simplify the arithmetic:

$- x = 9 - \frac{1}{3}$

Convert the integer into a fraction:

$- x = \frac{27}{3} + \frac{- 1}{3}$

Combine the fractions:

$- x = \frac{(27 - 1)}{3}$

Combine the numerators:

$- x = \frac{26}{3}$

Multiply both sides by :

$- x \cdot - 1 = (\frac{26}{3}) \cdot - 1$

Remove the one(s):

$x = (\frac{26}{3}) \cdot - 1$

Remove the one(s):

$x = \frac{- 26}{3}$

18 additional steps

$(- 2 x + \frac{1}{3}) = - (- (x - 9))$

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

$(- 2 x + \frac{1}{3}) = x - 9$

Subtract from both sides:

$(- 2 x + \frac{1}{3}) - x = (x - 9) - x$

Group like terms:

$(- 2 x - x) + \frac{1}{3} = (x - 9) - x$

Simplify the arithmetic:

$- 3 x + \frac{1}{3} = (x - 9) - x$

Group like terms:

$- 3 x + \frac{1}{3} = (x - x) - 9$

Simplify the arithmetic:

$- 3 x + \frac{1}{3} = - 9$

Subtract from both sides:

$(- 3 x + \frac{1}{3}) - \frac{1}{3} = - 9 - \frac{1}{3}$

Combine the fractions:

$- 3 x + \frac{(1 - 1)}{3} = - 9 - \frac{1}{3}$

Combine the numerators:

$- 3 x + \frac{0}{3} = - 9 - \frac{1}{3}$

Reduce the zero numerator:

$- 3 x + 0 = - 9 - \frac{1}{3}$

Simplify the arithmetic:

$- 3 x = - 9 - \frac{1}{3}$

Convert the integer into a fraction:

$- 3 x = \frac{- 27}{3} + \frac{- 1}{3}$

Combine the fractions:

$- 3 x = \frac{(- 27 - 1)}{3}$

Combine the numerators:

$- 3 x = \frac{- 28}{3}$

Divide both sides by :

$\frac{(- 3 x)}{- 3} = \frac{(\frac{- 28}{3})}{- 3}$

Cancel out the negatives:

$\frac{3 x}{3} = \frac{(\frac{- 28}{3})}{- 3}$

Simplify the fraction:

$x = \frac{(\frac{- 28}{3})}{- 3}$

Simplify the arithmetic:

$x = \frac{- 28}{(3 \cdot - 3)}$

$x = \frac{28}{9}$

4. List the solutions

$x = - \frac{26}{3}, \frac{28}{9}$
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
$y = | - 2 x + \frac{1}{3} |$
$y = - | x - 9 |$
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 with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

Why learn this

Terms and topics

Related links

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved