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Solution - Absolute value equations

Exact form:

x = \frac{7}{3}, \frac{4}{3}

x=\frac{7}{3} , \frac{4}{3}

Mixed number form:

x = 2 \frac{1}{3}, 1 \frac{1}{3}

x=2\frac{1}{3} , 1\frac{1}{3}

Decimal form:

x = 2.333, 1.333

x=2.333 , 1.333

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
$| 3 x - 5 | = | x - \frac{1}{3} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 3 x - 5 \| = \| x - \frac{1}{3} \|$
$x = + y$	$(3 x - 5) = (x - \frac{1}{3})$
$x = - y$	$(3 x - 5) = - (x - \frac{1}{3})$
$+ x = y$	$(3 x - 5) = (x - \frac{1}{3})$
$- x = y$	$- (3 x - 5) = (x - \frac{1}{3})$

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:

$\| x \| = \| y \|$	$\| 3 x - 5 \| = \| x - \frac{1}{3} \|$
$x = + y$ , $+ x = y$	$(3 x - 5) = (x - \frac{1}{3})$
$x = - y$ , $- x = y$	$(3 x - 5) = - (x - \frac{1}{3})$

2. Solve the two equations for x

13 additional steps

$(3 x - 5) = (x + \frac{- 1}{3})$

Subtract from both sides:

$(3 x - 5) - x = (x + \frac{- 1}{3}) - x$

Group like terms:

$(3 x - x) - 5 = (x + \frac{- 1}{3}) - x$

Simplify the arithmetic:

$2 x - 5 = (x + \frac{- 1}{3}) - x$

Group like terms:

$2 x - 5 = (x - x) + \frac{- 1}{3}$

Simplify the arithmetic:

$2 x - 5 = \frac{- 1}{3}$

Add to both sides:

$(2 x - 5) + 5 = (\frac{- 1}{3}) + 5$

Simplify the arithmetic:

$2 x = (\frac{- 1}{3}) + 5$

Convert the integer into a fraction:

$2 x = \frac{- 1}{3} + \frac{15}{3}$

Combine the fractions:

$2 x = \frac{(- 1 + 15)}{3}$

Combine the numerators:

$2 x = \frac{14}{3}$

Divide both sides by :

$\frac{(2 x)}{2} = \frac{(\frac{14}{3})}{2}$

Simplify the fraction:

$x = \frac{(\frac{14}{3})}{2}$

Simplify the arithmetic:

$x = \frac{14}{(3 \cdot 2)}$

$x = \frac{7}{3}$

14 additional steps

$(3 x - 5) = - (x + \frac{- 1}{3})$

Expand the parentheses:

$(3 x - 5) = - x + \frac{1}{3}$

Add to both sides:

$(3 x - 5) + x = (- x + \frac{1}{3}) + x$

Group like terms:

$(3 x + x) - 5 = (- x + \frac{1}{3}) + x$

Simplify the arithmetic:

$4 x - 5 = (- x + \frac{1}{3}) + x$

Group like terms:

$4 x - 5 = (- x + x) + \frac{1}{3}$

Simplify the arithmetic:

$4 x - 5 = \frac{1}{3}$

Add to both sides:

$(4 x - 5) + 5 = (\frac{1}{3}) + 5$

Simplify the arithmetic:

$4 x = (\frac{1}{3}) + 5$

Convert the integer into a fraction:

$4 x = \frac{1}{3} + \frac{15}{3}$

Combine the fractions:

$4 x = \frac{(1 + 15)}{3}$

Combine the numerators:

$4 x = \frac{16}{3}$

Divide both sides by :

$\frac{(4 x)}{4} = \frac{(\frac{16}{3})}{4}$

Simplify the fraction:

$x = \frac{(\frac{16}{3})}{4}$

Simplify the arithmetic:

$x = \frac{16}{(3 \cdot 4)}$

$x = \frac{4}{3}$

3. List the solutions

$x = \frac{7}{3}, \frac{4}{3}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 3 x - 5 |$
$y = | x - \frac{1}{3} |$
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 x

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 x

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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