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Solution - Absolute value equations

Exact form:

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

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

Mixed number form:

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

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

Decimal form:

x = 3, 1.5

x=3 , 1.5

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

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

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:

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

2. Solve the two equations for x

18 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Group the coefficients:

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

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

Combine the fractions:

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

Combine the numerators:

$\frac{2}{3} \cdot x - 2 = \frac{0}{3} x$

Reduce the zero numerator:

$\frac{2}{3} x - 2 = 0 x$

Simplify the arithmetic:

$\frac{2}{3} x - 2 = 0$

Add to both sides:

$(\frac{2}{3} x - 2) + 2 = 0 + 2$

Simplify the arithmetic:

$\frac{2}{3} x = 0 + 2$

Simplify the arithmetic:

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

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

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

Multiply the fraction(s):

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

Simplify the arithmetic:

$x = 3$

17 additional steps

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

Add to both sides:

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

Simplify the arithmetic:

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

Add to both sides:

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

Group the coefficients:

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

Convert the integer into a fraction:

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

Combine the fractions:

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

Combine the numerators:

$\frac{4}{3} \cdot x = (\frac{- 1}{3} \cdot x + 2) + \frac{1}{3} x$

Group like terms:

$\frac{4}{3} \cdot x = (\frac{- 1}{3} \cdot x + \frac{1}{3} x) + 2$

Combine the fractions:

$\frac{4}{3} \cdot x = \frac{(- 1 + 1)}{3} x + 2$

Combine the numerators:

$\frac{4}{3} \cdot x = \frac{0}{3} x + 2$

Reduce the zero numerator:

$\frac{4}{3} x = 0 x + 2$

Simplify the arithmetic:

$\frac{4}{3} x = 2$

Multiply both sides by inverse fraction :

$(\frac{4}{3} x) \cdot \frac{3}{4} = 2 \cdot \frac{3}{4}$

Group like terms:

$(\frac{4}{3} \cdot \frac{3}{4}) x = 2 \cdot \frac{3}{4}$

Multiply the coefficients:

$\frac{(4 \cdot 3)}{(3 \cdot 4)} x = 2 \cdot \frac{3}{4}$

Simplify the fraction:

$x = 2 \cdot \frac{3}{4}$

Multiply the fraction(s):

$x = \frac{(2 \cdot 3)}{4}$

Simplify the arithmetic:

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

3. List the solutions

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

4. Graph

Each line represents the function of one side of the equation:
$y = | x - 2 |$
$y = | \frac{1}{3} x |$
The equation is true where the two lines cross.

How did we do?

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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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