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Solution - Absolute value equations

Exact form:

x = \frac{11}{2}, \frac{13}{6}

x=\frac{11}{2} , \frac{13}{6}

Mixed number form:

x = 5 \frac{1}{2}, 2 \frac{1}{6}

x=5\frac{1}{2} , 2\frac{1}{6}

Decimal form:

x = 5.5, 2.167

x=5.5 , 2.167

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
$| 4 x - 2 | = 8 | x - 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 4 x - 2 \| = 8 \| x - 3 \|$
$x = + y$	$(4 x - 2) = 8 (x - 3)$
$x = - y$	$(4 x - 2) = 8 (- (x - 3))$
$+ x = y$	$(4 x - 2) = 8 (x - 3)$
$- x = y$	$- (4 x - 2) = 8 (x - 3)$

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:

$\| x \| = \| y \|$	$\| 4 x - 2 \| = 8 \| x - 3 \|$
$x = + y$ , $+ x = y$	$(4 x - 2) = 8 (x - 3)$
$x = - y$ , $- x = y$	$(4 x - 2) = 8 (- (x - 3))$

2. Solve the two equations for x

15 additional steps

$(4 x - 2) = 8 \cdot (x - 3)$

Expand the parentheses:

$(4 x - 2) = 8 x + 8 \cdot - 3$

Simplify the arithmetic:

$(4 x - 2) = 8 x - 24$

Subtract from both sides:

$(4 x - 2) - 8 x = (8 x - 24) - 8 x$

Group like terms:

$(4 x - 8 x) - 2 = (8 x - 24) - 8 x$

Simplify the arithmetic:

$- 4 x - 2 = (8 x - 24) - 8 x$

Group like terms:

$- 4 x - 2 = (8 x - 8 x) - 24$

Simplify the arithmetic:

$- 4 x - 2 = - 24$

Add to both sides:

$(- 4 x - 2) + 2 = - 24 + 2$

Simplify the arithmetic:

$- 4 x = - 24 + 2$

Simplify the arithmetic:

$- 4 x = - 22$

Divide both sides by :

$\frac{(- 4 x)}{- 4} = \frac{- 22}{- 4}$

Cancel out the negatives:

$\frac{4 x}{4} = \frac{- 22}{- 4}$

Simplify the fraction:

$x = \frac{- 22}{- 4}$

Cancel out the negatives:

$x = \frac{22}{4}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(11 \cdot 2)}{(2 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = \frac{11}{2}$

16 additional steps

$(4 x - 2) = 8 \cdot (- (x - 3))$

Expand the parentheses:

$(4 x - 2) = 8 \cdot (- x + 3)$

$(4 x - 2) = 8 \cdot - x + 8 \cdot 3$

Group like terms:

$(4 x - 2) = (8 \cdot - 1) x + 8 \cdot 3$

Multiply the coefficients:

$(4 x - 2) = - 8 x + 8 \cdot 3$

Simplify the arithmetic:

$(4 x - 2) = - 8 x + 24$

Add to both sides:

$(4 x - 2) + 8 x = (- 8 x + 24) + 8 x$

Group like terms:

$(4 x + 8 x) - 2 = (- 8 x + 24) + 8 x$

Simplify the arithmetic:

$12 x - 2 = (- 8 x + 24) + 8 x$

Group like terms:

$12 x - 2 = (- 8 x + 8 x) + 24$

Simplify the arithmetic:

$12 x - 2 = 24$

Add to both sides:

$(12 x - 2) + 2 = 24 + 2$

Simplify the arithmetic:

$12 x = 24 + 2$

Simplify the arithmetic:

$12 x = 26$

Divide both sides by :

$\frac{(12 x)}{12} = \frac{26}{12}$

Simplify the fraction:

$x = \frac{26}{12}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(13 \cdot 2)}{(6 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = \frac{13}{6}$

3. List the solutions

$x = \frac{11}{2}, \frac{13}{6}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 4 x - 2 |$
$y = 8 | x - 3 |$
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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