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Solution - Absolute value equations

Exact form:

x = \frac{2}{7}, \frac{2}{9}

x=\frac{2}{7} , \frac{2}{9}

Decimal form:

x = 0.286, 0.222

x=0.286 , 0.222

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
$\frac{1}{2} | x | = | 4 x - 1 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\frac{1}{2} \| x \| = \| 4 x - 1 \|$
$x = + y$	$\frac{1}{2} (x) = (4 x - 1)$
$x = - y$	$\frac{1}{2} (x) = - (4 x - 1)$
$+ x = y$	$\frac{1}{2} (x) = (4 x - 1)$
$- x = y$	$\frac{1}{2} (- (x)) = (4 x - 1)$

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:

$\| x \| = \| y \|$	$\frac{1}{2} \| x \| = \| 4 x - 1 \|$
$x = + y$ , $+ x = y$	$\frac{1}{2} (x) = (4 x - 1)$
$x = - y$ , $- x = y$	$\frac{1}{2} (x) = - (4 x - 1)$

2. Solve the two equations for x

13 additional steps

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

Subtract from both sides:

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

Group the coefficients:

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

Convert the integer into a fraction:

$(\frac{1}{2} + \frac{- 8}{2}) x = (4 x - 1) - 4 x$

Combine the fractions:

$\frac{(1 - 8)}{2} x = (4 x - 1) - 4 x$

Combine the numerators:

$\frac{- 7}{2} x = (4 x - 1) - 4 x$

Group like terms:

$\frac{- 7}{2} x = (4 x - 4 x) - 1$

Simplify the arithmetic:

$\frac{- 7}{2} x = - 1$

Multiply both sides by inverse fraction :

$(\frac{- 7}{2} x) \cdot \frac{2}{- 7} = - 1 \cdot \frac{2}{- 7}$

Move the negative sign from the denominator to the numerator:

$\frac{- 7}{2} x \cdot \frac{- 2}{7} = - 1 \cdot \frac{2}{- 7}$

Group like terms:

$(\frac{- 7}{2} \cdot \frac{- 2}{7}) x = - 1 \cdot \frac{2}{- 7}$

Multiply the coefficients:

$\frac{(- 7 \cdot - 2)}{(2 \cdot 7)} x = - 1 \cdot \frac{2}{- 7}$

Simplify the arithmetic:

$1 x = - 1 \cdot \frac{2}{- 7}$

$x = - 1 \cdot \frac{2}{- 7}$

Cancel out the negatives:

$x = \frac{2}{7}$

12 additional steps

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

Expand the parentheses:

$\frac{1}{2} x = - 4 x + 1$

Add to both sides:

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

Group the coefficients:

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

Convert the integer into a fraction:

$(\frac{1}{2} + \frac{8}{2}) x = (- 4 x + 1) + 4 x$

Combine the fractions:

$\frac{(1 + 8)}{2} x = (- 4 x + 1) + 4 x$

Combine the numerators:

$\frac{9}{2} x = (- 4 x + 1) + 4 x$

Group like terms:

$\frac{9}{2} x = (- 4 x + 4 x) + 1$

Simplify the arithmetic:

$\frac{9}{2} x = 1$

Multiply both sides by inverse fraction :

$(\frac{9}{2} x) \cdot \frac{2}{9} = 1 \cdot \frac{2}{9}$

Group like terms:

$(\frac{9}{2} \cdot \frac{2}{9}) x = 1 \cdot \frac{2}{9}$

Multiply the coefficients:

$\frac{(9 \cdot 2)}{(2 \cdot 9)} x = 1 \cdot \frac{2}{9}$

Simplify the fraction:

$x = 1 \cdot \frac{2}{9}$

Remove the one(s):

$x = \frac{2}{9}$

3. List the solutions

$x = \frac{2}{7}, \frac{2}{9}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = \frac{1}{2} | x |$
$y = | 4 x - 1 |$
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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