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Solution - Absolute value equations

Exact form:

x = 0

x=0

Decimal form:

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
$| 26 x - \frac{1}{27} | = | 26 x + \frac{1}{27} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 26 x - \frac{1}{27} \| = \| 26 x + \frac{1}{27} \|$
$x = + y$	$(26 x - \frac{1}{27}) = (26 x + \frac{1}{27})$
$x = - y$	$(26 x - \frac{1}{27}) = - (26 x + \frac{1}{27})$
$+ x = y$	$(26 x - \frac{1}{27}) = (26 x + \frac{1}{27})$
$- x = y$	$- (26 x - \frac{1}{27}) = (26 x + \frac{1}{27})$

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:

$\| x \| = \| y \|$	$\| 26 x - \frac{1}{27} \| = \| 26 x + \frac{1}{27} \|$
$x = + y$ , $+ x = y$	$(26 x - \frac{1}{27}) = (26 x + \frac{1}{27})$
$x = - y$ , $- x = y$	$(26 x - \frac{1}{27}) = - (26 x + \frac{1}{27})$

2. Solve the two equations for x

5 additional steps

$(26 x + \frac{- 1}{27}) = (26 x + \frac{1}{27})$

Subtract from both sides:

$(26 x + \frac{- 1}{27}) - 26 x = (26 x + \frac{1}{27}) - 26 x$

Group like terms:

$(26 x - 26 x) + \frac{- 1}{27} = (26 x + \frac{1}{27}) - 26 x$

Simplify the arithmetic:

$\frac{- 1}{27} = (26 x + \frac{1}{27}) - 26 x$

Group like terms:

$\frac{- 1}{27} = (26 x - 26 x) + \frac{1}{27}$

Simplify the arithmetic:

$\frac{- 1}{27} = \frac{1}{27}$

The statement is false:

$\frac{- 1}{27} = \frac{1}{27}$

The equation is false so it has no solution.

14 additional steps

$(26 x + \frac{- 1}{27}) = - (26 x + \frac{1}{27})$

Expand the parentheses:

$(26 x + \frac{- 1}{27}) = - 26 x + \frac{- 1}{27}$

Add to both sides:

$(26 x + \frac{- 1}{27}) + 26 x = (- 26 x + \frac{- 1}{27}) + 26 x$

Group like terms:

$(26 x + 26 x) + \frac{- 1}{27} = (- 26 x + \frac{- 1}{27}) + 26 x$

Simplify the arithmetic:

$52 x + \frac{- 1}{27} = (- 26 x + \frac{- 1}{27}) + 26 x$

Group like terms:

$52 x + \frac{- 1}{27} = (- 26 x + 26 x) + \frac{- 1}{27}$

Simplify the arithmetic:

$52 x + \frac{- 1}{27} = \frac{- 1}{27}$

Add to both sides:

$(52 x + \frac{- 1}{27}) + \frac{1}{27} = (\frac{- 1}{27}) + \frac{1}{27}$

Combine the fractions:

$52 x + \frac{(- 1 + 1)}{27} = (\frac{- 1}{27}) + \frac{1}{27}$

Combine the numerators:

$52 x + \frac{0}{27} = (\frac{- 1}{27}) + \frac{1}{27}$

Reduce the zero numerator:

$52 x + 0 = (\frac{- 1}{27}) + \frac{1}{27}$

Simplify the arithmetic:

$52 x = (\frac{- 1}{27}) + \frac{1}{27}$

Combine the fractions:

$52 x = \frac{(- 1 + 1)}{27}$

Combine the numerators:

$52 x = \frac{0}{27}$

Reduce the zero numerator:

$52 x = 0$

Divide both sides by the coefficient:

$x = 0$

3. Graph

Each line represents the function of one side of the equation:
$y = | 26 x - \frac{1}{27} |$
$y = | 26 x + \frac{1}{27} |$
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. 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. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved