Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

x = 12

x=12

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
$| \frac{1}{4} x + 1 | = | \frac{1}{4} x - 7 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{4} x + 1 \| = \| \frac{1}{4} x - 7 \|$
$x = + y$	$(\frac{1}{4} x + 1) = (\frac{1}{4} x - 7)$
$x = - y$	$(\frac{1}{4} x + 1) = - (\frac{1}{4} x - 7)$
$+ x = y$	$(\frac{1}{4} x + 1) = (\frac{1}{4} x - 7)$
$- x = y$	$- (\frac{1}{4} x + 1) = (\frac{1}{4} x - 7)$

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:

$\| x \| = \| y \|$	$\| \frac{1}{4} x + 1 \| = \| \frac{1}{4} x - 7 \|$
$x = + y$ , $+ x = y$	$(\frac{1}{4} x + 1) = (\frac{1}{4} x - 7)$
$x = - y$ , $- x = y$	$(\frac{1}{4} x + 1) = - (\frac{1}{4} x - 7)$

2. Solve the two equations for x

11 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

$\frac{0}{4} \cdot x + 1 = (\frac{1}{4} \cdot x - 7) - \frac{1}{4} x$

Reduce the zero numerator:

$0 x + 1 = (\frac{1}{4} \cdot x - 7) - \frac{1}{4} x$

Simplify the arithmetic:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

$1 = \frac{0}{4} x - 7$

Reduce the zero numerator:

$1 = 0 x - 7$

Simplify the arithmetic:

$1 = - 7$

The statement is false:

$1 = - 7$

The equation is false so it has no solution.

19 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

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

Group like terms:

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

Combine the fractions:

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

Combine the numerators:

$\frac{1}{2} \cdot x + 1 = \frac{0}{4} x + 7$

Reduce the zero numerator:

$\frac{1}{2} x + 1 = 0 x + 7$

Simplify the arithmetic:

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

Subtract from both sides:

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

Simplify the arithmetic:

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

Simplify the arithmetic:

$\frac{1}{2} x = 6$

Multiply both sides by inverse fraction :

$(\frac{1}{2} x) \cdot \frac{2}{1} = 6 \cdot \frac{2}{1}$

Group like terms:

$(\frac{1}{2} \cdot 2) x = 6 \cdot \frac{2}{1}$

Multiply the coefficients:

$\frac{(1 \cdot 2)}{2} x = 6 \cdot \frac{2}{1}$

Simplify the fraction:

$x = 6 \cdot \frac{2}{1}$

Simplify the arithmetic:

$x = 12$

3. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{4} x + 1 |$
$y = | \frac{1}{4} x - 7 |$
The equation is true where the two lines cross.

How did we do?

Please leave us feedback.

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