Enter an equation or problem

Camera input is not recognized!

Solution - Absolute value equations

Exact form:

x = 5, 1

x=5 , 1

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
$| \frac{1}{2} x + \frac{3}{2} | = | \frac{3}{2} x - \frac{7}{2} |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{1}{2} x + \frac{3}{2} \| = \| \frac{3}{2} x - \frac{7}{2} \|$
$x = + y$	$(\frac{1}{2} x + \frac{3}{2}) = (\frac{3}{2} x - \frac{7}{2})$
$x = - y$	$(\frac{1}{2} x + \frac{3}{2}) = - (\frac{3}{2} x - \frac{7}{2})$
$+ x = y$	$(\frac{1}{2} x + \frac{3}{2}) = (\frac{3}{2} x - \frac{7}{2})$
$- x = y$	$- (\frac{1}{2} x + \frac{3}{2}) = (\frac{3}{2} x - \frac{7}{2})$

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:

$\| x \| = \| y \|$	$\| \frac{1}{2} x + \frac{3}{2} \| = \| \frac{3}{2} x - \frac{7}{2} \|$
$x = + y$ , $+ x = y$	$(\frac{1}{2} x + \frac{3}{2}) = (\frac{3}{2} x - \frac{7}{2})$
$x = - y$ , $- x = y$	$(\frac{1}{2} x + \frac{3}{2}) = - (\frac{3}{2} x - \frac{7}{2})$

2. Solve the two equations for x

23 additional steps

$(\frac{1}{2} \cdot x + \frac{3}{2}) = (\frac{3}{2} x + \frac{- 7}{2})$

Subtract from both sides:

$(\frac{1}{2} x + \frac{3}{2}) - \frac{3}{2} \cdot x = (\frac{3}{2} x + \frac{- 7}{2}) - \frac{3}{2} x$

Group like terms:

$(\frac{1}{2} \cdot x + \frac{- 3}{2} \cdot x) + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 7}{2}) - \frac{3}{2} x$

Combine the fractions:

$\frac{(1 - 3)}{2} \cdot x + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 7}{2}) - \frac{3}{2} x$

Combine the numerators:

$\frac{- 2}{2} \cdot x + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 7}{2}) - \frac{3}{2} x$

Find the greatest common factor of the numerator and denominator:

$\frac{(- 1 \cdot 2)}{(1 \cdot 2)} \cdot x + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 7}{2}) - \frac{3}{2} x$

Factor out and cancel the greatest common factor:

$- 1 x + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 7}{2}) - \frac{3}{2} x$

Simplify the arithmetic:

$- x + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 7}{2}) - \frac{3}{2} x$

Group like terms:

$- x + \frac{3}{2} = (\frac{3}{2} \cdot x + \frac{- 3}{2} x) + \frac{- 7}{2}$

Combine the fractions:

$- x + \frac{3}{2} = \frac{(3 - 3)}{2} x + \frac{- 7}{2}$

Combine the numerators:

$- x + \frac{3}{2} = \frac{0}{2} x + \frac{- 7}{2}$

Reduce the zero numerator:

$- x + \frac{3}{2} = 0 x + \frac{- 7}{2}$

Simplify the arithmetic:

$- x + \frac{3}{2} = \frac{- 7}{2}$

Subtract from both sides:

$(- x + \frac{3}{2}) - \frac{3}{2} = (\frac{- 7}{2}) - \frac{3}{2}$

Combine the fractions:

$- x + \frac{(3 - 3)}{2} = (\frac{- 7}{2}) - \frac{3}{2}$

Combine the numerators:

$- x + \frac{0}{2} = (\frac{- 7}{2}) - \frac{3}{2}$

Reduce the zero numerator:

$- x + 0 = (\frac{- 7}{2}) - \frac{3}{2}$

Simplify the arithmetic:

$- x = (\frac{- 7}{2}) - \frac{3}{2}$

Combine the fractions:

$- x = \frac{(- 7 - 3)}{2}$

Combine the numerators:

$- x = \frac{- 10}{2}$

Find the greatest common factor of the numerator and denominator:

$- x = \frac{(- 5 \cdot 2)}{(1 \cdot 2)}$

Factor out and cancel the greatest common factor:

$- x = - 5$

Multiply both sides by :

$- x \cdot - 1 = - 5 \cdot - 1$

Remove the one(s):

$x = - 5 \cdot - 1$

Simplify the arithmetic:

$x = 5$

23 additional steps

$(\frac{1}{2} x + \frac{3}{2}) = - (\frac{3}{2} x + \frac{- 7}{2})$

Expand the parentheses:

$(\frac{1}{2} \cdot x + \frac{3}{2}) = \frac{- 3}{2} x + \frac{7}{2}$

Add to both sides:

$(\frac{1}{2} x + \frac{3}{2}) + \frac{3}{2} \cdot x = (\frac{- 3}{2} x + \frac{7}{2}) + \frac{3}{2} x$

Group like terms:

$(\frac{1}{2} \cdot x + \frac{3}{2} \cdot x) + \frac{3}{2} = (\frac{- 3}{2} \cdot x + \frac{7}{2}) + \frac{3}{2} x$

Combine the fractions:

$\frac{(1 + 3)}{2} \cdot x + \frac{3}{2} = (\frac{- 3}{2} \cdot x + \frac{7}{2}) + \frac{3}{2} x$

Combine the numerators:

$\frac{4}{2} \cdot x + \frac{3}{2} = (\frac{- 3}{2} \cdot x + \frac{7}{2}) + \frac{3}{2} x$

Find the greatest common factor of the numerator and denominator:

$\frac{(2 \cdot 2)}{(1 \cdot 2)} \cdot x + \frac{3}{2} = (\frac{- 3}{2} \cdot x + \frac{7}{2}) + \frac{3}{2} x$

Factor out and cancel the greatest common factor:

$2 x + \frac{3}{2} = (\frac{- 3}{2} \cdot x + \frac{7}{2}) + \frac{3}{2} x$

Group like terms:

$2 x + \frac{3}{2} = (\frac{- 3}{2} \cdot x + \frac{3}{2} x) + \frac{7}{2}$

Combine the fractions:

$2 x + \frac{3}{2} = \frac{(- 3 + 3)}{2} x + \frac{7}{2}$

Combine the numerators:

$2 x + \frac{3}{2} = \frac{0}{2} x + \frac{7}{2}$

Reduce the zero numerator:

$2 x + \frac{3}{2} = 0 x + \frac{7}{2}$

Simplify the arithmetic:

$2 x + \frac{3}{2} = \frac{7}{2}$

Subtract from both sides:

$(2 x + \frac{3}{2}) - \frac{3}{2} = (\frac{7}{2}) - \frac{3}{2}$

Combine the fractions:

$2 x + \frac{(3 - 3)}{2} = (\frac{7}{2}) - \frac{3}{2}$

Combine the numerators:

$2 x + \frac{0}{2} = (\frac{7}{2}) - \frac{3}{2}$

Reduce the zero numerator:

$2 x + 0 = (\frac{7}{2}) - \frac{3}{2}$

Simplify the arithmetic:

$2 x = (\frac{7}{2}) - \frac{3}{2}$

Combine the fractions:

$2 x = \frac{(7 - 3)}{2}$

Combine the numerators:

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

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$2 x = 2$

Divide both sides by :

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

Simplify the fraction:

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

Simplify the fraction:

$x = 1$

3. List the solutions

$x = 5, 1$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | \frac{1}{2} x + \frac{3}{2} |$
$y = | \frac{3}{2} x - \frac{7}{2} |$
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. 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

Latest Related Drills Solved