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Solution - Absolute value equations

Exact form:

x = - 13, - 1

x=-13 , -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
$| 2 x + 5 | = \frac{1}{2} | 3 x - 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 2 x + 5 \| = \frac{1}{2} \| 3 x - 3 \|$
$x = + y$	$(2 x + 5) = \frac{1}{2} (3 x - 3)$
$x = - y$	$(2 x + 5) = \frac{1}{2} (- (3 x - 3))$
$+ x = y$	$(2 x + 5) = \frac{1}{2} (3 x - 3)$
$- x = y$	$- (2 x + 5) = \frac{1}{2} (3 x - 3)$

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:

$\| x \| = \| y \|$	$\| 2 x + 5 \| = \frac{1}{2} \| 3 x - 3 \|$
$x = + y$ , $+ x = y$	$(2 x + 5) = \frac{1}{2} (3 x - 3)$
$x = - y$ , $- x = y$	$(2 x + 5) = \frac{1}{2} (- (3 x - 3))$

2. Solve the two equations for x

23 additional steps

$(2 x + 5) = \frac{1}{2} \cdot (3 x - 3)$

Multiply the fraction(s):

$(2 x + 5) = \frac{(1 \cdot (3 x - 3))}{2}$

Break up the fraction:

$(2 x + 5) = \frac{3 x}{2} + \frac{- 3}{2}$

Subtract from both sides:

$(2 x + 5) - \frac{3 x}{2} = (\frac{3 x}{2} + \frac{- 3}{2}) - \frac{3 x}{2}$

Group like terms:

$(2 x + \frac{- 3}{2} x) + 5 = (\frac{3 x}{2} + \frac{- 3}{2}) - \frac{3 x}{2}$

Group the coefficients:

$(2 + \frac{- 3}{2}) x + 5 = (\frac{3 x}{2} + \frac{- 3}{2}) - \frac{3 x}{2}$

Convert the integer into a fraction:

$(\frac{4}{2} + \frac{- 3}{2}) x + 5 = (\frac{3 x}{2} + \frac{- 3}{2}) - \frac{3 x}{2}$

Combine the fractions:

$\frac{(4 - 3)}{2} x + 5 = (\frac{3 x}{2} + \frac{- 3}{2}) - \frac{3 x}{2}$

Combine the numerators:

$\frac{1}{2} x + 5 = (\frac{3 x}{2} + \frac{- 3}{2}) - \frac{3 x}{2}$

Group like terms:

$\frac{1}{2} \cdot x + 5 = (\frac{3 x}{2} + \frac{- 3}{2} x) + \frac{- 3}{2}$

Combine the fractions:

$\frac{1}{2} \cdot x + 5 = \frac{(3 - 3)}{2} x + \frac{- 3}{2}$

Combine the numerators:

$\frac{1}{2} \cdot x + 5 = \frac{0}{2} x + \frac{- 3}{2}$

Reduce the zero numerator:

$\frac{1}{2} x + 5 = 0 x + \frac{- 3}{2}$

Simplify the arithmetic:

$\frac{1}{2} x + 5 = \frac{- 3}{2}$

Subtract from both sides:

$(\frac{1}{2} x + 5) - 5 = (\frac{- 3}{2}) - 5$

Simplify the arithmetic:

$\frac{1}{2} x = (\frac{- 3}{2}) - 5$

Convert the integer into a fraction:

$\frac{1}{2} x = \frac{- 3}{2} + \frac{- 10}{2}$

Combine the fractions:

$\frac{1}{2} x = \frac{(- 3 - 10)}{2}$

Combine the numerators:

$\frac{1}{2} x = \frac{- 13}{2}$

Multiply both sides by inverse fraction :

$(\frac{1}{2} x) \cdot \frac{2}{1} = (\frac{- 13}{2}) \cdot \frac{2}{1}$

Group like terms:

$(\frac{1}{2} \cdot 2) x = (\frac{- 13}{2}) \cdot \frac{2}{1}$

Multiply the coefficients:

$\frac{(1 \cdot 2)}{2} x = (\frac{- 13}{2}) \cdot \frac{2}{1}$

Simplify the fraction:

$x = (\frac{- 13}{2}) \cdot \frac{2}{1}$

Multiply the fraction(s):

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

Simplify the arithmetic:

$x = - 13$

24 additional steps

$(2 x + 5) = \frac{1}{2} \cdot (- (3 x - 3))$

Multiply the fraction(s):

$(2 x + 5) = \frac{(1 \cdot (- (3 x - 3)))}{2}$

Expand the parentheses:

$(2 x + 5) = \frac{(- 3 x + 3)}{2}$

Break up the fraction:

$(2 x + 5) = \frac{- 3 x}{2} + \frac{3}{2}$

Add to both sides:

$(2 x + 5) + \frac{3}{2} \cdot x = (\frac{- 3 x}{2} + \frac{3}{2}) + \frac{3}{2} x$

Group like terms:

$(2 x + \frac{3}{2} \cdot x) + 5 = (\frac{- 3 x}{2} + \frac{3}{2}) + \frac{3}{2} x$

Group the coefficients:

$(2 + \frac{3}{2}) x + 5 = (\frac{- 3 x}{2} + \frac{3}{2}) + \frac{3}{2} x$

Convert the integer into a fraction:

$(\frac{4}{2} + \frac{3}{2}) x + 5 = (\frac{- 3 x}{2} + \frac{3}{2}) + \frac{3}{2} x$

Combine the fractions:

$\frac{(4 + 3)}{2} \cdot x + 5 = (\frac{- 3 x}{2} + \frac{3}{2}) + \frac{3}{2} x$

Combine the numerators:

$\frac{7}{2} \cdot x + 5 = (\frac{- 3 x}{2} + \frac{3}{2}) + \frac{3}{2} x$

Group like terms:

$\frac{7}{2} \cdot x + 5 = (\frac{- 3 x}{2} + \frac{3}{2} x) + \frac{3}{2}$

Combine the fractions:

$\frac{7}{2} \cdot x + 5 = \frac{(- 3 + 3)}{2} x + \frac{3}{2}$

Combine the numerators:

$\frac{7}{2} \cdot x + 5 = \frac{0}{2} x + \frac{3}{2}$

Reduce the zero numerator:

$\frac{7}{2} x + 5 = 0 x + \frac{3}{2}$

Simplify the arithmetic:

$\frac{7}{2} x + 5 = \frac{3}{2}$

Subtract from both sides:

$(\frac{7}{2} x + 5) - 5 = (\frac{3}{2}) - 5$

Simplify the arithmetic:

$\frac{7}{2} x = (\frac{3}{2}) - 5$

Convert the integer into a fraction:

$\frac{7}{2} x = \frac{3}{2} + \frac{- 10}{2}$

Combine the fractions:

$\frac{7}{2} x = \frac{(3 - 10)}{2}$

Combine the numerators:

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

Multiply both sides by inverse fraction :

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

Group like terms:

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

Multiply the coefficients:

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

Simplify the fraction:

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

Multiply the fraction(s):

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

Simplify the arithmetic:

$x = - 1$

3. List the solutions

$x = - 13, - 1$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 2 x + 5 |$
$y = \frac{1}{2} | 3 x - 3 |$
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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