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Solution - Absolute value equations

Exact form:

x = 0.326, 0.278

x=0.326 , 0.278

See steps

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

$| \frac{2}{5} x | + | - 5 x + 1.5 | = 0$

Add $- | - 5 x + 1.5 |$ to both sides of the equation:

$| \frac{2}{5} x | + | - 5 x + 1.5 | - | - 5 x + 1.5 | = - | - 5 x + 1.5 |$

Simplify the arithmetic

$| \frac{2}{5} x | = - | - 5 x + 1.5 |$

2. 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
$| \frac{2}{5} x | = - | - 5 x + 1.5 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| \frac{2}{5} x \| = - \| - 5 x + 1.5 \|$
$x = + y$	$(\frac{2}{5} x) = - (- 5 x + 1.5)$
$x = - y$	$(\frac{2}{5} x) = - - (- 5 x + 1.5)$
$+ x = y$	$(\frac{2}{5} x) = - (- 5 x + 1.5)$
$- x = y$	$- (\frac{2}{5} x) = - (- 5 x + 1.5)$

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:

$\| x \| = \| y \|$	$\| \frac{2}{5} x \| = - \| - 5 x + 1.5 \|$
$x = + y$ , $+ x = y$	$(\frac{2}{5} x) = - (- 5 x + 1.5)$
$x = - y$ , $- x = y$	$(\frac{2}{5} x) = - - (- 5 x + 1.5)$

3. Solve the two equations for x

17 additional steps

$\frac{2}{5} x = - (- 5 x + 1.5)$

Expand the parentheses:

$\frac{2}{5} x = 5 x - 1.5$

Subtract from both sides:

$(\frac{2}{5} x) - 5 x = (5 x - 1.5) - 5 x$

Group the coefficients:

$(\frac{2}{5} - 5) x = (5 x - 1.5) - 5 x$

Convert the integer into a fraction:

$(\frac{2}{5} + \frac{- 25}{5}) x = (5 x - 1.5) - 5 x$

Combine the fractions:

$\frac{(2 - 25)}{5} x = (5 x - 1.5) - 5 x$

Combine the numerators:

$\frac{- 23}{5} x = (5 x - 1.5) - 5 x$

Group like terms:

$\frac{- 23}{5} x = (5 x - 5 x) - 1.5$

Simplify the arithmetic:

$\frac{- 23}{5} x = - 1.5$

Multiply both sides by inverse fraction :

$(\frac{- 23}{5} x) \cdot \frac{5}{- 23} = - 1.5 \cdot \frac{5}{- 23}$

Move the negative sign from the denominator to the numerator:

$\frac{- 23}{5} x \cdot \frac{- 5}{23} = - 1.5 \cdot \frac{5}{- 23}$

Group like terms:

$(\frac{- 23}{5} \cdot \frac{- 5}{23}) x = - 1.5 \cdot \frac{5}{- 23}$

Multiply the coefficients:

$\frac{(- 23 \cdot - 5)}{(5 \cdot 23)} x = - 1.5 \cdot \frac{5}{- 23}$

Simplify the arithmetic:

$1 x = - 1.5 \cdot \frac{5}{- 23}$

$x = - 1.5 \cdot \frac{5}{- 23}$

Move the negative sign from the denominator to the numerator:

$x = - 1.5 \cdot \frac{- 5}{23}$

Multiply the fraction(s):

$x = \frac{(- 1.5 \cdot - 5)}{23}$

Simplify the arithmetic:

$x = \frac{7.5}{23}$

$x = 0.3261$

14 additional steps

$\frac{2}{5} x = - (- (- 5 x + 1.5))$

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

$\frac{2}{5} x = - 5 x + 1.5$

Add to both sides:

$(\frac{2}{5} x) + 5 x = (- 5 x + 1.5) + 5 x$

Group the coefficients:

$(\frac{2}{5} + 5) x = (- 5 x + 1.5) + 5 x$

Convert the integer into a fraction:

$(\frac{2}{5} + \frac{25}{5}) x = (- 5 x + 1.5) + 5 x$

Combine the fractions:

$\frac{(2 + 25)}{5} x = (- 5 x + 1.5) + 5 x$

Combine the numerators:

$\frac{27}{5} x = (- 5 x + 1.5) + 5 x$

Group like terms:

$\frac{27}{5} x = (- 5 x + 5 x) + 1.5$

Simplify the arithmetic:

$\frac{27}{5} x = 1.5$

Multiply both sides by inverse fraction :

$(\frac{27}{5} x) \cdot \frac{5}{27} = 1.5 \cdot \frac{5}{27}$

Group like terms:

$(\frac{27}{5} \cdot \frac{5}{27}) x = 1.5 \cdot \frac{5}{27}$

Multiply the coefficients:

$\frac{(27 \cdot 5)}{(5 \cdot 27)} x = 1.5 \cdot \frac{5}{27}$

Simplify the fraction:

$x = 1.5 \cdot \frac{5}{27}$

Multiply the fraction(s):

$x = \frac{(1.5 \cdot 5)}{27}$

Simplify the arithmetic:

$x = \frac{7.5}{27}$

$x = 0.2778$

4. List the solutions

$x = 0.326, 0.278$
(2 solution(s))

5. Graph

Each line represents the function of one side of the equation:
$y = | \frac{2}{5} x |$
$y = - | - 5 x + 1.5 |$
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 with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

Why learn this

Terms and topics

Related links

Solution - Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

2. Rewrite the equation without absolute value bars

3. Solve the two equations for x

4. List the solutions

5. Graph

Why learn this

Terms and topics

Related links

Latest Related Drills Solved