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Solution - Absolute value equations

Exact form:

x = - 6, \frac{4}{5}

x=-6 , \frac{4}{5}

Decimal form:

x = - 6, 0.8

x=-6 , 0.8

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
$| 4 x - 10 | = 2 | 3 x + 1 |$
without the absolute value bars:

$\| x \| = \| y \|$	$\| 4 x - 10 \| = 2 \| 3 x + 1 \|$
$x = + y$	$(4 x - 10) = 2 (3 x + 1)$
$x = - y$	$(4 x - 10) = 2 (- (3 x + 1))$
$+ x = y$	$(4 x - 10) = 2 (3 x + 1)$
$- x = y$	$- (4 x - 10) = 2 (3 x + 1)$

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:

$\| x \| = \| y \|$	$\| 4 x - 10 \| = 2 \| 3 x + 1 \|$
$x = + y$ , $+ x = y$	$(4 x - 10) = 2 (3 x + 1)$
$x = - y$ , $- x = y$	$(4 x - 10) = 2 (- (3 x + 1))$

2. Solve the two equations for x

16 additional steps

$(4 x - 10) = 2 \cdot (3 x + 1)$

Expand the parentheses:

$(4 x - 10) = 2 \cdot 3 x + 2 \cdot 1$

Multiply the coefficients:

$(4 x - 10) = 6 x + 2 \cdot 1$

Simplify the arithmetic:

$(4 x - 10) = 6 x + 2$

Subtract from both sides:

$(4 x - 10) - 6 x = (6 x + 2) - 6 x$

Group like terms:

$(4 x - 6 x) - 10 = (6 x + 2) - 6 x$

Simplify the arithmetic:

$- 2 x - 10 = (6 x + 2) - 6 x$

Group like terms:

$- 2 x - 10 = (6 x - 6 x) + 2$

Simplify the arithmetic:

$- 2 x - 10 = 2$

Add to both sides:

$(- 2 x - 10) + 10 = 2 + 10$

Simplify the arithmetic:

$- 2 x = 2 + 10$

Simplify the arithmetic:

$- 2 x = 12$

Divide both sides by :

$\frac{(- 2 x)}{- 2} = \frac{12}{- 2}$

Cancel out the negatives:

$\frac{2 x}{2} = \frac{12}{- 2}$

Simplify the fraction:

$x = \frac{12}{- 2}$

Move the negative sign from the denominator to the numerator:

$x = \frac{- 12}{2}$

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$x = - 6$

15 additional steps

$(4 x - 10) = 2 \cdot (- (3 x + 1))$

Expand the parentheses:

$(4 x - 10) = 2 \cdot (- 3 x - 1)$

Expand the parentheses:

$(4 x - 10) = 2 \cdot - 3 x + 2 \cdot - 1$

Multiply the coefficients:

$(4 x - 10) = - 6 x + 2 \cdot - 1$

Simplify the arithmetic:

$(4 x - 10) = - 6 x - 2$

Add to both sides:

$(4 x - 10) + 6 x = (- 6 x - 2) + 6 x$

Group like terms:

$(4 x + 6 x) - 10 = (- 6 x - 2) + 6 x$

Simplify the arithmetic:

$10 x - 10 = (- 6 x - 2) + 6 x$

Group like terms:

$10 x - 10 = (- 6 x + 6 x) - 2$

Simplify the arithmetic:

$10 x - 10 = - 2$

Add to both sides:

$(10 x - 10) + 10 = - 2 + 10$

Simplify the arithmetic:

$10 x = - 2 + 10$

Simplify the arithmetic:

$10 x = 8$

Divide both sides by :

$\frac{(10 x)}{10} = \frac{8}{10}$

Simplify the fraction:

$x = \frac{8}{10}$

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$x = \frac{4}{5}$

3. List the solutions

$x = - 6, \frac{4}{5}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = | 4 x - 10 |$
$y = 2 | 3 x + 1 |$
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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