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Solution - Absolute value equations

Exact form:

x = - 3, \frac{1}{13}

x=-3 , \frac{1}{13}

Decimal form:

x = - 3, 0.077

x=-3 , 0.077

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
$4 | 2 x + 1 | = 5 | x - 1 |$
without the absolute value bars:

$\| x \| = \| y \|$	$4 \| 2 x + 1 \| = 5 \| x - 1 \|$
$x = + y$	$4 (2 x + 1) = 5 (x - 1)$
$x = - y$	$4 (2 x + 1) = 5 (- (x - 1))$
$+ x = y$	$4 (2 x + 1) = 5 (x - 1)$
$- x = y$	$4 (- (2 x + 1)) = 5 (x - 1)$

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:

$\| x \| = \| y \|$	$4 \| 2 x + 1 \| = 5 \| x - 1 \|$
$x = + y$ , $+ x = y$	$4 (2 x + 1) = 5 (x - 1)$
$x = - y$ , $- x = y$	$4 (2 x + 1) = 5 (- (x - 1))$

2. Solve the two equations for x

16 additional steps

$4 \cdot (2 x + 1) = 5 \cdot (x - 1)$

Expand the parentheses:

$4 \cdot 2 x + 4 \cdot 1 = 5 \cdot (x - 1)$

Multiply the coefficients:

$8 x + 4 \cdot 1 = 5 \cdot (x - 1)$

Simplify the arithmetic:

$8 x + 4 = 5 \cdot (x - 1)$

Expand the parentheses:

$8 x + 4 = 5 x + 5 \cdot - 1$

Simplify the arithmetic:

$8 x + 4 = 5 x - 5$

Subtract from both sides:

$(8 x + 4) - 5 x = (5 x - 5) - 5 x$

Group like terms:

$(8 x - 5 x) + 4 = (5 x - 5) - 5 x$

Simplify the arithmetic:

$3 x + 4 = (5 x - 5) - 5 x$

Group like terms:

$3 x + 4 = (5 x - 5 x) - 5$

Simplify the arithmetic:

$3 x + 4 = - 5$

Subtract from both sides:

$(3 x + 4) - 4 = - 5 - 4$

Simplify the arithmetic:

$3 x = - 5 - 4$

Simplify the arithmetic:

$3 x = - 9$

Divide both sides by :

$\frac{(3 x)}{3} = \frac{- 9}{3}$

Simplify the fraction:

$x = \frac{- 9}{3}$

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$x = - 3$

17 additional steps

$4 \cdot (2 x + 1) = 5 \cdot (- (x - 1))$

Expand the parentheses:

$4 \cdot 2 x + 4 \cdot 1 = 5 \cdot (- (x - 1))$

Multiply the coefficients:

$8 x + 4 \cdot 1 = 5 \cdot (- (x - 1))$

Simplify the arithmetic:

$8 x + 4 = 5 \cdot (- (x - 1))$

Expand the parentheses:

$8 x + 4 = 5 \cdot (- x + 1)$

$8 x + 4 = 5 \cdot - x + 5 \cdot 1$

Group like terms:

$8 x + 4 = (5 \cdot - 1) x + 5 \cdot 1$

Multiply the coefficients:

$8 x + 4 = - 5 x + 5 \cdot 1$

Simplify the arithmetic:

$8 x + 4 = - 5 x + 5$

Add to both sides:

$(8 x + 4) + 5 x = (- 5 x + 5) + 5 x$

Group like terms:

$(8 x + 5 x) + 4 = (- 5 x + 5) + 5 x$

Simplify the arithmetic:

$13 x + 4 = (- 5 x + 5) + 5 x$

Group like terms:

$13 x + 4 = (- 5 x + 5 x) + 5$

Simplify the arithmetic:

$13 x + 4 = 5$

Subtract from both sides:

$(13 x + 4) - 4 = 5 - 4$

Simplify the arithmetic:

$13 x = 5 - 4$

Simplify the arithmetic:

$13 x = 1$

Divide both sides by :

$\frac{(13 x)}{13} = \frac{1}{13}$

Simplify the fraction:

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

3. List the solutions

$x = - 3, \frac{1}{13}$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = 4 | 2 x + 1 |$
$y = 5 | x - 1 |$
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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