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Solution - Absolute value equations

Exact form:

x = 28, 16

x=28 , 16

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
$2 | x - 19 | = | x - 10 |$
without the absolute value bars:

$\| x \| = \| y \|$	$2 \| x - 19 \| = \| x - 10 \|$
$x = + y$	$2 (x - 19) = (x - 10)$
$x = - y$	$2 (x - 19) = - (x - 10)$
$+ x = y$	$2 (x - 19) = (x - 10)$
$- x = y$	$2 (- (x - 19)) = (x - 10)$

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:

$\| x \| = \| y \|$	$2 \| x - 19 \| = \| x - 10 \|$
$x = + y$ , $+ x = y$	$2 (x - 19) = (x - 10)$
$x = - y$ , $- x = y$	$2 (x - 19) = - (x - 10)$

2. Solve the two equations for x

9 additional steps

$2 \cdot (x - 19) = (x - 10)$

Expand the parentheses:

$2 x + 2 \cdot - 19 = (x - 10)$

Simplify the arithmetic:

$2 x - 38 = (x - 10)$

Subtract from both sides:

$(2 x - 38) - x = (x - 10) - x$

Group like terms:

$(2 x - x) - 38 = (x - 10) - x$

Simplify the arithmetic:

$x - 38 = (x - 10) - x$

Group like terms:

$x - 38 = (x - x) - 10$

Simplify the arithmetic:

$x - 38 = - 10$

Add to both sides:

$(x - 38) + 38 = - 10 + 38$

Simplify the arithmetic:

$x = - 10 + 38$

Simplify the arithmetic:

$x = 28$

14 additional steps

$2 \cdot (x - 19) = - (x - 10)$

Expand the parentheses:

$2 x + 2 \cdot - 19 = - (x - 10)$

Simplify the arithmetic:

$2 x - 38 = - (x - 10)$

Expand the parentheses:

$2 x - 38 = - x + 10$

Add to both sides:

$(2 x - 38) + x = (- x + 10) + x$

Group like terms:

$(2 x + x) - 38 = (- x + 10) + x$

Simplify the arithmetic:

$3 x - 38 = (- x + 10) + x$

Group like terms:

$3 x - 38 = (- x + x) + 10$

Simplify the arithmetic:

$3 x - 38 = 10$

Add to both sides:

$(3 x - 38) + 38 = 10 + 38$

Simplify the arithmetic:

$3 x = 10 + 38$

Simplify the arithmetic:

$3 x = 48$

Divide both sides by :

$\frac{(3 x)}{3} = \frac{48}{3}$

Simplify the fraction:

$x = \frac{48}{3}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(16 \cdot 3)}{(1 \cdot 3)}$

Factor out and cancel the greatest common factor:

$x = 16$

3. List the solutions

$x = 28, 16$
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
$y = 2 | x - 19 |$
$y = | x - 10 |$
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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