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Solution - Absolute value equations

Exact form:

x = \frac{1}{2}, \frac{1}{2}

x=\frac{1}{2} , \frac{1}{2}

Decimal form:

x = 0.5, 0.5

x=0.5 , 0.5

See steps

Step-by-step explanation

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

$9 | 2 x - 1 | - 2 | 6 x - 3 | = 0$

Add $2 | 6 x - 3 |$ to both sides of the equation:

$9 | 2 x - 1 | - 2 | 6 x - 3 | + 2 | 6 x - 3 | = 2 | 6 x - 3 |$

Simplify the arithmetic

$9 | 2 x - 1 | = 2 | 6 x - 3 |$

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
$9 | 2 x - 1 | = 2 | 6 x - 3 |$
without the absolute value bars:

$\| x \| = \| y \|$	$9 \| 2 x - 1 \| = 2 \| 6 x - 3 \|$
$x = + y$	$9 (2 x - 1) = 2 (6 x - 3)$
$x = - y$	$9 (2 x - 1) = 2 (- (6 x - 3))$
$+ x = y$	$9 (2 x - 1) = 2 (6 x - 3)$
$- x = y$	$9 (- (2 x - 1)) = 2 (6 x - 3)$

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:

$\| x \| = \| y \|$	$9 \| 2 x - 1 \| = 2 \| 6 x - 3 \|$
$x = + y$ , $+ x = y$	$9 (2 x - 1) = 2 (6 x - 3)$
$x = - y$ , $- x = y$	$9 (2 x - 1) = 2 (- (6 x - 3))$

3. Solve the two equations for x

17 additional steps

$9 \cdot (2 x - 1) = 2 \cdot (6 x - 3)$

Expand the parentheses:

$9 \cdot 2 x + 9 \cdot - 1 = 2 \cdot (6 x - 3)$

Multiply the coefficients:

$18 x + 9 \cdot - 1 = 2 \cdot (6 x - 3)$

Simplify the arithmetic:

$18 x - 9 = 2 \cdot (6 x - 3)$

Expand the parentheses:

$18 x - 9 = 2 \cdot 6 x + 2 \cdot - 3$

Multiply the coefficients:

$18 x - 9 = 12 x + 2 \cdot - 3$

Simplify the arithmetic:

$18 x - 9 = 12 x - 6$

Subtract from both sides:

$(18 x - 9) - 12 x = (12 x - 6) - 12 x$

Group like terms:

$(18 x - 12 x) - 9 = (12 x - 6) - 12 x$

Simplify the arithmetic:

$6 x - 9 = (12 x - 6) - 12 x$

Group like terms:

$6 x - 9 = (12 x - 12 x) - 6$

Simplify the arithmetic:

$6 x - 9 = - 6$

Add to both sides:

$(6 x - 9) + 9 = - 6 + 9$

Simplify the arithmetic:

$6 x = - 6 + 9$

Simplify the arithmetic:

$6 x = 3$

Divide both sides by :

$\frac{(6 x)}{6} = \frac{3}{6}$

Simplify the fraction:

$x = \frac{3}{6}$

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

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

18 additional steps

$9 \cdot (2 x - 1) = 2 \cdot (- (6 x - 3))$

Expand the parentheses:

$9 \cdot 2 x + 9 \cdot - 1 = 2 \cdot (- (6 x - 3))$

Multiply the coefficients:

$18 x + 9 \cdot - 1 = 2 \cdot (- (6 x - 3))$

Simplify the arithmetic:

$18 x - 9 = 2 \cdot (- (6 x - 3))$

Expand the parentheses:

$18 x - 9 = 2 \cdot (- 6 x + 3)$

Expand the parentheses:

$18 x - 9 = 2 \cdot - 6 x + 2 \cdot 3$

Multiply the coefficients:

$18 x - 9 = - 12 x + 2 \cdot 3$

Simplify the arithmetic:

$18 x - 9 = - 12 x + 6$

Add to both sides:

$(18 x - 9) + 12 x = (- 12 x + 6) + 12 x$

Group like terms:

$(18 x + 12 x) - 9 = (- 12 x + 6) + 12 x$

Simplify the arithmetic:

$30 x - 9 = (- 12 x + 6) + 12 x$

Group like terms:

$30 x - 9 = (- 12 x + 12 x) + 6$

Simplify the arithmetic:

$30 x - 9 = 6$

Add to both sides:

$(30 x - 9) + 9 = 6 + 9$

Simplify the arithmetic:

$30 x = 6 + 9$

Simplify the arithmetic:

$30 x = 15$

Divide both sides by :

$\frac{(30 x)}{30} = \frac{15}{30}$

Simplify the fraction:

$x = \frac{15}{30}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(1 \cdot 15)}{(2 \cdot 15)}$

Factor out and cancel the greatest common factor:

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

4. List the solutions

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

5. Graph

Each line represents the function of one side of the equation:
$y = 9 | 2 x - 1 |$
$y = 2 | 6 x - 3 |$
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 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