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Solution - Absolute value equations

Exact form:

x = \frac{3}{4}, 0

x=\frac{3}{4} , 0

Decimal form:

x = 0.75, 0

x=0.75 , 0

See steps

Step-by-step explanation

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

$| 2 x - 3 | - 3 | - 2 x + 1 | = 0$

Add $3 | - 2 x + 1 |$ to both sides of the equation:

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

Simplify the arithmetic

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

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

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

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:

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

3. Solve the two equations for x

14 additional steps

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

Expand the parentheses:

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

Multiply the coefficients:

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

Simplify the arithmetic:

$(2 x - 3) = - 6 x + 3$

Add to both sides:

$(2 x - 3) + 6 x = (- 6 x + 3) + 6 x$

Group like terms:

$(2 x + 6 x) - 3 = (- 6 x + 3) + 6 x$

Simplify the arithmetic:

$8 x - 3 = (- 6 x + 3) + 6 x$

Group like terms:

$8 x - 3 = (- 6 x + 6 x) + 3$

Simplify the arithmetic:

$8 x - 3 = 3$

Add to both sides:

$(8 x - 3) + 3 = 3 + 3$

Simplify the arithmetic:

$8 x = 3 + 3$

Simplify the arithmetic:

$8 x = 6$

Divide both sides by :

$\frac{(8 x)}{8} = \frac{6}{8}$

Simplify the fraction:

$x = \frac{6}{8}$

Find the greatest common factor of the numerator and denominator:

$x = \frac{(3 \cdot 2)}{(4 \cdot 2)}$

Factor out and cancel the greatest common factor:

$x = \frac{3}{4}$

12 additional steps

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

Expand the parentheses:

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

Expand the parentheses:

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

Multiply the coefficients:

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

Simplify the arithmetic:

$(2 x - 3) = 6 x - 3$

Subtract from both sides:

$(2 x - 3) - 6 x = (6 x - 3) - 6 x$

Group like terms:

$(2 x - 6 x) - 3 = (6 x - 3) - 6 x$

Simplify the arithmetic:

$- 4 x - 3 = (6 x - 3) - 6 x$

Group like terms:

$- 4 x - 3 = (6 x - 6 x) - 3$

Simplify the arithmetic:

$- 4 x - 3 = - 3$

Add to both sides:

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

Simplify the arithmetic:

$- 4 x = - 3 + 3$

Simplify the arithmetic:

$- 4 x = 0$

Divide both sides by the coefficient:

$x = 0$

4. List the solutions

$x = \frac{3}{4}, 0$
(2 solution(s))

5. Graph

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