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Solution - Absolute value equations

Exact form:

y = - 2, 0

y=-2 , 0

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
$| 6 y + 2 | = 2 | 2 y - 1 |$
without the absolute value bars:

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

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:

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

2. Solve the two equations for y

14 additional steps

$(6 y + 2) = 2 \cdot (2 y - 1)$

Expand the parentheses:

$(6 y + 2) = 2 \cdot 2 y + 2 \cdot - 1$

Multiply the coefficients:

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

Simplify the arithmetic:

$(6 y + 2) = 4 y - 2$

Subtract from both sides:

$(6 y + 2) - 4 y = (4 y - 2) - 4 y$

Group like terms:

$(6 y - 4 y) + 2 = (4 y - 2) - 4 y$

Simplify the arithmetic:

$2 y + 2 = (4 y - 2) - 4 y$

Group like terms:

$2 y + 2 = (4 y - 4 y) - 2$

Simplify the arithmetic:

$2 y + 2 = - 2$

Subtract from both sides:

$(2 y + 2) - 2 = - 2 - 2$

Simplify the arithmetic:

$2 y = - 2 - 2$

Simplify the arithmetic:

$2 y = - 4$

Divide both sides by :

$\frac{(2 y)}{2} = \frac{- 4}{2}$

Simplify the fraction:

$y = \frac{- 4}{2}$

Find the greatest common factor of the numerator and denominator:

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

Factor out and cancel the greatest common factor:

$y = - 2$

12 additional steps

$(6 y + 2) = 2 \cdot (- (2 y - 1))$

Expand the parentheses:

$(6 y + 2) = 2 \cdot (- 2 y + 1)$

Expand the parentheses:

$(6 y + 2) = 2 \cdot - 2 y + 2 \cdot 1$

Multiply the coefficients:

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

Simplify the arithmetic:

$(6 y + 2) = - 4 y + 2$

Add to both sides:

$(6 y + 2) + 4 y = (- 4 y + 2) + 4 y$

Group like terms:

$(6 y + 4 y) + 2 = (- 4 y + 2) + 4 y$

Simplify the arithmetic:

$10 y + 2 = (- 4 y + 2) + 4 y$

Group like terms:

$10 y + 2 = (- 4 y + 4 y) + 2$

Simplify the arithmetic:

$10 y + 2 = 2$

Subtract from both sides:

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

Simplify the arithmetic:

$10 y = 2 - 2$

Simplify the arithmetic:

$10 y = 0$

Divide both sides by the coefficient:

$y = 0$

3. List the solutions

$y = - 2, 0$
(2 solution(s))

4. Graph

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

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 y

3. List the solutions

4. Graph

Why learn this

Terms and topics

Related links

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