Precalculus Examples

$| 2 y | + 3 = 2 + 3$

Step 1

Add $2$ and $3$ .

$| 2 y | + 3 = 5$

Step 2

Move all terms not containing $| 2 y |$ to the right side of the equation.

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Subtract $3$ from both sides of the equation.

$| 2 y | = 5 - 3$

Subtract $3$ from $5$ .

$| 2 y | = 2$

Step 3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 y = \pm 2$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$2 y = 2$

Divide each term in $2 y = 2$ by $2$ and simplify.

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Divide each term in $2 y = 2$ by $2$ .

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

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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Divide $2$ by $2$ .

$y = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$2 y = - 2$

Divide each term in $2 y = - 2$ by $2$ and simplify.

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Divide each term in $2 y = - 2$ by $2$ .

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

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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Divide $- 2$ by $2$ .

$y = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 1, - 1$

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