Precalculus Examples

$y^{2} + 4 y - x + 3 = 0$

Step 1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.1

Subtract $y^{2}$ from both sides of the equation.

$4 y - x + 3 = - y^{2}$

Step 1.2

Subtract $4 y$ from both sides of the equation.

$- x + 3 = - y^{2} - 4 y$

Step 1.3

Subtract $3$ from both sides of the equation.

$- x = - y^{2} - 4 y - 3$

Step 2

Divide each term in $- x = - y^{2} - 4 y - 3$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- x = - y^{2} - 4 y - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- y^{2}}{- 1} + \frac{- 4 y}{- 1} + \frac{- 3}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- y^{2}}{- 1} + \frac{- 4 y}{- 1} + \frac{- 3}{- 1}$

Step 2.2.2

Divide $x$ by $1$ .

$x = \frac{- y^{2}}{- 1} + \frac{- 4 y}{- 1} + \frac{- 3}{- 1}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Dividing two negative values results in a positive value.

$x = \frac{y^{2}}{1} + \frac{- 4 y}{- 1} + \frac{- 3}{- 1}$

Step 2.3.1.2

Divide $y^{2}$ by $1$ .

$x = y^{2} + \frac{- 4 y}{- 1} + \frac{- 3}{- 1}$

Step 2.3.1.3

Move the negative one from the denominator of $\frac{- 4 y}{- 1}$ .

$x = y^{2} - 1 \cdot (- 4 y) + \frac{- 3}{- 1}$

Step 2.3.1.4

Rewrite $- 1 \cdot (- 4 y)$ as $- (- 4 y)$ .

$x = y^{2} - (- 4 y) + \frac{- 3}{- 1}$

Step 2.3.1.5

Multiply $- 4$ by $- 1$ .

$x = y^{2} + 4 y + \frac{- 3}{- 1}$

Step 2.3.1.6

Divide $- 3$ by $- 1$ .

$x = y^{2} + 4 y + 3$