Precalculus Examples

$- 12 y + 10 = x^{2} - 4 x + 14$

Step 1

Solve for $y$ .

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Step 1.1

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

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Step 1.1.1

Subtract $10$ from both sides of the equation.

$- 12 y = x^{2} - 4 x + 14 - 10$

Step 1.1.2

Subtract $10$ from $14$ .

$- 12 y = x^{2} - 4 x + 4$

Step 1.2

Divide each term in $- 12 y = x^{2} - 4 x + 4$ by $- 12$ and simplify.

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Step 1.2.1

Divide each term in $- 12 y = x^{2} - 4 x + 4$ by $- 12$ .

$\frac{- 12 y}{- 12} = \frac{x^{2}}{- 12} + \frac{- 4 x}{- 12} + \frac{4}{- 12}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $- 12$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{- 12 y}{- 12} = \frac{x^{2}}{- 12} + \frac{- 4 x}{- 12} + \frac{4}{- 12}$

Step 1.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{2}}{- 12} + \frac{- 4 x}{- 12} + \frac{4}{- 12}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Move the negative in front of the fraction.

$y = - \frac{x^{2}}{12} + \frac{- 4 x}{- 12} + \frac{4}{- 12}$

Step 1.2.3.1.2

Cancel the common factor of $- 4$ and $- 12$ .

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Step 1.2.3.1.2.1

Factor $- 4$ out of $- 4 x$ .

$y = - \frac{x^{2}}{12} + \frac{- 4 (x)}{- 12} + \frac{4}{- 12}$

Step 1.2.3.1.2.2

Cancel the common factors.

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Step 1.2.3.1.2.2.1

Factor $- 4$ out of $- 12$ .

$y = - \frac{x^{2}}{12} + \frac{- 4 (x)}{- 4 (3)} + \frac{4}{- 12}$

Step 1.2.3.1.2.2.2

Cancel the common factor.

$y = - \frac{x^{2}}{12} + \frac{- 4 x}{- 4 \cdot 3} + \frac{4}{- 12}$

Step 1.2.3.1.2.2.3

Rewrite the expression.

$y = - \frac{x^{2}}{12} + \frac{x}{3} + \frac{4}{- 12}$

Step 1.2.3.1.3

Cancel the common factor of $4$ and $- 12$ .

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Step 1.2.3.1.3.1

Factor $4$ out of $4$ .

$y = - \frac{x^{2}}{12} + \frac{x}{3} + \frac{4 (1)}{- 12}$

Step 1.2.3.1.3.2

Cancel the common factors.

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Step 1.2.3.1.3.2.1

Factor $4$ out of $- 12$ .

$y = - \frac{x^{2}}{12} + \frac{x}{3} + \frac{4 \cdot 1}{4 \cdot - 3}$

Step 1.2.3.1.3.2.2

Cancel the common factor.

$y = - \frac{x^{2}}{12} + \frac{x}{3} + \frac{4 \cdot 1}{4 \cdot - 3}$

Step 1.2.3.1.3.2.3

Rewrite the expression.

$y = - \frac{x^{2}}{12} + \frac{x}{3} + \frac{1}{- 3}$

Step 1.2.3.1.4

Move the negative in front of the fraction.

$y = - \frac{x^{2}}{12} + \frac{x}{3} - \frac{1}{3}$

Step 2

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$y = a x^{2} + b x + c$

Step 3

Reorder terms.

$y = - (\frac{1}{12} x^{2}) + \frac{1}{3} x - \frac{1}{3}$

Step 4

Remove parentheses.

$y = - \frac{1}{12} x^{2} + \frac{1}{3} x - \frac{1}{3}$

Step 5