Precalculus Examples

$\frac{y^{2} - 16}{y^{2} - 10 y + 25} \div \frac{3 y - 12}{y^{2} - 3 y - 10}$

Step 1

Set the denominator in $\frac{y^{2} - 16}{y^{2} - 10 y + 25}$ equal to $0$ to find where the expression is undefined.

$y^{2} - 10 y + 25 = 0$

Step 2

Solve for $y$ .

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Step 2.1

Factor using the perfect square rule.

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Step 2.1.1

Rewrite $25$ as $5^{2}$ .

$y^{2} - 10 y + 5^{2} = 0$

Step 2.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 y = 2 \cdot y \cdot 5$

Step 2.1.3

Rewrite the polynomial.

$y^{2} - 2 \cdot y \cdot 5 + 5^{2} = 0$

Step 2.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 5$ .

${(y - 5)}^{2} = 0$

Step 2.2

Set the $y - 5$ equal to $0$ .

$y - 5 = 0$

Step 2.3

Add $5$ to both sides of the equation.

$y = 5$

Step 3

Set the denominator in $\frac{3 y - 12}{y^{2} - 3 y - 10}$ equal to $0$ to find where the expression is undefined.

$y^{2} - 3 y - 10 = 0$

Step 4

Solve for $y$ .

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Step 4.1

Factor $y^{2} - 3 y - 10$ using the AC method.

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Step 4.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Step 4.1.2

Write the factored form using these integers.

$(y - 5) (y + 2) = 0$

Step 4.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 5 = 0$

$y + 2 = 0$

Step 4.3

Set $y - 5$ equal to $0$ and solve for $y$ .

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Step 4.3.1

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

Step 4.3.2

Add $5$ to both sides of the equation.

$y = 5$

Step 4.4

Set $y + 2$ equal to $0$ and solve for $y$ .

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Step 4.4.1

Set $y + 2$ equal to $0$ .

$y + 2 = 0$

Step 4.4.2

Subtract $2$ from both sides of the equation.

$y = - 2$

Step 4.5

The final solution is all the values that make $(y - 5) (y + 2) = 0$ true.

$y = 5, - 2$

Step 5

Set the denominator in $\frac{y^{2} - 16}{y^{2} - 10 y + 25} \div \frac{3 y - 12}{y^{2} - 3 y - 10}$ equal to $0$ to find where the expression is undefined.

$\frac{3 y - 12}{y^{2} - 3 y - 10} = 0$

Step 6

Solve for $y$ .

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Step 6.1

Set the numerator equal to zero.

$3 y - 12 = 0$

Step 6.2

Solve the equation for $y$ .

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Step 6.2.1

Add $12$ to both sides of the equation.

$3 y = 12$

Step 6.2.2

Divide each term in $3 y = 12$ by $3$ and simplify.

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Step 6.2.2.1

Divide each term in $3 y = 12$ by $3$ .

$\frac{3 y}{3} = \frac{12}{3}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $3$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{12}{3}$

Step 6.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{12}{3}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $12$ by $3$ .

$y = 4$

Step 7

The domain is all values of $y$ that make the expression defined.

Interval Notation:

$(- \infty, - 2) \cup (- 2, 4) \cup (4, 5) \cup (5, \infty)$

Set-Builder Notation:

${y | y \neq - 2, 4, 5}$

Step 8