Precalculus Examples

$\frac{y + 3}{y + 1} \div \frac{y^{2} + 4 y + 3}{- 5 - 5 y}$

Step 1

Set the denominator in $\frac{y + 3}{y + 1}$ equal to $0$ to find where the expression is undefined.

$y + 1 = 0$

Step 2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 3

Set the denominator in $\frac{y^{2} + 4 y + 3}{- 5 - 5 y}$ equal to $0$ to find where the expression is undefined.

$- 5 - 5 y = 0$

Step 4

Solve for $y$ .

Tap for more steps...

Step 4.1

Add $5$ to both sides of the equation.

$- 5 y = 5$

Step 4.2

Divide each term in $- 5 y = 5$ by $- 5$ and simplify.

Tap for more steps...

Step 4.2.1

Divide each term in $- 5 y = 5$ by $- 5$ .

$\frac{- 5 y}{- 5} = \frac{5}{- 5}$

Step 4.2.2

Simplify the left side.

Tap for more steps...

Step 4.2.2.1

Cancel the common factor of $- 5$ .

Tap for more steps...

Step 4.2.2.1.1

Cancel the common factor.

$\frac{- 5 y}{- 5} = \frac{5}{- 5}$

Step 4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{5}{- 5}$

Step 4.2.3

Simplify the right side.

Tap for more steps...

Step 4.2.3.1

Divide $5$ by $- 5$ .

$y = - 1$

Step 5

Set the denominator in $\frac{y + 3}{y + 1} \div \frac{y^{2} + 4 y + 3}{- 5 - 5 y}$ equal to $0$ to find where the expression is undefined.

$\frac{y^{2} + 4 y + 3}{- 5 - 5 y} = 0$

Step 6

Solve for $y$ .

Tap for more steps...

Step 6.1

Set the numerator equal to zero.

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

Step 6.2

Solve the equation for $y$ .

Tap for more steps...

Step 6.2.1

Factor $y^{2} + 4 y + 3$ using the AC method.

Tap for more steps...

Step 6.2.1.1

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 $3$ and whose sum is $4$ .

$1, 3$

Step 6.2.1.2

Write the factored form using these integers.

$(y + 1) (y + 3) = 0$

Step 6.2.2

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

$y + 1 = 0$

$y + 3 = 0$

Step 6.2.3

Set $y + 1$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 6.2.3.1

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Step 6.2.3.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 6.2.4

Set $y + 3$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 6.2.4.1

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

Step 6.2.4.2

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 6.2.5

The final solution is all the values that make $(y + 1) (y + 3) = 0$ true.

$y = - 1, - 3$

Step 6.3

Exclude the solutions that do not make $\frac{y^{2} + 4 y + 3}{- 5 - 5 y} = 0$ true.

$y = - 3$

Step 7

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

Interval Notation:

$(- \infty, - 3) \cup (- 3, - 1) \cup (- 1, \infty)$

Set-Builder Notation:

${y | y \neq - 1, - 3}$

Step 8