Precalculus Examples

$\frac{50}{4 x - 12} + \frac{x - 4}{x^{2} + x - 12} = \frac{x}{2}$

Step 1

Set the denominator in $\frac{50}{4 x - 12}$ equal to $0$ to find where the expression is undefined.

$4 x - 12 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Add $12$ to both sides of the equation.

$4 x = 12$

Step 2.2

Divide each term in $4 x = 12$ by $4$ and simplify.

Tap for more steps...

Step 2.2.1

Divide each term in $4 x = 12$ by $4$ .

$\frac{4 x}{4} = \frac{12}{4}$

Step 2.2.2

Simplify the left side.

Tap for more steps...

Step 2.2.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{12}{4}$

Step 2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{12}{4}$

Step 2.2.3

Simplify the right side.

Tap for more steps...

Step 2.2.3.1

Divide $12$ by $4$ .

$x = 3$

Step 3

Set the denominator in $\frac{x - 4}{x^{2} + x - 12}$ equal to $0$ to find where the expression is undefined.

$x^{2} + x - 12 = 0$

Step 4

Solve for $x$ .

Tap for more steps...

Step 4.1

Factor $x^{2} + x - 12$ using the AC method.

Tap for more steps...

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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Step 4.1.2

Write the factored form using these integers.

$(x - 3) (x + 4) = 0$

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 $0$ .

$x - 3 = 0$

$x + 4 = 0$

Step 4.3

Set $x - 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 4.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 4.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.4

Set $x + 4$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 4.4.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 4.4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 4.5

The final solution is all the values that make $(x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

Step 5

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3, - 4}$

Step 6