Find Where Undefined/Discontinuous x/(x+4)=1/(x+1)-(3x)/(x^2+5x+4)

$\frac{x}{x + 4} = \frac{1}{x + 1} - \frac{3 x}{x^{2} + 5 x + 4}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $\frac{1}{x + 1}$ from both sides of the equation.

$\frac{x}{x + 4} - \frac{1}{x + 1} = - \frac{3 x}{x^{2} + 5 x + 4}$

Add $\frac{3 x}{x^{2} + 5 x + 4}$ to both sides of the equation.

$\frac{x}{x + 4} - \frac{1}{x + 1} + \frac{3 x}{x^{2} + 5 x + 4} = 0$

Step 2

Simplify $\frac{x}{x + 4} - \frac{1}{x + 1} + \frac{3 x}{x^{2} + 5 x + 4}$ .

Factor $x^{2} + 5 x + 4$ using the AC method.

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 $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

$\frac{x}{x + 4} - \frac{1}{x + 1} + \frac{3 x}{(x + 1) (x + 4)} = 0$

To write $\frac{x}{x + 4}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{x}{x + 4} \cdot \frac{x + 1}{x + 1} - \frac{1}{x + 1} + \frac{3 x}{(x + 1) (x + 4)} = 0$

To write $- \frac{1}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{x + 4}{x + 4}$ .

$\frac{x}{x + 4} \cdot \frac{x + 1}{x + 1} - \frac{1}{x + 1} \cdot \frac{x + 4}{x + 4} + \frac{3 x}{(x + 1) (x + 4)} = 0$

Write each expression with a common denominator of $(x + 4) (x + 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{x + 4}$ by $\frac{x + 1}{x + 1}$ .

$\frac{x (x + 1)}{(x + 4) (x + 1)} - \frac{1}{x + 1} \cdot \frac{x + 4}{x + 4} + \frac{3 x}{(x + 1) (x + 4)} = 0$

Multiply $\frac{1}{x + 1}$ by $\frac{x + 4}{x + 4}$ .

$\frac{x (x + 1)}{(x + 4) (x + 1)} - \frac{x + 4}{(x + 1) (x + 4)} + \frac{3 x}{(x + 1) (x + 4)} = 0$

Reorder the factors of $(x + 4) (x + 1)$ .

$\frac{x (x + 1)}{(x + 1) (x + 4)} - \frac{x + 4}{(x + 1) (x + 4)} + \frac{3 x}{(x + 1) (x + 4)} = 0$

Combine the numerators over the common denominator.

$\frac{x (x + 1) - (x + 4)}{(x + 1) (x + 4)} + \frac{3 x}{(x + 1) (x + 4)} = 0$

Combine the numerators over the common denominator.

$\frac{x (x + 1) - (x + 4) + 3 x}{(x + 1) (x + 4)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x \cdot x + x \cdot 1 - (x + 4) + 3 x}{(x + 1) (x + 4)} = 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 1 - (x + 4) + 3 x}{(x + 1) (x + 4)} = 0$

Multiply $x$ by $1$ .

$\frac{x^{2} + x - (x + 4) + 3 x}{(x + 1) (x + 4)} = 0$

Apply the distributive property.

$\frac{x^{2} + x - x - 1 \cdot 4 + 3 x}{(x + 1) (x + 4)} = 0$

Multiply $- 1$ by $4$ .

$\frac{x^{2} + x - x - 4 + 3 x}{(x + 1) (x + 4)} = 0$

Combine the opposite terms in $x^{2} + x - x - 4 + 3 x$ .

Subtract $x$ from $x$ .

$\frac{x^{2} + 0 - 4 + 3 x}{(x + 1) (x + 4)} = 0$

Add $x^{2}$ and $0$ .

$\frac{x^{2} - 4 + 3 x}{(x + 1) (x + 4)} = 0$

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

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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$\frac{(x - 1) (x + 4)}{(x + 1) (x + 4)} = 0$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$\frac{(x - 1) (x + 4)}{(x + 1) (x + 4)} = 0$

Rewrite the expression.

$\frac{x - 1}{x + 1} = 0$

Step 3

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

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Find Where Undefined/Discontinuous x/(x+4)=1/(x+1)-(3x)/(x^2+5x+4)

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