Find Where Undefined/Discontinuous 2/(x+3)+5/(x-4)=35/((x+3)(x-4))

$\frac{2}{x + 3} + \frac{5}{x - 4} = \frac{35}{(x + 3) (x - 4)}$

Step 1

Subtract $\frac{35}{(x + 3) (x - 4)}$ from both sides of the equation.

$\frac{2}{x + 3} + \frac{5}{x - 4} - \frac{35}{(x + 3) (x - 4)} = 0$

Step 2

Simplify $\frac{2}{x + 3} + \frac{5}{x - 4} - \frac{35}{(x + 3) (x - 4)}$ .

To write $\frac{2}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$\frac{2}{x + 3} \cdot \frac{x - 4}{x - 4} + \frac{5}{x - 4} - \frac{35}{(x + 3) (x - 4)} = 0$

To write $\frac{5}{x - 4}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{2}{x + 3} \cdot \frac{x - 4}{x - 4} + \frac{5}{x - 4} \cdot \frac{x + 3}{x + 3} - \frac{35}{(x + 3) (x - 4)} = 0$

Write each expression with a common denominator of $(x + 3) (x - 4)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{x + 3}$ by $\frac{x - 4}{x - 4}$ .

$\frac{2 (x - 4)}{(x + 3) (x - 4)} + \frac{5}{x - 4} \cdot \frac{x + 3}{x + 3} - \frac{35}{(x + 3) (x - 4)} = 0$

Multiply $\frac{5}{x - 4}$ by $\frac{x + 3}{x + 3}$ .

$\frac{2 (x - 4)}{(x + 3) (x - 4)} + \frac{5 (x + 3)}{(x - 4) (x + 3)} - \frac{35}{(x + 3) (x - 4)} = 0$

Reorder the factors of $(x - 4) (x + 3)$ .

$\frac{2 (x - 4)}{(x + 3) (x - 4)} + \frac{5 (x + 3)}{(x + 3) (x - 4)} - \frac{35}{(x + 3) (x - 4)} = 0$

Combine the numerators over the common denominator.

$\frac{2 (x - 4) + 5 (x + 3)}{(x + 3) (x - 4)} - \frac{35}{(x + 3) (x - 4)} = 0$

Combine the numerators over the common denominator.

$\frac{2 (x - 4) + 5 (x + 3) - 35}{(x + 3) (x - 4)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{2 x + 2 \cdot - 4 + 5 (x + 3) - 35}{(x + 3) (x - 4)} = 0$

Multiply $2$ by $- 4$ .

$\frac{2 x - 8 + 5 (x + 3) - 35}{(x + 3) (x - 4)} = 0$

Apply the distributive property.

$\frac{2 x - 8 + 5 x + 5 \cdot 3 - 35}{(x + 3) (x - 4)} = 0$

Multiply $5$ by $3$ .

$\frac{2 x - 8 + 5 x + 15 - 35}{(x + 3) (x - 4)} = 0$

Add $2 x$ and $5 x$ .

$\frac{7 x - 8 + 15 - 35}{(x + 3) (x - 4)} = 0$

Add $- 8$ and $15$ .

$\frac{7 x + 7 - 35}{(x + 3) (x - 4)} = 0$

Subtract $35$ from $7$ .

$\frac{7 x - 28}{(x + 3) (x - 4)} = 0$

Factor $7$ out of $7 x - 28$ .

Factor $7$ out of $7 x$ .

$\frac{7 (x) - 28}{(x + 3) (x - 4)} = 0$

Factor $7$ out of $- 28$ .

$\frac{7 x + 7 \cdot - 4}{(x + 3) (x - 4)} = 0$

Factor $7$ out of $7 x + 7 \cdot - 4$ .

$\frac{7 (x - 4)}{(x + 3) (x - 4)} = 0$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{7 (x - 4)}{(x + 3) (x - 4)} = 0$

Rewrite the expression.

$\frac{7}{x + 3} = 0$

Step 3

Set the denominator in $\frac{7}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Find Where Undefined/Discontinuous 2/(x+3)+5/(x-4)=35/((x+3)(x-4))

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