Find the Excluded Values 1/(x-2)+1/(x+3)=1/(x^2+x-6)

$\frac{1}{x - 2} + \frac{1}{x + 3} = \frac{1}{x^{2} + x - 6}$

Subtract $\frac{1}{x^{2} + x - 6}$ from both sides of the equation.

$\frac{1}{x - 2} + \frac{1}{x + 3} - \frac{1}{x^{2} + x - 6} = 0$

Simplify $\frac{1}{x - 2} + \frac{1}{x + 3} - \frac{1}{x^{2} + x - 6}$ .

Factor $x^{2} + x - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

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

To write $\frac{1}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

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

To write $\frac{1}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

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

Write each expression with a common denominator of $(x - 2) (x + 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x - 2}$ by $\frac{x + 3}{x + 3}$ .

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

Multiply $\frac{1}{x + 3}$ by $\frac{x - 2}{x - 2}$ .

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

Reorder the factors of $(x - 2) (x + 3)$ .

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

Combine the numerators over the common denominator.

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

Combine the numerators over the common denominator.

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

Add $x$ and $x$ .

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

Subtract $2$ from $3$ .

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

Combine the opposite terms in $2 x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

Set the denominator in $\frac{2 x}{(x + 3) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(x + 3) (x - 2) = 0$

Solve for $x$ .

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

$x + 3 = 0$

$x - 2 = 0$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x + 3) (x - 2) = 0$ true.

$x = - 3, 2$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 3, x = 2$

Find the Excluded Values 1/(x-2)+1/(x+3)=1/(x^2+x-6)

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