Question 1,1.6.13 ,
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Points: 0
Decide what values of the variable cannot possibly be solutions for the equation.
\[
\frac{1}{x-3}+\frac{1}{x+2}=\frac{1}{x^{2}-x-6}
\]
What values of $x$ cannot be solutions of the equation?

Step 1 ：Decide what values of the variable cannot possibly be solutions for the equation \(\frac{1}{x-3}+\frac{1}{x+2}=\frac{1}{x^{2}-x-6}\).

Step 2 ：The values of \(x\) that cannot be solutions of the equation are the values that make the denominator of any of the fractions equal to zero, because division by zero is undefined in mathematics.

Step 3 ：Therefore, we need to find the values of \(x\) that make \(x-3=0\), \(x+2=0\), and \(x^{2}-x-6=0\).

Step 4 ：The solutions to these equations are \(x=3\), \(x=-2\), and \(x=-2, 3\) respectively.

Step 5 ：The values of \(x\) that cannot be solutions of the equation are \(3\) and \(-2\).

Step 6 ：Final Answer: The values of \(x\) that cannot be solutions of the equation are \(\boxed{3}\) and \(\boxed{-2}\).