Final answer:
To solve the given equation, one must factor the denominator, eliminate the fractions by multiplying through by the common denominator, expand and simplify the equation to a quadratic, and finally use the quadratic formula to find potential solutions, checking for validity with the original domain restrictions.
Step-by-step explanation:
The student's question involves finding the solution(s) to the equation \( \frac{1}{x+7} + \frac{x}{x-2} = \frac{18}{x^2+5x-14} \). To solve this, we can first observe that the denominator on the right side, \(x^2+5x-14\), can be factored into \((x+7)(x-2)\). This means we can write the equation as:
\( \frac{1}{x+7} + \frac{x}{x-2} = \frac{18}{(x+7)(x-2)} \).
Next, we can multiply each term by the common denominator \((x+7)(x-2)\) to eliminate the fractions:
\((x-2) + x(x+7) = 18\).
Expanding and simplifying, we get:
\(x^2+7x+x-2=18\),
which simplifies to:
\(x^2+8x-20=0\).
Now, we can apply the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), to find the roots of the equation.
Upon doing this, we determine the values of \(x\) that satisfy the equation, considering that any solution must be valid in the original problem's domain (i.e., \(x \\eq -7\) and \(x \\eq 2\)).