Question

solve for f.
\frac{15}{f - 11} = \frac{-f}{2}
there may be 1 or 2 solutions.
f = \square or f = \square

Explanation:

Step1: Cross - multiply to eliminate fractions

Given the equation \(\frac{15}{f - 11}=\frac{-f}{2}\), cross - multiplying (which is valid when \(f
eq11\) and \(2
eq0\)) gives us \(15\times2=-f\times(f - 11)\).
So, \(30=-f^{2}+11f\).

Step2: Rearrange into standard quadratic form

Rearrange the equation \(30=-f^{2}+11f\) to the standard quadratic form \(ax^{2}+bx + c = 0\).
We get \(f^{2}-11f + 30 = 0\) (by multiplying both sides of \(30=-f^{2}+11f\) by \(- 1\)).

Step3: Factor the quadratic equation

Factor the quadratic \(f^{2}-11f + 30\). We need two numbers that multiply to \(30\) and add up to \(-11\) (the coefficient of \(f\) with a sign change because the middle term is \(-11f\) and the constant term is \(30\)). The numbers are \(-5\) and \(-6\).
So, \(f^{2}-11f + 30=(f - 5)(f - 6)=0\).

Step4: Solve for \(f\)

Set each factor equal to zero:

• If \(f - 5 = 0\), then \(f = 5\).
• If \(f - 6 = 0\), then \(f = 6\).

We also need to check if these solutions make the original denominator non - zero. For \(f = 5\), the denominator \(f-11=5 - 11=-6
eq0\). For \(f = 6\), the denominator \(f - 11=6-11=-5
eq0\). So both solutions are valid.

Answer:

\(f = 5\) or \(f = 6\)