Question

solve for c.
\\(\frac{1}{c + 7} = \frac{-c}{8c + 50}\\)
there may be 1 or 2 solutions.
\\(c = \square\\) or \\(c = \square\\)

Explanation:

Step1: Cross - multiply

To solve the equation \(\frac{1}{c + 7}=\frac{-c}{8c+50}\), we cross - multiply. Cross - multiplying gives us \(1\times(8c + 50)=-c\times(c + 7)\).

Step2: Expand both sides

Expanding the left - hand side: \(8c+50\).
Expanding the right - hand side: \(-c\times c-c\times7=-c^{2}-7c\).
So our equation becomes \(8c + 50=-c^{2}-7c\).

Step3: Rearrange into standard quadratic form

Bring all terms to one side to get a quadratic equation. Add \(c^{2}\) and \(7c\) to both sides: \(c^{2}+7c + 8c+50 = 0\).
Combine like terms: \(c^{2}+15c + 50=0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(50\) and add up to \(15\). The numbers are \(10\) and \(5\). So we can factor the quadratic as \((c + 10)(c+5)=0\).

Step5: Solve for c

Set each factor equal to zero:

• If \(c + 10=0\), then \(c=-10\).
• If \(c + 5=0\), then \(c=-5\).

We also need to check for extraneous solutions. For the original equation \(\frac{1}{c + 7}=\frac{-c}{8c+50}\), the denominators \(c + 7
eq0\) (so \(c
eq - 7\)) and \(8c+50
eq0\) (so \(c
eq-\frac{50}{8}=-\frac{25}{4}=-6.25\)). Both \(-10\) and \(-5\) do not make the denominators zero, so they are valid solutions.

Answer:

\(c=-10\) or \(c = - 5\)