Question

solve for s.\\(\frac{s + 6}{s + 3} = \frac{s + 2}{s - 4}\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{s + 6}{s+3}=\frac{s + 2}{s - 4}\), we cross - multiply. Cross - multiplying gives us \((s + 6)(s - 4)=(s + 2)(s + 3)\).

Step2: Expand both sides

Expand the left - hand side: \((s+6)(s - 4)=s\times s-4\times s+6\times s - 6\times4=s^{2}-4s + 6s-24=s^{2}+2s-24\).
Expand the right - hand side: \((s + 2)(s + 3)=s\times s+3\times s+2\times s+2\times3=s^{2}+3s + 2s + 6=s^{2}+5s+6\).
So our equation becomes \(s^{2}+2s-24=s^{2}+5s + 6\).

Step3: Subtract \(s^{2}\) from both sides

Subtract \(s^{2}\) from both sides of the equation \(s^{2}+2s-24=s^{2}+5s + 6\). We get \(2s-24 = 5s+6\).

Step4: Move the terms with s to one side

Subtract \(2s\) from both sides: \(2s-2s-24=5s-2s + 6\), which simplifies to \(-24 = 3s+6\).

Step5: Solve for s

Subtract 6 from both sides: \(-24-6=3s+6 - 6\), so \(-30 = 3s\).
Divide both sides by 3: \(s=\frac{-30}{3}=- 10\).
We need to check if this solution makes the denominators zero. For \(s=-10\), \(s + 3=-10 + 3=-7
eq0\) and \(s-4=-10-4=-14
eq0\). So \(s=-10\) is a valid solution.

Answer:

\(s=-10\)