Question

solve for n. \\(\frac{3}{n + 8} = \frac{-n}{5}\\) there may be 1 or 2 solutions. \\(n = \square\\) or \\(n = \square\\)

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{3}{n + 8}=\frac{-n}{5}\), cross - multiplying (which is based on the property that if \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\)) gives us \(3\times5=(n + 8)\times(-n)\).
So, \(15=-n^{2}-8n\).

Step2: Rearrange into standard quadratic form

Rearrange the equation \(15=-n^{2}-8n\) to the standard quadratic form \(ax^{2}+bx + c = 0\).
Add \(n^{2}\) and \(8n\) to both sides of the equation: \(n^{2}+8n + 15 = 0\).

Step3: Factor the quadratic equation

Factor the quadratic expression \(n^{2}+8n + 15\). We need two numbers that multiply to \(15\) and add up to \(8\). The numbers are \(3\) and \(5\).
So, \(n^{2}+8n + 15=(n + 3)(n + 5)=0\).

Step4: Solve for \(n\)

Set each factor equal to zero and solve for \(n\).

• For \(n+3 = 0\), subtracting \(3\) from both sides gives \(n=-3\).
• For \(n + 5=0\), subtracting \(5\) from both sides gives \(n=-5\).

We also need to check if these solutions make the original denominator non - zero. For \(n=-3\), the denominator \(n + 8=-3 + 8 = 5
eq0\). For \(n=-5\), the denominator \(n + 8=-5 + 8 = 3
eq0\). So both solutions are valid.

Answer:

\(n=-3\) or \(n=-5\)