Question

solve for a.
\frac{a + 10}{a + 6} = \frac{a - 4}{a}

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{a + 10}{a + 6}=\frac{a - 4}{a}\), we use the cross - multiplication property of fractions. If \(\frac{m}{n}=\frac{p}{q}\), then \(m\times q=n\times p\). So we get \((a + 10)\times a=(a - 4)\times(a + 6)\).

Step2: Expand both sides

Expand the left - hand side: \((a + 10)a=a^{2}+10a\).
Expand the right - hand side: \((a - 4)(a + 6)=a^{2}+6a-4a - 24=a^{2}+2a-24\).
Now our equation becomes \(a^{2}+10a=a^{2}+2a - 24\).

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

Subtract \(a^{2}\) from both sides of the equation \(a^{2}+10a=a^{2}+2a - 24\). We have \(a^{2}+10a-a^{2}=a^{2}+2a - 24-a^{2}\), which simplifies to \(10a=2a - 24\).

Step4: Subtract \(2a\) from both sides

Subtract \(2a\) from both sides of the equation \(10a=2a - 24\). We get \(10a-2a=2a - 24-2a\), so \(8a=-24\).

Step5: Solve for \(a\)

Divide both sides of the equation \(8a=-24\) by 8. \(\frac{8a}{8}=\frac{-24}{8}\), which gives \(a=- 3\).
We need to check if this solution makes the original denominators zero. For the denominator \(a + 6\), when \(a=-3\), \(a + 6=-3 + 6 = 3
eq0\). For the denominator \(a\), when \(a=-3\), \(a=-3
eq0\). So \(a=-3\) is a valid solution.

Answer:

\(a=-3\)