Q77. - Algebra 2

Question

Solve the equation $\frac{s-3}{s+4}=\frac{6}{s^2-16}$Check your solution.

Step-by-Step Solution

Answer

The solution for the equation is $s=\left\{1,6\right\}$.

Step 1. Write down the given expression.

The given equation is $\frac{s-3}{s+4}=\frac{6}{s^2-16}$.

Step 2. Simplify the equation.

The given equation $\frac{s-3}{s+4}=\frac{6}{s^2-16}$ is solved as:

$\begin{array}{c}\frac{s-3}{s+4}=\frac{6}{s^2-16}\mathrm{....}\left(\text{Given}\right)\\ \left(s-3\right)\left(s^2-16\right)=6\left(s+4\right)\\ \left(s-3\right)\left(s+4\right)\left(s-4\right)=6\left(s+4\right)\\ \left(s-3\right)\left(s-4\right)=6\\ s^2-7s+6=0\end{array}$

Further simplifying,

$\begin{array}{c}\left(s-1\right)\left(s-6\right)=\mathrm{0....}\left(\text{Factoring}\right)\\ s-1=0\text{ or }s-6=0\\ s=1\text{ or }s=6\\ s=\left\{1,6\right\}\end{array}$

Step 3. Verify the result for obtained value.

Plugging $s=1$ in given equation and verify the result.

$\begin{array}{c}\frac{s-3}{s+4}=\frac{6}{s^2-16}\mathrm{....}\left(\text{Given}\right)\\ \frac{\left(1\right)-3}{\left(1\right)+4}=\frac{6}{{\left(1\right)}^2-16}\mathrm{....}\left(\text{Plugging }s=1\right)\\ \frac{-2}{5}=\frac{6}{-15}\\ -\frac{2}{5}=-\frac{2}{5}\mathrm{....}\left(\text{True}\right)\end{array}$

Hence the solution is verified for $s=1$  .

Plugging  in given equation and verify the result.

$\begin{array}{c}\frac{s-3}{s+4}=\frac{6}{s^2-16}\mathrm{....}\left(\text{Given}\right)\\ \frac{\left(6\right)-3}{\left(6\right)+4}=\frac{6}{{\left(6\right)}^2-16}\mathrm{....}\left(\text{Plugging }s=6\right)\\ \frac{3}{10}=\frac{6}{20}\\ \frac{3}{10}=\frac{3}{10}\mathrm{....}\left(\text{True}\right)\end{array}$

Hence the solution is verified for$s=6$

Step 4. Conclusion.

The solution for the given equation is $s=\left\{1,6\right\}$.