Q 36. - Algebra 1

Question

Solve each equation. State any extraneous solutions.

$\frac{\mathbf{t}\mathbf{-}\mathbf{3}}{\mathbf{t}\mathbf{+}\mathbf{3}}\mathbf{-}\frac{\mathbf{2}\mathbf{t}}{\mathbf{t}\mathbf{-}\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The solution of the equation is $\mathbf{0}$.

Step 1. State the concept for observational studies.

The linear equations in one variable is an equation which is expressed in the form of $ax+b=0$, where a and b are two integers, and x is a variable and has only one solution.

Step 2. Solve the equation.

The equation can be solved as follows:

$\begin{array}{c}\frac{t-3}{t+3}-\frac{2t}{t-3}=-1\\ \frac{\left(t-3\right)\left(t-3\right)-2t\left(t+3\right)}{\left(t-3\right)\left(t+3\right)}=-1\\ \left(t-3\right)\left(t-3\right)-2t\left(t+3\right)=-1\left(t-3\right)\left(t+3\right)\\ t^2-3t-3t+9-2t^2-6t=-t^2+3t-3t+9\\ -t^2-12t+9=-t^2+9\\ -12t=0\\ t=-\frac{0}{12}\\ t=0\end{array}$

Step 3. Check the solution.

To check the solution, substitute $t=0$ for in the original equation,

$\begin{array}{c}\frac{t-3}{t+3}-\frac{2t}{t-3}=-1\\ \frac{0-3}{0+3}-\frac{2\left(0\right)}{0-3}=-1\\ \frac{-3}{3}-\frac{0}{-3}=-1\\ -1-0=-1\\ -1=-1\\ LHS=RHS\end{array}$

Since both sides are equal, the equation is correct.