Q31. - Algebra 1

Question

Solve each equation. State any extraneous solutions.

$\frac{\mathbf{j}}{\mathbf{j}\mathbf{+}\mathbf{2}}\mathbf{=}\frac{\mathbf{j}\mathbf{-}\mathbf{6}}{\mathbf{j}\mathbf{-}\mathbf{2}}$

Step-by-Step Solution

Answer

The solution of the equation is $\mathbf{-}\mathbf{6}$

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:

$\begin{array}{c}\frac{j}{j+2}=\frac{j-6}{j-2}\\ j\left(j-2\right)=\left(j+2\right)\left(j-6\right)\\ j^2-2j=j\left(j-6\right)+2\left(j-6\right)\\ j^2-2j=j^2-6j+2j-12\\ -2j=-4j-12\\ -2j+4j=-12\\ 2j=-12\\ \frac{2j}{2}=\frac{-12}{2}\\ j=-6\end{array}$

Step 3. Check the solution.

To check the solution, substitute $j=-6$ for in the original equation,

$\begin{array}{c}\frac{j}{j+2}=\frac{j-6}{j-2}\\ \frac{\left(-6\right)}{\left(-6\right)+2}=\frac{\left(-6\right)-6}{\left(-6\right)-2}\\ \frac{-6}{-4}=\frac{-12}{-8}\\ \frac{3}{2}=\frac{3}{2}\\ LHS=RHS\end{array}$

Since both sides are equal, the equation is correct.