Check whether the points $y=-6,y=4$are the solutions of the equation $\frac{y}{y+6}=\frac{72}{y^2-36}+4$or not.

The equation is given as $\frac{y}{y+6}=\frac{72}{y^2-36}+4$to check whether the points $y=-6,4$are the solutions or not.

Apply the definition of the solution to find the required answer to the problem.

• A point is said to be the solution of the equation if the substitution of this point satisfies the given problem.
• Substitute $y=-6$into the equation (1) and check.

$\begin{array}{rcl}\frac{y}{y+6}& =& \frac{72}{y^2-36}+4\\ \frac{y}{y+6}& =& \frac{72+4(y^2-36)}{(y^2-36)}\\ & =& \frac{72+4(y^2-36)}{(y-6)(y+6)}\\ \frac{y}{\cancel{(y+6)}}& =& \frac{72+4(y^2-36)}{\cancel{(y+6)}(y-6)}\\ y& =& \frac{72+4(y^2-36)}{(y-6)}........................\left(1\right)\\ -6& =& \frac{72+4({(-6)}^2-36)}{(-6-6)}\\ & =& \frac{72+4(36-36)}{-12}\\ & =& \frac{72+4\left(0\right)}{-12}\\ & =& \frac{72+0}{-12}\\ & =& \frac{72}{-12}\\ & =& -6...............................\left(2\right)\end{array}$

Hence as per equation (2), the given point is the solution of the given equation.

Apply the definition of the solution to find the required answer to the problem.

• A point is said to be the solution of the equation if the substitution of this point satisfies the given problem.
• Substitute $y=4$the equation (1) and check.

$\begin{array}{rcl}y& =& \frac{72+4(y^2-36)}{(y-6)}........................\left(1\right)\\ 4& =& \frac{72+4(4^2-36)}{(4-6)}\\ & =& \frac{72+4(16-36)}{-2}\\ & =& \frac{72+4(-20)}{-2}\\ & =& \frac{72-80}{-2}\\ & =& \frac{-8}{-2}\\ & =& 4...................\left(3\right)\end{array}$

Hence as per equation (3), the point $y=4$is the solution as it satisfies the given equation.