Q. 48 - Precalculus Enhanced with Graphing Utilities

Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}{x}^2-xy-2y^2=0\\ xy+x+6=0\end{array}\right.$

Step-by-Step Solution

Answer

The solution of system of equations are $\left\{3,-3\right\},\left\{-2,2\right\}$.

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}{x}^2-xy-2y^2=0 ...... \left(i\right)\\ xy+x+6=0 ...... \left(ii\right)\end{array}\right.$

Divide equation (i) by $y^2$,

$$\begin{gathered} \frac{x^2}{y^2}-\frac{xy}{y^2}-\frac{2y^2}{y^2}=0 \\ \frac{x^2}{y^2}-\frac{x}{y}-2=0 \\ \left(\frac{x}{y}-2\right)\left(\frac{x}{y}+1\right)=0 \\ x=2y,y \end{gathered}$$

Step 3. Substitute the values of x in equation (ii).

Substitute $x=y$ in equation (ii),

$$\begin{gathered} \left(y\right)y+\left(y\right)+6=0 \\ {y}^2+y+6=0 \\ y=-3,2 \end{gathered}$$

Substitute $x=2y$ in equation (ii),

$$\begin{gathered} \left(2y\right)y+\left(2y\right)+6=0 \\ 2y^2+2y+6=0 \\ {y}^2+y+3=0 \end{gathered}$$

There is no solution for this value.

Therefore, the solution sets are $\left\{3,-3\right\},\left\{-2,2\right\}$.