Q. 47 - Precalculus Enhanced with Graphing Utilities

Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}{x}^2-3xy+2y^2=0\\ x^2+xy=6\end{array}\right.$

Step-by-Step Solution

Answer

The solution of system of equations are $\left\{-\sqrt{3},-\sqrt{3}\right\},\left\{\sqrt{3},\sqrt{3}\right\},\left\{-2,-1\right\},\left\{2,1\right\}$.

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}{x}^2-3xy+2y^2=0 ...... \left(i\right)\\ x^2+xy=6 ...... \left(ii\right)\end{array}\right.$

Subtract equation (i) and (ii),

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

Step 2. Write in terms of x.

Writing in terms of x,

$$\begin{gathered} 2xy=3+y^2 \\ x=\frac{3+y^2}{2y} ...... \left(iii\right) \end{gathered}$$

Substitute the value of x in equation (ii),

$$\begin{gathered} {\left(\frac{3+y^2}{2y}\right)}^2+\left(\frac{3+y^2}{2y}\right)y=6 \\ \frac{9+y^4+6y^2}{4y^2}+\frac{3+y^2}{2}=6 \\ \left(5y^2+3\right)\left(y^2-3\right)=0 \\ y=-\sqrt{3},\sqrt{3},1,-1 \end{gathered}$$

Step 3. Substitute y = - 1 , 1 in equation (iii) and solve the equation.

Substitute $y=1$ in equation (iii),

$$\begin{gathered} x=\frac{3+1^2}{2\left(1\right)} \\ x=\frac{4}{2} \\ x=2 \end{gathered}$$

Substitute $y=-1$ in equation (iii),

$$\begin{gathered} x=\frac{3+{\left(-1\right)}^2}{2\left(-1\right)} \\ x=\frac{4}{-2} \\ x=-2 \end{gathered}$$

Step 4. Substitute y = - 3 , 3 in equation (iii) and solve the equation.

Substitute $y=\sqrt{3}$ in equation (iii),

 $$\begin{gathered} x=\frac{3+{\left(\sqrt{3}\right)}^2}{2\sqrt{3}} \\ x=\frac{12}{2\sqrt{3}} \\ x=\sqrt{3} \end{gathered}$$

Substitute $y=-\sqrt{3}$ in equation (iii),

$$\begin{gathered} x=\frac{3+{\left(-\sqrt{3}\right)}^2}{2\left(-\sqrt{3}\right)} \\ x=\frac{12}{-2\sqrt{3}} \\ x=-\sqrt{3} \end{gathered}$$

Therefore, the solution sets are $\left\{-\sqrt{3},-\sqrt{3}\right\},\left\{\sqrt{3},\sqrt{3}\right\},\left\{-2,-1\right\},\left\{2,1\right\}$.