Q. 42 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 39–43, solve each system of equations.

$\left\{\begin{array}{l}3x^2+4xy+5y^2=8\\ x^2+3xy+2y^2=0\end{array}\right.$

Step-by-Step Solution

Answer

Solutions of the system of equations $\left\{\begin{array}{l}3x^2+4xy+5y^2=8\\ x^2+3xy+2y^2=0\end{array}\right.$are

$(\sqrt{2},-\sqrt{2}), (-\sqrt{2},\sqrt{2}), \left(\frac{4}{3}\sqrt{2},\frac{-2}{3}\sqrt{2}\right), \& \left(\frac{-4}{3}\sqrt{2},\frac{2}{3}\sqrt{2}\right)$

Step 1. Given data

The given system of equation is 

$\left\{\begin{array}{l}3x^2+4xy+5y^2=8\\ x^2+3xy+2y^2=0\end{array}\right.$

Step 2. Determine the relationship between variables

Multiply $3$ to both sides of the equation $x^2+3xy+2y^2=0$

$3x^2+9xy+6y^2=0$

so the new system of equation is $\left\{\begin{array}{l}3x^2+4xy+5y^2=8\\ 3x^2+9xy+6y^2=0\end{array}\right.$

Subtract the first equation from the second equation

$$\begin{gathered} \left(3x^2+9xy+6y^2\right)-\left(3x^2+4xy+5y^2\right)=0-8 \\ 5xy+y^2=-8 \\ x=\frac{-y^2-8}{5y} \end{gathered}$$

Step 3. Formation of the single-variable equation

Substitute $x=\frac{-y^2-8}{5y}$in equation $x^2+3xy+2y^2=0$

$$\begin{gathered} x^2+3xy+2y^2=0 \\ {\left(\frac{-y^2-8}{5y}\right)}^2+3y\left(\frac{-y^2-8}{5y}\right)+2y^2=0 \\ \frac{64+16y^2+y^4}{25y^2}+\frac{-3y^3-24y}{5y}+2y^2=0 \\ 64+16y^2+y^4+5\left(-3y^3-24y\right)+25y^2\left(2y^2\right)=0 \\ 9y^4-26y^2+16=0 \\ 9y^4-18y^2-8y^2+16=0 \\ 9y^2\left(y^2-2\right)-8(y^2-2)=0 \\ \left(9y^2-8\right)(y^2-2)=0 \end{gathered}$$

Step 4. Solution of the single variable equation

Use zero product property for$\left(9y^2-8\right)(y^2-2)=0$

$$\begin{gathered} 9y^2-8=0 \\ {y}^2=\frac{8}{9} \\ y=\pm \frac{2\sqrt{2}}{3} \end{gathered}$$

and

$$\begin{gathered} y^2-2=0 \\ {y}^2=2 \\ y=\pm \sqrt{2} \end{gathered}$$

Step 5. Solution of the system of equations

Substitute$y=\sqrt{2}$ in equation$x=\frac{-y^2-8}{5y}$

$$\begin{gathered} x=\frac{-{\left(\sqrt{2}\right)}^2-8}{5\sqrt{2}} \\ x=-\sqrt{2} \end{gathered}$$

Substitute $y=-\sqrt{2}$in equation$x=\frac{-y^2-8}{5y}$

$$\begin{gathered} x=\frac{-{\left(-\sqrt{2}\right)}^2-8}{5\left(-\sqrt{2}\right)} \\ x=\sqrt{2} \end{gathered}$$

Substitute $y=\frac{2\sqrt{2}}{3}$in equation $x=\frac{-y^2-8}{5y}$

$$\begin{gathered} x=\frac{-{\left(\frac{2\sqrt{2}}{3}\right)}^2-8}{5\left(\frac{2\sqrt{2}}{3}\right)} \\ x=\frac{-4}{3}\sqrt{2} \end{gathered}$$

Substitute $y=\frac{-2\sqrt{2}}{3}$in equation$x=\frac{-y^2-8}{5y}$

$$\begin{gathered} x=\frac{-{\left(\frac{-2\sqrt{2}}{3}\right)}^2-8}{5\left(\frac{-2\sqrt{2}}{3}\right)} \\ x=\frac{4}{3}\sqrt{2} \end{gathered}$$

So solutions of the system of equations are $(\sqrt{2},-\sqrt{2}), (-\sqrt{2},\sqrt{2}), \left(\frac{4}{3}\sqrt{2},\frac{-2}{3}\sqrt{2}\right), \& \left(\frac{-4}{3}\sqrt{2},\frac{2}{3}\sqrt{2}\right)$