Chapter 11, Problem 75RE

To determine

Solve each system of equations.

Answer to Problem 75RE

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

Explanation of Solution

Given information:

Solve each system of equations.

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

Calculation:

Use elimination method to solve the equation.

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

  $3x^{2}9xy+6y^2=0$

Now, the system of equations is $\{\begin{array}{l}3x^2+9xy+6y^2=0\\ 3x^2+4xy+5y^2=8\end{array}$

Subtract the second equation from the first.

  $\begin{array}{l}\{\begin{array}{l}3x^2+9xy+6y^2=0\\ \begin{array}{c}3x^2+4xy+5y^2=8\end{array}\end{array}\\ \begin{array}{cccccc}& & 5xy+y^2=-8& & & \end{array}\end{array}$

Solve for $x$ .

  $x=\frac{-8-y^2}{5y}$

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

  ${\left[\frac{-8-y^2}{5y}\right]}^2+3y\left[\frac{-8-y^2}{5y}\right]+2y^2=0$

  $\frac{64+16y^2+y^4}{25y^2}+3y\left[\frac{-8-y^2}{5y}\right]+2y^2=0$

Multiply both sides of the equation by $25y^2$ and simplify.

  $64+16y^2+y^4+\left[15y^2\left(-8-y^2\right)\right]+2y^2=0$

  $64+16y^2+y^4-120y^2-15y^4+50y^4=0$

  $36y^4-104y^2+64=0$

  $9y^4-26y^2+16=0$

Substitute $u$ for $y^2$ and $u^2$ for $y^4$ in the equation $9y^4-26y^2+16=0$ and solve for $u$ .

  $v$ .

  $9u^2-26u+16=0$

  $9u^2-18u-18u+16=0$

  $9u(u-2)-8(u-2)=0$

  $(9u-8)(u-2)=0$

Apply the zero-product rule.

  $9u-8=0$ Or $u-2=0$

  $u=\frac{8}{9}$ Or $u=2$

Back-substitute $y^2$ for $u$ .

  $y^2=\frac{8}{9}$ Or $y^2=2$

  $y=\pm \frac{2\sqrt{2}}{3}$ Or $y=\pm \sqrt{2}$

Thus, we have $y=\pm \frac{2\sqrt{2}}{3},\pm \sqrt{2}$ .

Substitute $\pm \frac{2\sqrt{2}}{3}$ for $y$ in the equation $x=\frac{-8-y^2}{5y}$ and solve for $x$ .

When $y=\frac{2}{3}\sqrt{2}$

  $x=\frac{-8-y^2}{5y}=\frac{-8-\left(\frac{2}{3}\sqrt{2}\right)}{5\left(\frac{2}{3}\sqrt{2}\right)}=\frac{-8-\frac{8}{9}}{\frac{10\sqrt{2}}{3}}=-\frac{4}{3}\sqrt{2}$

When $y=-\frac{2}{3}\sqrt{2}$

  $x=\frac{-8-y^2}{5y}=\frac{-8-\left(-\frac{2}{3}\sqrt{2}\right)}{5\left(-\frac{2}{3}\sqrt{2}\right)}=\frac{-8-\frac{8}{9}}{-\frac{10\sqrt{2}}{3}}=\frac{4}{3}\sqrt{2}$

Substitute $\pm \sqrt{2}$ for $y$ in the equation $x=\frac{-8-y^2}{5y}$ and solve for $x$ .

When $y=\sqrt{2}$

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

When $y=-\sqrt{2}$

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

Hence, the solutions written as ordered pairs are

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