Chapter 11, Problem 76RE

To determine

Solve each system of equations.

Answer to Problem 76RE

  $\boxed{\left(\frac{7\pm \sqrt{217}}{\pm \sqrt{3\left(19\pm \sqrt{217}\right)}},\pm \sqrt{\frac{19\pm \sqrt{217}}{3}}\right)}$

Explanation of Solution

Given information:

Solve each system of equations.

  $\{\begin{array}{l}3x^2+2xy-2y^2=6\\ xy-2y^2+4=0\end{array}$

Calculation:

The given system of equations in three variables is given by the expressions given below.

  $\{\begin{array}{l}3x^2+2xy-2y^2=6\\ xy-2y^2+4=0\end{array}$

The given system of equations can be solved by the method of substitution. We obtain a relation of x in terms of y from the second equation and substitute in the first equation. The relation can be obtained as follows:

  $xy-y^2+4=0$

  $\begin{array}{l}xy=y^2-4\\ x=\frac{y^2-4}{y}\end{array}$ .....(1)

Now we can substitute the above obtained relation in the first equation to obtain the possible values of $y$ .

  $3{\left(\frac{y^2-4}{y}\right)}^2+2\left(\frac{y^2-4}{y}\right)y-2y^2=6$

  $3\left(\frac{y^4-8y^2+16}{y^2}\right)+2\left(y^2-4\right)-2y^2=6$

  $3y^{4}24y^2+48+y^2\left(2\left(y^4-4\right)-2y^2\right)=6y^2$

  $3y^4-38y^2+48=0$

  $y^2=\frac{38\pm \sqrt{{\left(-38\right)}^2-4\left(3\right)}\left(48\right)}{2\left(3\right)}$

  $y^2=\frac{38\pm 2\sqrt{217}}{6}$

  $y=\pm \sqrt{\frac{19\pm \sqrt{217}}{3}}$

The obtained value of $x$ can be substituted in the equation $(1)$ to obtain the value of $x$ .

  $x=\frac{{\left(\pm \sqrt{\frac{19\pm \sqrt{217}}{3}}\right)}^2-4}{\pm \sqrt{\frac{19\pm \sqrt{217}}{3}}}$

  $=\frac{\frac{19\pm \sqrt{217}-12}{3}}{\pm \sqrt{\frac{19\pm \sqrt{217}}{3}}}$

  $=\frac{7\pm \sqrt{217}}{\pm \sqrt{3\left(19\pm \sqrt{217}\right)}}$

Hence, the solution for the given equations are $\left(\frac{7\pm \sqrt{217}}{\pm \sqrt{3\left(19\pm \sqrt{217}\right)}},\pm \sqrt{\frac{19\pm \sqrt{217}}{3}}\right)$