Chapter 11, Problem 72RE

To determine

Solve each system of equations.

Answer to Problem 72RE

  $\boxed{\left(\pm 2\sqrt{2},\pm \sqrt{3}\right)}$

Explanation of Solution

Given information:

Solve each system of equations.

  $\{\begin{array}{l}3x^2-y^2=1\\ 7x^2-2y^2-5=0\end{array}$

Calculation:

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

  $\{\begin{array}{l}3x^2-y^2=1\\ 7x^2-2y^2-5=0\end{array}$

The given system of equations can be solved by the method of elimination. The first equation cam be multiplied by $2$ and second equation can be multiplied by $1$ in order make the coefficient of $y^2$ same in both the equations.

  $\{\begin{array}{l}\left(3x^2-y^2=1\right)\times 2\\ \left(7x^2-2y^2-5=0\right)\times 1\end{array}$

  $\{\begin{array}{l}6x^2-2y^2=2\\ 7x^2-2y^2-5=0\end{array}$

The equation can now be solved for $x$ since the coefficient of $y^2$ is same in both the equations.

  $6x^2-2y^2-7x^2+2y^2=2-5$

  $\begin{array}{l}-x^2=-3\\ x^2=3\\ x=\pm \sqrt{3}\end{array}$

The value of $y$ can be determined by substituting the obtained value of $x$ .

  $6{(\pm \sqrt{3})}^2-2y^2=2$

  $6\times 3-2y^2=2$

  $2y^2=18-2$

  $y^2=\frac{16}{2}$

  $y=\pm 2\sqrt{2}$

Hence, the solution to the given system of equations is $\left(\pm 2\sqrt{2},\pm \sqrt{3}\right)$