Chapter 9.7, Problem 23PPS

To determine

To find: The solution for the given system of the equations.

Answer to Problem 23PPS

The solution for the given equation is $\left(\sqrt{2},4\right),\left(-\sqrt{2},4\right)$ .

Explanation of Solution

Given:

The given system of equation is,

  $\begin{array}{l}{y}^2-x^2+3y=26\\ 2y^2+x^2=34\end{array}$

Calculation:

Consider the given equation are,

  $\begin{array}{l}{y}^2-x^2+3y=26\\ 2y^2+x^2=34\end{array}$

Consider the second equation as,

  $x^2=34-2y^2$

Then substitute the above equation in the first equation as,

  $\begin{array}{l}{y}^2-\left(34-2y^2\right)+3y=26\\ y^2-34+2y^2+3y=26\\ 3y^2+3y-60=0\\ y^2+y-20=0\\ y^2+5y-4y-20=0\\ \left(y+5\right)\left(y-4\right)=0\\ y=-5,4\end{array}$

Substitute $y=-5$ in $x^2=34-2y^2$ .

  $\begin{array}{l}{x}^2=34-2{\left(-5\right)}^2\\ x^2=34-50\\ x^2=-16\\ x=\pm \sqrt{-16}\end{array}$

No real solution for $y=-5$ .

Substitute $y=4$ in $x^2=34-2y^2$ .

  $\begin{array}{l}{x}^2=34-2{\left(4\right)}^2\\ x^2=34-32\\ x^2=2\\ x=\pm \sqrt{2}\end{array}$

Thus, the solution of the equations are $\left(\sqrt{2},4\right),\left(-\sqrt{2},4\right)$ .