Chapter 11, Problem 22CT

To determine

To find: The solution of the system of equations.

Answer to Problem 22CT

The solutions of the system of equations are $\left(1,2\right)$ and $\left(3,4\right)$ .

Explanation of Solution

Given information:

The system of the equations is

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

Calculation:

Consider the system of the equations as

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

Calculate the value of $y$ from second equation.

  $\begin{array}{c}y-x=1\\ y=1+x\end{array}$

Substitute the value in the first equation.

  $\begin{array}{c}2{\left(1+x\right)}^2-3x^2=5\\ -x^2+4x+2-5=0\\ x^2-4x+3=0\\ x=1,3\end{array}$

Calculate the values of $y$ .

  $\begin{array}{c}y=1+1\\ =2\end{array}$

And

  $\begin{array}{c}y=1+3\\ =4\end{array}$

Therefore, the solutions of the system of equations are $\left(1,2\right)$ and $\left(3,4\right)$ .