Answer
$\{(1,2),(3,4)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2y^2-3x^2=5\\
y-x=1
\end{cases}$
Substitute $y=x+1$ from the second equation into the first:
$2(x+1)^2-3x^2=5$
$2(x^2+2x+1)-3x^2=5$
$2x^2+4x+2-3x^2-5=0$
$-x^2+4x-3=0$
$x^2-4x+3=0$
Solve the equation:
$x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}=\dfrac{4\pm 2}{2}$
$x_1=\dfrac{4-2}{2}=1$
$x_2=\dfrac{4+2}{2}=3$
Determine $y$:
$x_1=1\Rightarrow y_1=1+1=2$
$x_2=3\Rightarrow y_2=3+1=4$
The solution set is:
$\{(1,2),(3,4)\}$
