Answer
The solution is $\left\{ \left( \frac{19}{29},-\frac{11}{29} \right),\left( 1,1 \right) \right\}$
Work Step by Step
The given equations are,
$\begin{align}
& 3{{x}^{2}}-2{{y}^{2}}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& 4x-y=3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\
\end{align}$
For solving the equations, use the substitution method:
Solve the equation (ii) for y:
$ y=4x-3$ (III)
Put the value of $ y=4x-3$ into equation (I):
By using the formula ${{\left( A-B \right)}^{2}}={{A}^{2}}+{{B}^{2}}-2AB $
$\begin{align}
& 3{{x}^{2}}-2{{\left( 4x-3 \right)}^{2}}=1 \\
& 3{{x}^{2}}-2\left( 16{{x}^{2}}-24x+9 \right)-1=0 \\
& 3{{x}^{2}}-32{{x}^{2}}+48x-18-1=0
\end{align}$
Finally,
$-29{{x}^{2}}+48x-19=0$ (IV)
By multiplying equation (IV) by $\left( -1 \right)$ on both sides: we get,
$\begin{align}
& 29{{x}^{2}}-48x+19=0 \\
& 29{{x}^{2}}-29x-19x+19=0 \\
& 29x\left( x-1 \right)-19(x-1)=0 \\
& \left( 29x-19 \right)\left( x-1 \right)=0
\end{align}$
Finally,
$29x-19=0$ or $ x-1=0$
$29x=19$ or $ x=1$
$ x=\frac{19}{29}$ or $ x=1$
If $ x=\frac{19}{29}$ then substitute $ x=\frac{19}{29}$ in equation (III):
$\begin{align}
& y=4\left( \frac{19}{29} \right)-3 \\
& y=-\frac{11}{29} \\
\end{align}$
If $ x=1$ then substitute $ x=1$ in equation (III):
$\begin{align}
& y=4\left( 1 \right)-3 \\
& y=1 \\
\end{align}$
Therefore, the solution set is,
$\left\{ \left( \frac{19}{29},-\frac{11}{29} \right),\left( 1,1 \right) \right\}$
Hence, the solution set of the system is, $\left\{ \left( \frac{19}{29},-\frac{11}{29} \right),\left( 1,1 \right) \right\}$.
