Answer
$(1,1), \left(\dfrac{19}{29},-\dfrac{11}{29}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
4x-y=3\\
3x^2-2y^2=1
\end{cases}$
We will use the substitution method. Solve Equation 1 for $y$ and substitute the expression of $y$ in Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
y=4x-3\\
3x^2-2(4x-3)^2=1
\end{cases}$
$3x^2-2(16x^2-24x+9)=1$
$3x^2-32x^2+48x-18=1$
$-29x^2+48x-19=0$
$29x^2-48x+19=0$
$29x^2-29x-19x+19=0$
$29x(x-1)-19(x-1)=0$
$(x-1)(29x-19)=0$
$x-1=0\Rightarrow x_1=1$
$29x-19=0\Rightarrow x_2=\dfrac{19}{29}$
Substitute each of the values of $x$ in the expression of $y$ to determine $y$:
$y=4x-3$
$x_1=1\Rightarrow y_1=4(1)-3=1$
$x_2=\dfrac{19}{29}\Rightarrow y_2=4\left(\dfrac{19}{29}\right)-3=-\dfrac{11}{29}$
The system's solutions are:
$(1,1), \left(\dfrac{19}{29},-\dfrac{11}{29}\right)$
