## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( \frac{19}{29},-\frac{11}{29} \right),\left( 1,1 \right) \right\}$
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, $\left\{ \left( \frac{19}{29},-\frac{11}{29} \right),\left( 1,1 \right) \right\}$.