Solve Simaltoinecusly \( 3 x-4 y=5 \) and \( 2 x^{2}-5 x y+3 y^{2}=4 \)

Solve the system of equations \( 3x-4y=5;2x^{2}-5xy+3y^{2}=4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x-4y=5\\2x^{2}-5xy+3y^{2}=4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{5+4y}{3}\\2x^{2}-5xy+3y^{2}=4\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(\frac{5+4y}{3}\right)^{2}-5\times \frac{5+4y}{3}\times y+3y^{2}=4\) - step3: Simplify: \(\frac{50}{9}+\frac{80}{9}y+\frac{59}{9}y^{2}-\frac{5\left(5+4y\right)y}{3}=4\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{50}{9}+\frac{80}{9}y+\frac{59}{9}y^{2}-\frac{5\left(5+4y\right)y}{3}\right)\times 9=4\times 9\) - step5: Simplify the equation: \(50+5y-y^{2}=36\) - step6: Move the expression to the left side: \(50+5y-y^{2}-36=0\) - step7: Subtract the numbers: \(14+5y-y^{2}=0\) - step8: Factor the expression: \(\left(7-y\right)\left(2+y\right)=0\) - step9: Separate into possible cases: \(\begin{align}&7-y=0\\&2+y=0\end{align}\) - step10: Solve the equation: \(\begin{align}&y=7\\&y=-2\end{align}\) - step11: Calculate: \(y=7\cup y=-2\) - step12: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{5+4y}{3}\\y=7\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5+4y}{3}\\y=-2\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=11\\y=7\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=-2\end{array}\right.\) - step14: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=11\\y=7\end{array}\right.\) - step15: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=11\\y=7\end{array}\right.\) - step16: Rewrite: \(\left(x,y\right) = \left(-1,-2\right)\cup \left(x,y\right) = \left(11,7\right)\) The solutions to the system of equations \(3x-4y=5\) and \(2x^{2}-5xy+3y^{2}=4\) are \((x,y) = (-1,-2)\) and \((x,y) = (11,7)\).