1.3 Solve simultaneously for \( x \) and \( y \) : \( 4 x^{2}+y=7 \) and \( 3 x^{2}+2 x y=y^{2} \)

Solve the system of equations \( 4x^2+y=7;3x^2+2xy=y^2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}4x^{2}+y=7\\3x^{2}+2xy=y^{2}\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=7-4x^{2}\\3x^{2}+2xy=y^{2}\end{array}\right.\) - step2: Substitute the value of \(y:\) \(3x^{2}+2x\left(7-4x^{2}\right)=\left(7-4x^{2}\right)^{2}\) - step3: Expand the expression: \(3x^{2}+14x-8x^{3}=\left(7-4x^{2}\right)^{2}\) - step4: Expand the expression: \(3x^{2}+14x-8x^{3}=49-56x^{2}+16x^{4}\) - step5: Move the expression to the left side: \(3x^{2}+14x-8x^{3}-\left(49-56x^{2}+16x^{4}\right)=0\) - step6: Calculate: \(59x^{2}+14x-8x^{3}-49-16x^{4}=0\) - step7: Factor the expression: \(\left(-1+x\right)\left(7+4x\right)\left(x+7-4x^{2}\right)=0\) - step8: Separate into possible cases: \(-1+x=0\cup 7+4x=0\cup x+7-4x^{2}=0\) - step9: Solve the equation: \(x=1\cup x=-\frac{7}{4}\cup x=\frac{1+\sqrt{113}}{8}\cup x=\frac{1-\sqrt{113}}{8}\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=1\\y=7-4x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{7}{4}\\y=7-4x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1+\sqrt{113}}{8}\\y=7-4x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1-\sqrt{113}}{8}\\y=7-4x^{2}\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{7}{4}\\y=-\frac{21}{4}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1+\sqrt{113}}{8}\\y=-\frac{1+\sqrt{113}}{8}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1-\sqrt{113}}{8}\\y=\frac{-1+\sqrt{113}}{8}\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-\frac{7}{4}\\y=-\frac{21}{4}\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1+\sqrt{113}}{8}\\y=-\frac{1+\sqrt{113}}{8}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1-\sqrt{113}}{8}\\y=\frac{-1+\sqrt{113}}{8}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{7}{4}\\y=-\frac{21}{4}\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1+\sqrt{113}}{8}\\y=-\frac{1+\sqrt{113}}{8}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1-\sqrt{113}}{8}\\y=\frac{-1+\sqrt{113}}{8}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-\frac{7}{4},-\frac{21}{4}\right)\cup \left(x,y\right) = \left(1,3\right)\cup \left(x,y\right) = \left(\frac{1+\sqrt{113}}{8},-\frac{1+\sqrt{113}}{8}\right)\cup \left(x,y\right) = \left(\frac{1-\sqrt{113}}{8},\frac{-1+\sqrt{113}}{8}\right)\) The solutions to the system of equations \(4x^{2}+y=7\) and \(3x^{2}+2xy=y^{2}\) are: 1. \(x = -\frac{7}{4}, y = -\frac{21}{4}\) 2. \(x = 1, y = 3\) 3. \(x = \frac{1+\sqrt{113}}{8}, y = -\frac{1+\sqrt{113}}{8}\) 4. \(x = \frac{1-\sqrt{113}}{8}, y = \frac{-1+\sqrt{113}}{8}\)