(g) \( x+2 y=0 \quad \) and \( y-x y=\frac{1}{2} x^{2}-2 \) (h) \( y=\frac{-6}{x+2}-1 \) and \( y+2 x-6=0 \) (i) \( y=\frac{3}{x}+3 \quad \) and \( 3 y-x=1 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-2y\\y-xy=\frac{1}{2}x^{2}-2\end{array}\right.\) - step2: Substitute the value of \(x:\) \(y-\left(-2y\times y\right)=\frac{1}{2}\left(-2y\right)^{2}-2\) - step3: Simplify: \(y+2y^{2}=2y^{2}-2\) - step4: Cancel equal terms: \(y=-2\) - step5: Substitute the value of \(y:\) \(x=-2\left(-2\right)\) - step6: Calculate: \(x=4\) - step7: Calculate: \(\left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\) - step8: Check the solution: \(\left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\) - step9: Rewrite: \(\left(x,y\right) = \left(4,-2\right)\) Solve the system of equations \( y=\frac{3}{x}+3; 3y-x=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\) - step1: Substitute the value of \(y:\) \(3\left(\frac{3}{x}+3\right)-x=1\) - step2: Multiply the terms: \(\frac{9+9x}{x}-x=1\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{9+9x}{x}-x\right)x=1\times x\) - step4: Simplify the equation: \(9+9x-x^{2}=x\) - step5: Move the expression to the left side: \(9+9x-x^{2}-x=0\) - step6: Subtract the terms: \(9+8x-x^{2}=0\) - step7: Factor the expression: \(\left(9-x\right)\left(1+x\right)=0\) - step8: Separate into possible cases: \(\begin{align}&9-x=0\\&1+x=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=9\\&x=-1\end{align}\) - step10: Calculate: \(x=9\cup x=-1\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=9\\y=\frac{3}{x}+3\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=\frac{3}{x}+3\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=9\\y=\frac{10}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=0\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=9\\y=\frac{10}{3}\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=9\\y=\frac{10}{3}\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-1,0\right)\cup \left(x,y\right) = \left(9,\frac{10}{3}\right)\) Solve the system of equations \( y=\frac{-6}{x+2}-1; y+2x-6=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=\frac{-6}{x+2}-1\\y+2x-6=0\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}y=-\frac{6}{x+2}-1\\y+2x-6=0\end{array}\right.\) - step2: Substitute the value of \(y:\) \(-\frac{6}{x+2}-1+2x-6=0\) - step3: Simplify: \(-\frac{6}{x+2}-7+2x=0\) - step4: Multiply both sides of the equation by LCD: \(\left(-\frac{6}{x+2}-7+2x\right)\left(x+2\right)=0\times \left(x+2\right)\) - step5: Simplify the equation: \(-20-3x+2x^{2}=0\) - step6: Factor the expression: \(\left(-4+x\right)\left(5+2x\right)=0\) - step7: Separate into possible cases: \(\begin{align}&-4+x=0\\&5+2x=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=4\\&x=-\frac{5}{2}\end{align}\) - step9: Calculate: \(x=4\cup x=-\frac{5}{2}\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=4\\y=-\frac{6}{x+2}-1\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5}{2}\\y=-\frac{6}{x+2}-1\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5}{2}\\y=11\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-\frac{5}{2}\\y=11\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{5}{2}\\y=11\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=-2\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-\frac{5}{2},11\right)\cup \left(x,y\right) = \left(4,-2\right)\) Here are the solutions for the given systems of equations: ### (g) Equations: 1. \( x + 2y = 0 \) 2. \( y - xy = \frac{1}{2}x^2 - 2 \) **Solution:** The solution to this system is: \[ (x, y) = (4, -2) \] ### (h) Equations: 1. \( y = \frac{-6}{x+2} - 1 \) 2. \( y + 2x - 6 = 0 \) **Solution:** The solutions to this system are: \[ (x, y) = \left(-\frac{5}{2}, 11\right) \cup (x, y) = (4, -2) \] This means there are two pairs of solutions: one at \((-2.5, 11)\) and another at \((4, -2)\). ### (i) Equations: 1. \( y = \frac{3}{x} + 3 \) 2. \( 3y - x = 1 \) **Solution:** The solutions to this system are: \[ (x, y) = (-1, 0) \cup (x, y) = (9, \frac{10}{3}) \] This means there are two pairs of solutions: one at \((-1, 0)\) and another at \((9, 3.\dot{3})\). If you have any further questions or need additional assistance, feel free to ask!