\( \begin{array}{lll}y-2 x=2 & \text { and } & x^{2}-2 x=3-y \\ y=-x-3 & \text { and } & y=2 x^{2}-3 x-3 \\ 3 x-y=2 & \text { and } & 3 y+9 x^{2}=4\end{array} \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y-2x=2\\x^{2}-2x=3-y\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=2+2x\\x^{2}-2x=3-y\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x^{2}-2x=3-\left(2+2x\right)\) - step3: Simplify: \(x^{2}-2x=1-2x\) - step4: Cancel equal terms: \(x^{2}=1\) - step5: Simplify the expression: \(x=\pm \sqrt{1}\) - step6: Simplify: \(x=\pm 1\) - step7: Separate into possible cases: \(x=1\cup x=-1\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=1\\y=2+2x\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=2+2x\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=1\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=0\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=4\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=4\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-1,0\right)\cup \left(x,y\right) = \left(1,4\right)\) Solve the system of equations \( y=-x-3; y=2 x^{2}-3 x-3 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-x-3\\y=2x^{2}-3x-3\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-x-3=2x^{2}-3x-3\) - step2: Cancel equal terms: \(-x=2x^{2}-3x\) - step3: Move the expression to the left side: \(-x-\left(2x^{2}-3x\right)=0\) - step4: Subtract the terms: \(2x-2x^{2}=0\) - step5: Factor the expression: \(2x\left(1-x\right)=0\) - step6: Separate into possible cases: \(\begin{align}&2x=0\\&1-x=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=0\\&x=1\end{align}\) - step8: Calculate: \(x=0\cup x=1\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=0\\y=-x-3\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-x-3\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=0\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-4\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-4\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(0,-3\right)\cup \left(x,y\right) = \left(1,-4\right)\) Solve the system of equations \( 3 x-y=2; 3 y+9 x^{2}=4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x-y=2\\3y+9x^{2}=4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=-2+3x\\3y+9x^{2}=4\end{array}\right.\) - step2: Substitute the value of \(y:\) \(3\left(-2+3x\right)+9x^{2}=4\) - step3: Expand the expression: \(-6+9x+9x^{2}=4\) - step4: Move the expression to the left side: \(-6+9x+9x^{2}-4=0\) - step5: Subtract the numbers: \(-10+9x+9x^{2}=0\) - step6: Factor the expression: \(\left(-2+3x\right)\left(5+3x\right)=0\) - step7: Separate into possible cases: \(\begin{align}&-2+3x=0\\&5+3x=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=\frac{2}{3}\\&x=-\frac{5}{3}\end{align}\) - step9: Calculate: \(x=\frac{2}{3}\cup x=-\frac{5}{3}\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{2}{3}\\y=-2+3x\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5}{3}\\y=-2+3x\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=\frac{2}{3}\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5}{3}\\y=-7\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-\frac{5}{3}\\y=-7\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{3}\\y=0\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{5}{3}\\y=-7\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2}{3}\\y=0\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-\frac{5}{3},-7\right)\cup \left(x,y\right) = \left(\frac{2}{3},0\right)\) Here are the solutions for the given systems of equations: 1. For the system: \[ \begin{align*} y - 2x &= 2 \\ x^2 - 2x &= 3 - y \end{align*} \] The solutions are: \[ (x,y) = (-1,0) \cup (x,y) = (1,4) \] 2. For the system: \[ \begin{align*} y &= -x - 3 \\ y &= 2x^2 - 3x - 3 \end{align*} \] The solutions are: \[ (x,y) = (0,-3) \cup (x,y) = (1,-4) \] 3. For the system: \[ \begin{align*} 3x - y &= 2 \\ 3y + 9x^2 &= 4 \end{align*} \] The solutions are: \[ (x,y) = \left(-\frac{5}{3}, -7\right) \cup (x,y) = \left(\frac{2}{3}, 0\right) \] Alternatively, in decimal form: \[ (x,y) = (-1.6, -7) \cup (x,y) = (0.6, 0) \] These results provide the pairs \((x,y)\) that satisfy each respective system of equations.