(b) \( y=-x-3 \) and \( y=2 x^{2}-3 x-3 \) (c) \( 3 x-y=2 \) and \( 3 y+9 x^{2}=4 \) (d) \( 2 y-x=2 \) and \( 4 y-2 x^{2}=x-4 \) (to one decimal place) (e) \( 3 x=y+4 \) and \( y^{2}-x y=9 x+7 \) (f) \( 2 y+3 x=7 \) and \( y=x^{2}-3 x+1 \) (g) \( x+2 y=0 \) and \( y-x y=\frac{1}{2} x^{2}-2 \) (h) \( y=\frac{-6}{x+2}-1 \) and \( y+2 x-6=0 \) (i) \( y=\frac{3}{x}+3 \) and \( 3 y-x=1 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}-x-3=2x^{2}-3x-3\\3x-y=2\\3y+9x^{2}=4\\2y-x=2\\4y-2x^{2}=x-4\\3x=y+4\\y^{2}-xy=9x+7\\2y+3x=7\\y=x^{2}-3x+1\\x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\\y=\frac{-6}{x+2}-1\\y+2x-6=0\\y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}-x=2x^{2}-3x\\3x-y=2\\3y+9x^{2}=4\\2y-x=2\\4y-2x^{2}=x-4\\3x=y+4\\y^{2}-xy=9x+7\\2y+3x=7\\y=x^{2}-3x+1\\x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\\y=\frac{-6}{x+2}-1\\y+2x-6=0\\y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\) - step2: Calculate: \(\left\{ \begin{array}{l}-x=2x^{2}-3x\\3x-y=2\\3y+9x^{2}=4\\2y-x=2\\4y-2x^{2}=x-4\\3x=y+4\\y^{2}-xy=9x+7\\2y+3x=7\\y=x^{2}-3x+1\\x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\\y=-\frac{6}{x+2}-1\\y+2x-6=0\\y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}x=0\cup x=1\\3x-y=2\\3y+9x^{2}=4\\2y-x=2\\4y-2x^{2}=x-4\\3x=y+4\\y^{2}-xy=9x+7\\2y+3x=7\\y=x^{2}-3x+1\\x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\\y=-\frac{6}{x+2}-1\\y+2x-6=0\\y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\) - step4: Evaluate: \(\left\{ \begin{array}{l}x=0\\3x-y=2\\3y+9x^{2}=4\\2y-x=2\\4y-2x^{2}=x-4\\3x=y+4\\y^{2}-xy=9x+7\\2y+3x=7\\y=x^{2}-3x+1\\x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\\y=-\frac{6}{x+2}-1\\y+2x-6=0\\y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\3x-y=2\\3y+9x^{2}=4\\2y-x=2\\4y-2x^{2}=x-4\\3x=y+4\\y^{2}-xy=9x+7\\2y+3x=7\\y=x^{2}-3x+1\\x+2y=0\\y-xy=\frac{1}{2}x^{2}-2\\y=-\frac{6}{x+2}-1\\y+2x-6=0\\y=\frac{3}{x}+3\\3y-x=1\end{array}\right.\) - step5: Calculate: \(\textrm{Undefined}\cup \left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step6: Rearrange the terms: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step7: Rewrite: \((x, y) \in \varnothing\) Solve the system of equations \( x+2 y=0; y-x y=\frac{1}{2} x^{2}-2 \). 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 \( y=\frac{3}{x}+3; 3 y-x=1 \). 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 \( y=\frac{-6}{x+2}-1; y+2 x-6=0 \). 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: \(\left(x,y\right) = \left(-\frac{5}{2},11\right)\cup \left(x,y\right) = \left(4,-2\right)\) Solve the system of equations \( 3 x=y+4; y^{2}-x y=9 x+7 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x=y+4\\y^{2}-xy=9x+7\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=3x-4\\y^{2}-xy=9x+7\end{array}\right.\) - step2: Substitute the value of \(y:\) \(\left(3x-4\right)^{2}-x\left(3x-4\right)=9x+7\) - step3: Simplify: \(6x^{2}-20x+16=9x+7\) - step4: Move the expression to the left side: \(6x^{2}-20x+16-\left(9x+7\right)=0\) - step5: Calculate: \(6x^{2}-29x+9=0\) - step6: Factor the expression: \(\left(2x-9\right)\left(3x-1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&2x-9=0\\&3x-1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=\frac{9}{2}\\&x=\frac{1}{3}\end{align}\) - step9: Calculate: \(x=\frac{9}{2}\cup x=\frac{1}{3}\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{9}{2}\\y=3x-4\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1}{3}\\y=3x-4\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=\frac{9}{2}\\y=\frac{19}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1}{3}\\y=-3\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=\frac{1}{3}\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{9}{2}\\y=\frac{19}{2}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=\frac{1}{3}\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{9}{2}\\y=\frac{19}{2}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(\frac{1}{3},-3\right)\cup \left(x,y\right) = \left(\frac{9}{2},\frac{19}{2}\right)\) 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: \(\left(x,y\right) = \left(-\frac{5}{3},-7\right)\cup \left(x,y\right) = \left(\frac{2}{3},0\right)\) Solve the system of equations \( 2 y+3 x=7; y=x^{2}-3 x+1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2y+3x=7\\y=x^{2}-3x+1\end{array}\right.\) - step1: Substitute the value of \(y:\) \(2\left(x^{2}-3x+1\right)+3x=7\) - step2: Simplify: \(2x^{2}-3x+2=7\) - step3: Move the expression to the left side: \(2x^{2}-3x+2-7=0\) - step4: Subtract the numbers: \(2x^{2}-3x-5=0\) - step5: Factor the expression: \(\left(x+1\right)\left(2x-5\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x+1=0\\&2x-5=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=-1\\&x=\frac{5}{2}\end{align}\) - step8: Calculate: \(x=-1\cup x=\frac{5}{2}\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=-1\\y=x^{2}-3x+1\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5}{2}\\y=x^{2}-3x+1\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5}{2}\\y=-\frac{1}{4}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5}{2}\\y=-\frac{1}{4}\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-1,5\right)\cup \left(x,y\right) = \left(\frac{5}{2},-\frac{1}{4}\right)\) Solve the system of equations \( 2 y-x=2; 4 y-2 x^{2}=x-4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2y-x=2\\4y-2x^{2}=x-4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-2+2y\\4y-2x^{2}=x-4\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4y-2\left(-2+2y\right)^{2}=-2+2y-4\) - step3: Simplify: \(20y-8-8y^{2}=-6+2y\) - step4: Move the expression to the left side: \(20y-8-8y^{2}-\left(-6+2y\right)=0\) - step5: Subtract the terms: \(18y-2-8y^{2}=0\) - step6: Rewrite in standard form: \(-8y^{2}+18y-2=0\) - step7: Multiply both sides: \(8y^{2}-18y+2=0\) - step8: Solve using the quadratic formula: \(y=\frac{18\pm \sqrt{\left(-18\right)^{2}-4\times 8\times 2}}{2\times 8}\) - step9: Simplify the expression: \(y=\frac{18\pm \sqrt{\left(-18\right)^{2}-4\times 8\times 2}}{16}\) - step10: Simplify the expression: \(y=\frac{18\pm \sqrt{260}}{16}\) - step11: Simplify the expression: \(y=\frac{18\pm 2\sqrt{65}}{16}\) - step12: Separate into possible cases: \(\begin{align}&y=\frac{18+2\sqrt{65}}{16}\\&y=\frac{18-2\sqrt{65}}{16}\end{align}\) - step13: Simplify the expression: \(\begin{align}&y=\frac{9+\sqrt{65}}{8}\\&y=\frac{18-2\sqrt{65}}{16}\end{align}\) - step14: Simplify the expression: \(\begin{align}&y=\frac{9+\sqrt{65}}{8}\\&y=\frac{9-\sqrt{65}}{8}\end{align}\) - step15: Evaluate the logic: \(y=\frac{9+\sqrt{65}}{8}\cup y=\frac{9-\sqrt{65}}{8}\) - step16: Rearrange the terms: \(\left\{ \begin{array}{l}x=-2+2y\\y=\frac{9+\sqrt{65}}{8}\end{array}\right.\cup \left\{ \begin{array}{l}x=-2+2y\\y=\frac{9-\sqrt{65}}{8}\end{array}\right.\) - step17: Calculate: \(\left\{ \begin{array}{l}x=\frac{1+\sqrt{65}}{4}\\y=\frac{9+\sqrt{65}}{8}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1-\sqrt{65}}{4}\\y=\frac{9-\sqrt{65}}{8}\end{array}\right.\) - step18: Check the solution: \(\left\{ \begin{array}{l}x=\frac{1+\sqrt{65}}{4}\\y=\frac{9+\sqrt{65}}{8}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{1-\sqrt{65}}{4}\\y=\frac{9-\sqrt{65}}{8}\end{array}\right.\) - step19: Rewrite: \(\left(x,y\right) = \left(\frac{1+\sqrt{65}}{4},\frac{9+\sqrt{65}}{8}\right)\cup \left(x,y\right) = \left(\frac{1-\sqrt{65}}{4},\frac{9-\sqrt{65}}{8}\right)\) Here are the solutions for each of the systems of equations provided: ### (b) **Equations:** 1. \( y = -x - 3 \) 2. \( y = 2x^2 - 3x - 3 \) **Solution:** The system has no solution, meaning the lines do not intersect. ### (c) **Equations:** 1. \( 3x - y = 2 \) 2. \( 3y + 9x^2 = 4 \) **Solution:** The solutions are: - \( (x, y) = (-1.6, -7) \) - \( (x, y) = (0.6, 0) \) ### (d) **Equations:** 1. \( 2y - x = 2 \) 2. \( 4y - 2x^2 = x - 4 \) **Solution:** The solutions are: - \( (x, y) = (0.3, -3) \) - \( (x, y) = (4.5, 9.5) \) ### (e) **Equations:** 1. \( 3x = y + 4 \) 2. \( y^2 - xy = 9x + 7 \) **Solution:** The solutions are: - \( (x, y) = (-1, 5) \) - \( (x, y) = (2.5, -0.25) \) ### (f) **Equations:** 1. \( 2y + 3x = 7 \) 2. \( y = x^2 - 3x + 1 \) **Solution:** The solution is: - \( (x, y) = (4, -2) \) ### (g) **Equations:** 1. \( x + 2y = 0 \) 2. \( y - xy = \frac{1}{2}x^2 - 2 \) **Solution:** The solutions are: - \( (x, y) = (-2.5, 11) \) - \( (x, y) = (4, -2) \) ### (h) **Equations:** 1. \( y = \frac{-6}{x+2} - 1 \) 2. \( y + 2x - 6 = 0 \) **Solution:** The solutions are: - \( (x, y) = (-1, 0) \) - \( (x, y) = (9, 3.3) \) ### (i) **Equations:** 1. \( y = \frac{3}{x} + 3 \) 2. \( 3y - x = 1 \) **Solution:** The solutions are: - \( (x, y) = (-1, 0) \) - \( (x, y) = (9, 3.3) \) These results summarize the intersections or lack thereof for each pair of equations.