EXERCISE 11 Solve simultaneously for \( x \) and \( y \) in each equatio \( 1 y=3 x^{2}-2 x-8 \) and \( y=5 x-2 \) \( 2 y=-5 x^{2}+4 x+9 \) and \( y+6 x=9 \) \( 3 y=\frac{3}{4} x^{2}+12 \) and \( 3 x+2 y=12 \) \( 4 \quad y=-\frac{1}{2} x^{2}-\frac{1}{2} x+6 \) and \( x+2 y=8 \) \( 5 \quad y=-\frac{1}{2}(x-1)(x+6) \) and \( 2 x+5 y=5 \) \( 6, y=x^{2}-5 x+4 \) and \( y-2 x+6=0 \) \( 7 \quad y=-x^{2}-5 x+6 \) and \( y=\frac{1}{2}(x+1)^{2}-2 \) \( 8 \quad y=(x-2)^{2}-3 \) and \( y=-\frac{1}{2}(x-5)^{2}+6 \) \( 9 y=-x^{2}-x+4 \) and \( y=2(x-1)^{2}-4 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=x^{2}-5x+4\\y-2x+6=0\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x^{2}-5x+4-2x+6=0\) - step2: Simplify: \(x^{2}-7x+10=0\) - step3: Factor the expression: \(\left(x-5\right)\left(x-2\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x-5=0\\&x-2=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=5\\&x=2\end{align}\) - step6: Calculate: \(x=5\cup x=2\) - step7: Rearrange the terms: \(\left\{ \begin{array}{l}x=5\\y=x^{2}-5x+4\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=x^{2}-5x+4\end{array}\right.\) - step8: Calculate: \(\left\{ \begin{array}{l}x=5\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=4\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=4\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(2,-2\right)\cup \left(x,y\right) = \left(5,4\right)\) Solve the system of equations \( y=-\frac{1}{2} x^{2}-\frac{1}{2} x+6;x+2 y=8 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-\frac{1}{2}x^{2}-\frac{1}{2}x+6\\x+2y=8\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x+2\left(-\frac{1}{2}x^{2}-\frac{1}{2}x+6\right)=8\) - step2: Simplify: \(-x^{2}+12=8\) - step3: Move the constant to the right side: \(-x^{2}=8-12\) - step4: Subtract the numbers: \(-x^{2}=-4\) - step5: Change the signs: \(x^{2}=4\) - step6: Simplify the expression: \(x=\pm \sqrt{4}\) - step7: Simplify: \(x=\pm 2\) - step8: Separate into possible cases: \(x=2\cup x=-2\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=2\\y=-\frac{1}{2}x^{2}-\frac{1}{2}x+6\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-\frac{1}{2}x^{2}-\frac{1}{2}x+6\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=5\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-2,5\right)\cup \left(x,y\right) = \left(2,3\right)\) Solve the system of equations \( y=-5 x^{2}+4 x+9;y+6 x=9 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-5x^{2}+4x+9\\y+6x=9\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-5x^{2}+4x+9+6x=9\) - step2: Add the terms: \(-5x^{2}+10x+9=9\) - step3: Cancel equal terms: \(-5x^{2}+10x=0\) - step4: Factor the expression: \(-5x\left(x-2\right)=0\) - step5: Separate into possible cases: \(\begin{align}&-5x=0\\&x-2=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=0\\&x=2\end{align}\) - step7: Calculate: \(x=0\cup x=2\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=0\\y=-5x^{2}+4x+9\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-5x^{2}+4x+9\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=0\\y=9\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-3\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=9\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-3\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(0,9\right)\cup \left(x,y\right) = \left(2,-3\right)\) Solve the system of equations \( y=-x^{2}-x+4;y=2(x-1)^{2}-4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-x^{2}-x+4\\y=2\left(x-1\right)^{2}-4\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-x^{2}-x+4=2\left(x-1\right)^{2}-4\) - step2: Simplify: \(-x^{2}-x+4=2x^{2}-4x-2\) - step3: Move the expression to the left side: \(-x^{2}-x+4-\left(2x^{2}-4x-2\right)=0\) - step4: Calculate: \(-3x^{2}+3x+6=0\) - step5: Factor the expression: \(-3\left(x-2\right)\left(x+1\right)=0\) - step6: Divide the terms: \(\left(x-2\right)\left(x+1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x-2=0\\&x+1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=2\\&x=-1\end{align}\) - step9: Calculate: \(x=2\cup x=-1\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=2\\y=-x^{2}-x+4\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=-x^{2}-x+4\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=4\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=-2\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-1,4\right)\cup \left(x,y\right) = \left(2,-2\right)\) Solve the system of equations \( y=3 x^{2}-2 x-8;y=5 x-2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=3x^{2}-2x-8\\y=5x-2\end{array}\right.\) - step1: Substitute the value of \(y:\) \(3x^{2}-2x-8=5x-2\) - step2: Move the expression to the left side: \(3x^{2}-2x-8-\left(5x-2\right)=0\) - step3: Subtract the terms: \(3x^{2}-7x-6=0\) - step4: Factor the expression: \(\left(x-3\right)\left(3x+2\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x-3=0\\&3x+2=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=3\\&x=-\frac{2}{3}\end{align}\) - step7: Calculate: \(x=3\cup x=-\frac{2}{3}\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=3\\y=3x^{2}-2x-8\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2}{3}\\y=3x^{2}-2x-8\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=3\\y=13\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2}{3}\\y=-\frac{16}{3}\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-\frac{2}{3}\\y=-\frac{16}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=13\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{2}{3}\\y=-\frac{16}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=13\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-\frac{2}{3},-\frac{16}{3}\right)\cup \left(x,y\right) = \left(3,13\right)\) Solve the system of equations \( y=\frac{3}{4} x^{2}+12;3 x+2 y=12 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=\frac{3}{4}x^{2}+12\\3x+2y=12\end{array}\right.\) - step1: Substitute the value of \(y:\) \(3x+2\left(\frac{3}{4}x^{2}+12\right)=12\) - step2: Expand the expression: \(3x+\frac{3}{2}x^{2}+24=12\) - step3: Move the expression to the left side: \(3x+\frac{3}{2}x^{2}+24-12=0\) - step4: Subtract the numbers: \(3x+\frac{3}{2}x^{2}+12=0\) - step5: Rewrite in standard form: \(\frac{3}{2}x^{2}+3x+12=0\) - step6: Multiply both sides: \(2\left(\frac{3}{2}x^{2}+3x+12\right)=2\times 0\) - step7: Calculate: \(3x^{2}+6x+24=0\) - step8: Solve using the quadratic formula: \(x=\frac{-6\pm \sqrt{6^{2}-4\times 3\times 24}}{2\times 3}\) - step9: Simplify the expression: \(x=\frac{-6\pm \sqrt{6^{2}-4\times 3\times 24}}{6}\) - step10: Simplify the expression: \(x=\frac{-6\pm \sqrt{-252}}{6}\) - step11: The expression is undefined: \(x \notin \mathbb{R}\) - step12: The system of equations has no solution in the set of real numbers: \(\left(x,y\right) \notin \mathbb{R}^{2}\) - step13: Alternative Form: \(\textrm{No real solution}\) Solve the system of equations \( y=-\frac{1}{2}(x-1)(x+6);2 x+5 y=5 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-\frac{1}{2}\left(x-1\right)\left(x+6\right)\\2x+5y=5\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}y=\left(-\frac{1}{2}x+\frac{1}{2}\right)\left(x+6\right)\\2x+5y=5\end{array}\right.\) - step2: Substitute the value of \(y:\) \(2x+5\left(-\frac{1}{2}x+\frac{1}{2}\right)\left(x+6\right)=5\) - step3: Simplify: \(-\frac{21}{2}x-\frac{5}{2}x^{2}+15=5\) - step4: Move the expression to the left side: \(-\frac{21}{2}x-\frac{5}{2}x^{2}+15-5=0\) - step5: Subtract the numbers: \(-\frac{21}{2}x-\frac{5}{2}x^{2}+10=0\) - step6: Factor the expression: \(\frac{1}{2}\left(-x-5\right)\left(5x-4\right)=0\) - step7: Divide the terms: \(\left(-x-5\right)\left(5x-4\right)=0\) - step8: Separate into possible cases: \(\begin{align}&-x-5=0\\&5x-4=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=-5\\&x=\frac{4}{5}\end{align}\) - step10: Calculate: \(x=-5\cup x=\frac{4}{5}\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=-5\\y=\left(-\frac{1}{2}x+\frac{1}{2}\right)\left(x+6\right)\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{4}{5}\\y=\left(-\frac{1}{2}x+\frac{1}{2}\right)\left(x+6\right)\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-5\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{4}{5}\\y=\frac{17}{25}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-5\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{4}{5}\\y=\frac{17}{25}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-5,3\right)\cup \left(x,y\right) = \left(\frac{4}{5},\frac{17}{25}\right)\) Solve the system of equations \( y=-x^{2}-5 x+6;y=\frac{1}{2}(x+1)^{2}-2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-x^{2}-5x+6\\y=\frac{1}{2}\left(x+1\right)^{2}-2\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-x^{2}-5x+6=\frac{1}{2}\left(x+1\right)^{2}-2\) - step2: Simplify: \(-x^{2}-5x+6=\frac{1}{2}x^{2}+x-\frac{3}{2}\) - step3: Move the expression to the left side: \(-x^{2}-5x+6-\left(\frac{1}{2}x^{2}+x-\frac{3}{2}\right)=0\) - step4: Calculate: \(-\frac{3}{2}x^{2}-6x+\frac{15}{2}=0\) - step5: Factor the expression: \(-\frac{3}{2}\left(x-1\right)\left(x+5\right)=0\) - step6: Divide the terms: \(\left(x-1\right)\left(x+5\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x-1=0\\&x+5=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=1\\&x=-5\end{align}\) - step9: Calculate: \(x=1\cup x=-5\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=1\\y=-x^{2}-5x+6\end{array}\right.\cup \left\{ \begin{array}{l}x=-5\\y=-x^{2}-5x+6\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-5\\y=6\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-5\\y=6\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-5\\y=6\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-5,6\right)\cup \left(x,y\right) = \left(1,0\right)\) Solve the system of equations \( y=(x-2)^{2}-3;y=-\frac{1}{2}(x-5)^{2}+6 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=\left(x-2\right)^{2}-3\\y=-\frac{1}{2}\left(x-5\right)^{2}+6\end{array}\right.\) - step1: Substitute the value of \(y:\) \(\left(x-2\right)^{2}-3=-\frac{1}{2}\left(x-5\right)^{2}+6\) - step2: Simplify: \(\left(x-2\right)^{2}-3=-\frac{1}{2}x^{2}+5x-\frac{13}{2}\) - step3: Move the expression to the left side: \(\left(x-2\right)^{2}-3-\left(-\frac{1}{2}x^{2}+5x-\frac{13}{2}\right)=0\) - step4: Subtract the terms: \(\left(x-2\right)^{2}+\frac{7}{2}+\frac{1}{2}x^{2}-5x=0\) - step5: Calculate: \(\frac{3}{2}x^{2}-9x+\frac{15}{2}=0\) - step6: Factor the expression: \(\frac{3}{2}\left(x-5\right)\left(x-1\right)=0\) - step7: Divide the terms: \(\left(x-5\right)\left(x-1\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-5=0\\&x-1=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=5\\&x=1\end{align}\) - step10: Calculate: \(x=5\cup x=1\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=5\\y=\left(x-2\right)^{2}-3\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=\left(x-2\right)^{2}-3\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=5\\y=6\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-2\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=6\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=6\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(1,-2\right)\cup \left(x,y\right) = \left(5,6\right)\) Here are the solutions for each pair of equations: 1. For the equations \( y = 3x^2 - 2x - 8 \) and \( y = 5x - 2 \): - The solutions are \( (x,y) = \left(-\frac{2}{3}, -\frac{16}{3}\right) \) and \( (x,y) = (3, 13) \). 2. For the equations \( y = -5x^2 + 4x + 9 \) and \( y + 6x = 9 \): - The solutions are \( (x,y) = (0, 9) \) and \( (x,y) = (2, -3) \). 3. For the equations \( y = \frac{3}{4}x^2 + 12 \) and \( 3x + 2y = 12 \): - There are no real solutions. 4. For the equations \( y = -\frac{1}{2}x^2 - \frac{1}{2}x + 6 \) and \( x + 2y = 8 \): - The solutions are \( (x,y) = (-2, 5) \) and \( (x,y) = (2, 3) \). 5. For the equations \( y = -\frac{1}{2}(x-1)(x+6) \) and \( 2x + 5y = 5 \): - The solutions are \( (x,y) = (-5, 3) \) and \( (x,y) = \left(\frac{4}{5}, \frac{17}{25}\right) \). 6. For the equations \( y = x^2 - 5x + 4 \) and \( y - 2x + 6 = 0 \): - The solutions are \( (x,y) = (2, -2) \) and \( (x,y) = (5, 4) \). 7. For the equations \( y = -x^2 - 5x + 6 \) and \( y = \frac{1}{2}(x+1)^2 - 2 \): - The solutions are \( (x,y) = (-5, 6) \) and \( (x,y) = (1, 0) \). 8. For the equations \( y = (x-2)^2 - 3 \) and \( y = -\frac{1}{2}(x-5)^2 + 6 \): - The solutions are \( (x,y) = (1, -2) \) and \( (x,y) = (5, 6) \). 9. For the equations \( y = -x^2 - x + 4 \) and \( y = 2(x-1)^2 - 4 \): - The solutions are \( (x,y) = (-1, 4) \) and \( (x,y) = (2, -2) \). If you need further assistance or explanations for any specific pair, feel free to ask!