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 $$ y=2(x-1)^{2}-4 $$

Question

UpStudy AI · Accepted Answer

Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}y=x^{2}-5x+4\y-2x+6=0\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$x^{2}-5x+4-2x+6=0$$
- step2: Simplify:
  $$x^{2}-7x+10=0$$
- step3: Factor the expression:
  $$\left(x-5ight)\left(x-2ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=5\y=x^{2}-5x+4\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=x^{2}-5x+4\end{array}ight.$$
- step8: Calculate:
  $$\left\{ \begin{array}{l}x=5\y=4\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=-2\end{array}ight.$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=5\y=4\end{array}ight.$$
- step10: Check the solution:
  $$\left\{ \begin{array}{l}x=2\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=5\y=4\end{array}ight.$$
- step11: Rewrite:
  $$\left(x,yight) = \left(2,-2ight)\cup \left(x,yight) = \left(5,4ight)$$
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:
  $$\left\{ \begin{array}{l}y=-\frac{1}{2}x^{2}-\frac{1}{2}x+6\x+2y=8\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$x+2\left(-\frac{1}{2}x^{2}-\frac{1}{2}x+6ight)=8$$
- 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:
  $$\left\{ \begin{array}{l}x=2\y=-\frac{1}{2}x^{2}-\frac{1}{2}x+6\end{array}ight.\cup \left\{ \begin{array}{l}x=-2\y=-\frac{1}{2}x^{2}-\frac{1}{2}x+6\end{array}ight.$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=3\end{array}ight.\cup \left\{ \begin{array}{l}x=-2\y=5\end{array}ight.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=-2\y=5\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=3\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=-2\y=5\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=3\end{array}ight.$$
- step13: Rewrite:
  $$\left(x,yight) = \left(-2,5ight)\cup \left(x,yight) = \left(2,3ight)$$
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:
  $$\left\{ \begin{array}{l}y=-5x^{2}+4x+9\y+6x=9\end{array}ight.$$
- 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:
  $$-5x\left(x-2ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=0\y=-5x^{2}+4x+9\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=-5x^{2}+4x+9\end{array}ight.$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}x=0\y=9\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=-3\end{array}ight.$$
- step10: Check the solution:
  $$\left\{ \begin{array}{l}x=0\y=9\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=-3\end{array}ight.$$
- step11: Rewrite:
  $$\left(x,yight) = \left(0,9ight)\cup \left(x,yight) = \left(2,-3ight)$$
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:
  $$\left\{ \begin{array}{l}y=-x^{2}-x+4\y=2\left(x-1ight)^{2}-4\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$-x^{2}-x+4=2\left(x-1ight)^{2}-4$$
- step2: Simplify:
  $$-x^{2}-x+4=2x^{2}-4x-2$$
- step3: Move the expression to the left side:
  $$-x^{2}-x+4-\left(2x^{2}-4x-2ight)=0$$
- step4: Calculate:
  $$-3x^{2}+3x+6=0$$
- step5: Factor the expression:
  $$-3\left(x-2ight)\left(x+1ight)=0$$
- step6: Divide the terms:
  $$\left(x-2ight)\left(x+1ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=2\y=-x^{2}-x+4\end{array}ight.\cup \left\{ \begin{array}{l}x=-1\y=-x^{2}-x+4\end{array}ight.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=-1\y=4\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-1\y=4\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=-2\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-1\y=4\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=-2\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(-1,4ight)\cup \left(x,yight) = \left(2,-2ight)$$
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:
  $$\left\{ \begin{array}{l}y=3x^{2}-2x-8\y=5x-2\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$3x^{2}-2x-8=5x-2$$
- step2: Move the expression to the left side:
  $$3x^{2}-2x-8-\left(5x-2ight)=0$$
- step3: Subtract the terms:
  $$3x^{2}-7x-6=0$$
- step4: Factor the expression:
  $$\left(x-3ight)\left(3x+2ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=3\y=3x^{2}-2x-8\end{array}ight.\cup \left\{ \begin{array}{l}x=-\frac{2}{3}\y=3x^{2}-2x-8\end{array}ight.$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}x=3\y=13\end{array}ight.\cup \left\{ \begin{array}{l}x=-\frac{2}{3}\y=-\frac{16}{3}\end{array}ight.$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{2}{3}\y=-\frac{16}{3}\end{array}ight.\cup \left\{ \begin{array}{l}x=3\y=13\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{2}{3}\y=-\frac{16}{3}\end{array}ight.\cup \left\{ \begin{array}{l}x=3\y=13\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(-\frac{2}{3},-\frac{16}{3}ight)\cup \left(x,yight) = \left(3,13ight)$$
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:
  $$\left\{ \begin{array}{l}y=\frac{3}{4}x^{2}+12\3x+2y=12\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$3x+2\left(\frac{3}{4}x^{2}+12ight)=12$$
- 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:
  $$2\left(\frac{3}{2}x^{2}+3x+12ight)=2	imes 0$$
- step7: Calculate:
  $$3x^{2}+6x+24=0$$
- step8: Solve using the quadratic formula:
  $$x=\frac{-6\pm \sqrt{6^{2}-4	imes 3	imes 24}}{2	imes 3}$$
- step9: Simplify the expression:
  $$x=\frac{-6\pm \sqrt{6^{2}-4	imes 3	imes 24}}{6}$$
- step10: Simplify the expression:
  $$x=\frac{-6\pm \sqrt{-252}}{6}$$
- step11: The expression is undefined:
  $$x 
otin \mathbb{R}$$
- step12: The system of equations has no solution in the set of real numbers:
  $$\left(x,yight) 
otin \mathbb{R}^{2}$$
- step13: Alternative Form:
  $$	extrm{No real solution}$$
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:
  $$\left\{ \begin{array}{l}y=-\frac{1}{2}\left(x-1ight)\left(x+6ight)\2x+5y=5\end{array}ight.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}y=\left(-\frac{1}{2}x+\frac{1}{2}ight)\left(x+6ight)\2x+5y=5\end{array}ight.$$
- step2: Substitute the value of $$y:$$
  $$2x+5\left(-\frac{1}{2}x+\frac{1}{2}ight)\left(x+6ight)=5$$
- 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:
  $$\frac{1}{2}\left(-x-5ight)\left(5x-4ight)=0$$
- step7: Divide the terms:
  $$\left(-x-5ight)\left(5x-4ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=-5\y=\left(-\frac{1}{2}x+\frac{1}{2}ight)\left(x+6ight)\end{array}ight.\cup \left\{ \begin{array}{l}x=\frac{4}{5}\y=\left(-\frac{1}{2}x+\frac{1}{2}ight)\left(x+6ight)\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-5\y=3\end{array}ight.\cup \left\{ \begin{array}{l}x=\frac{4}{5}\y=\frac{17}{25}\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-5\y=3\end{array}ight.\cup \left\{ \begin{array}{l}x=\frac{4}{5}\y=\frac{17}{25}\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(-5,3ight)\cup \left(x,yight) = \left(\frac{4}{5},\frac{17}{25}ight)$$
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:
  $$\left\{ \begin{array}{l}y=-x^{2}-5x+6\y=\frac{1}{2}\left(x+1ight)^{2}-2\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$-x^{2}-5x+6=\frac{1}{2}\left(x+1ight)^{2}-2$$
- step2: Simplify:
  $$-x^{2}-5x+6=\frac{1}{2}x^{2}+x-\frac{3}{2}$$
- step3: Move the expression to the left side:
  $$-x^{2}-5x+6-\left(\frac{1}{2}x^{2}+x-\frac{3}{2}ight)=0$$
- step4: Calculate:
  $$-\frac{3}{2}x^{2}-6x+\frac{15}{2}=0$$
- step5: Factor the expression:
  $$-\frac{3}{2}\left(x-1ight)\left(x+5ight)=0$$
- step6: Divide the terms:
  $$\left(x-1ight)\left(x+5ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=1\y=-x^{2}-5x+6\end{array}ight.\cup \left\{ \begin{array}{l}x=-5\y=-x^{2}-5x+6\end{array}ight.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=0\end{array}ight.\cup \left\{ \begin{array}{l}x=-5\y=6\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-5\y=6\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=0\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-5\y=6\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=0\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(-5,6ight)\cup \left(x,yight) = \left(1,0ight)$$
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:
  $$\left\{ \begin{array}{l}y=\left(x-2ight)^{2}-3\y=-\frac{1}{2}\left(x-5ight)^{2}+6\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$\left(x-2ight)^{2}-3=-\frac{1}{2}\left(x-5ight)^{2}+6$$
- step2: Simplify:
  $$\left(x-2ight)^{2}-3=-\frac{1}{2}x^{2}+5x-\frac{13}{2}$$
- step3: Move the expression to the left side:
  $$\left(x-2ight)^{2}-3-\left(-\frac{1}{2}x^{2}+5x-\frac{13}{2}ight)=0$$
- step4: Subtract the terms:
  $$\left(x-2ight)^{2}+\frac{7}{2}+\frac{1}{2}x^{2}-5x=0$$
- step5: Calculate:
  $$\frac{3}{2}x^{2}-9x+\frac{15}{2}=0$$
- step6: Factor the expression:
  $$\frac{3}{2}\left(x-5ight)\left(x-1ight)=0$$
- step7: Divide the terms:
  $$\left(x-5ight)\left(x-1ight)=0$$
- 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:
  $$\left\{ \begin{array}{l}x=5\y=\left(x-2ight)^{2}-3\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=\left(x-2ight)^{2}-3\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=5\y=6\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=-2\end{array}ight.$$
- step13: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=5\y=6\end{array}ight.$$
- step14: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=5\y=6\end{array}ight.$$
- step15: Rewrite:
  $$\left(x,yight) = \left(1,-2ight)\cup \left(x,yight) = \left(5,6ight)$$
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 $$ (x,y) = \left(-\frac{2}{3}, -\frac{16}{3}ight) $$ 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 $$ (x,y) = \left(\frac{4}{5}, \frac{17}{25}ight) $$.

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!
Here are the solutions for each pair of equations: 1. $$ y = 3x^2 - 2x - 8 $$ and $$ y = 5x - 2 $$: - Solutions: $$ (-\frac{2}{3}, -\frac{16}{3}) $$ and $$ (3, 13) $$. 2. $$ y = -5x^2 + 4x + 9 $$ and $$ y + 6x = 9 $$: - Solutions: $$ (0, 9) $$ and $$ (2, -3) $$. 3. $$ y = \frac{3}{4}x^2 + 12 $$ and $$ 3x + 2y = 12 $$: - No real solutions. 4. $$ y = -\frac{1}{2}x^2 - \frac{1}{2}x + 6 $$ and $$ x + 2y = 8 $$: - Solutions: $$ (-2, 5) $$ and $$ (2, 3) $$. 5. $$ y = -\frac{1}{2}(x-1)(x+6) $$ and $$ 2x + 5y = 5 $$: - Solutions: $$ (-5, 3) $$ and $$ \left(\frac{4}{5}, \frac{17}{25}ight) $$. 6. $$ y = x^2 - 5x + 4 $$ and $$ y - 2x + 6 = 0 $$: - Solutions: $$ (2, -2) $$ and $$ (5, 4) $$. 7. $$ y = -x^2 - 5x + 6 $$ and $$ y = \frac{1}{2}(x+1)^2 - 2 $$: - Solutions: $$ (-5, 6) $$ and $$ (1, 0) $$. 8. $$ y = (x-2)^2 - 3 $$ and $$ y = -\frac{1}{2}(x-5)^2 + 6 $$: - Solutions: $$ (1, -2) $$ and $$ (5, 6) $$. 9. $$ y = -x^2 - x + 4 $$ and $$ y = 2(x-1)^2 - 4 $$: - Solutions: $$ (-1, 4) $$ and $$ (2, -2) $$. If you need more help with any specific pair, let me know!

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