Question

matching match the system of equations with its solution. write the letter of the answer in the correspo

1. {y = x² - 5x - 3, y = -7x - 3}
2. {y = x² + 6x + 9, y = 3x + 19}
3. {y = (x - 1)² - 5, y = -4x - 4}
4. {y = x² - x + 3, y = 3x + 3}
5. {y = -3x² + 9x, y = -3x + 9}
6. {y = 0.25x² + 2, y = 3}
7. {y = -(x + 2)², y = -2x - 3}
8. {y = -4x² + 9, y = -4x + 1}
9. {y = x² + 4x + 7, y = 4x + 8}
10. {y = -10x² + 4, 20x + y = 14}
11. {y = (x + 3)² - 5, y = 11x}
12. {y = 2(x + 3)² + 3, y = 6x + 17}
13. {y = (x - 5)² + 10, y = -x + 17}
14. {y = -2x² - 8x, y = 8x + 30}
15. {y = -2(x + 2)² + 2, y = 2}
16. {y = -3x², y = -3x}
17. {y = x² + 4x - 2, y = -9x - 42}
18. {y = (x - 2)², y = -x + 4}
19. {y = x², y = 2x - 1}
20. {y + 10x + 12 = x², 16x + y = -20}
21. {y = 2x² + 15, y = 6x + 15}
22. {y + x² = 3x + 2, y = 7x + 6}
23. {y + 5x + 4 = x², y + 14 = 2x}
24. {y = 6x² - 8, 6x + y = 4}
25. {y = -1/2(x - 3)² + 4, y = 2}

Explanation:

Step1: Set the two equations equal

Since in each system $y$ is expressed in two different ways, we set the right - hand sides equal to each other. For example, for the system $\begin{cases}y = x^{2}-5x - 3\\y=-7x - 3\end{cases}$, we have $x^{2}-5x - 3=-7x - 3$.

Step2: Rearrange to form a quadratic equation

For $x^{2}-5x - 3=-7x - 3$, we move all terms to one side: $x^{2}-5x + 7x-3 + 3 = 0$, which simplifies to $x^{2}+2x=0$. Factor out $x$: $x(x + 2)=0$. So $x = 0$ or $x=-2$.

Step3: Find the corresponding $y$ - values

When $x = 0$ in $y=-7x - 3$, $y=-3$. When $x=-2$ in $y=-7x - 3$, $y=-7\times(-2)-3=14 - 3 = 11$.
We repeat these steps for each system of equations:
1. For $\begin{cases}y = x^{2}-5x - 3\\y=-7x - 3\end{cases}$:
- Set $x^{2}-5x - 3=-7x - 3$, then $x^{2}+2x = 0$, $x(x + 2)=0$, $x = 0$ or $x=-2$.
- When $x = 0$, $y=-3$; when $x=-2$, $y = 11$.
2. For $\begin{cases}y = x^{2}+6x + 9\\y=3x + 19\end{cases}$:
- Set $x^{2}+6x + 9=3x + 19$, $x^{2}+3x - 10=0$. Factor: $(x + 5)(x - 2)=0$. So $x=-5$ or $x = 2$.
- When $x=-5$, $y=3\times(-5)+19 = 4$; when $x = 2$, $y=3\times2+19=25$.
3. For $\begin{cases}y=(x - 1)^{2}-5\\y=-4x - 4\end{cases}$:
- Expand $(x - 1)^{2}-5=x^{2}-2x+1 - 5=x^{2}-2x - 4$. Set $x^{2}-2x - 4=-4x - 4$, $x^{2}+2x=0$, $x(x + 2)=0$, $x = 0$ or $x=-2$.
- When $x = 0$, $y=-4$; when $x=-2$, $y=0$.
4. For $\begin{cases}y = x^{2}-x + 3\\y=3x + 3\end{cases}$:
- Set $x^{2}-x + 3=3x + 3$, $x^{2}-4x=0$, $x(x - 4)=0$, $x = 0$ or $x = 4$.
- When $x = 0$, $y=3$; when $x = 4$, $y=3\times4+3 = 15$.
5. For $\begin{cases}y=-3x^{2}+9x\\y=-3x + 9\end{cases}$:
- Set $-3x^{2}+9x=-3x + 9$, $-3x^{2}+12x - 9=0$, divide by $-3$ to get $x^{2}-4x + 3=0$. Factor: $(x - 1)(x - 3)=0$. So $x = 1$ or $x = 3$.
- When $x = 1$, $y=-3\times1+9 = 6$; when $x = 3$, $y=-3\times3+9 = 0$.
6. For $\begin{cases}y = 0.25x^{2}+2\\y=3\end{cases}$:
- Set $0.25x^{2}+2=3$, $0.25x^{2}=1$, $x^{2}=4$, $x=\pm2$.
7. For $\begin{cases}y=-(x + 2)^{2}\\y=-2x - 3\end{cases}$:
- Expand $-(x + 2)^{2}=-x^{2}-4x - 4$. Set $-x^{2}-4x - 4=-2x - 3$, $-x^{2}-2x - 1=0$, or $x^{2}+2x + 1=0$, $(x + 1)^{2}=0$, $x=-1$. Then $y=-2\times(-1)-3=-1$.
8. For $\begin{cases}y=-4x^{2}+9\\y=-4x + 1\end{cases}$:
- Set $-4x^{2}+9=-4x + 1$, $-4x^{2}+4x + 8=0$, divide by $-4$ to get $x^{2}-x - 2=0$. Factor: $(x - 2)(x+1)=0$. So $x = 2$ or $x=-1$.
- When $x = 2$, $y=-4\times2 + 1=-7$; when $x=-1$, $y=-4\times(-1)+1 = 5$.
9. For $\begin{cases}y = x^{2}+4x + 7\\y=4x + 8\end{cases}$:
- Set $x^{2}+4x + 7=4x + 8$, $x^{2}=1$, $x=\pm1$.
10. For $\begin{cases}y=-10x^{2}+4\\20x + y=14\end{cases}$:
- Substitute $y=14 - 20x$ into $y=-10x^{2}+4$, $14 - 20x=-10x^{2}+4$, $10x^{2}-20x + 10=0$, $x^{2}-2x + 1=0$, $(x - 1)^{2}=0$, $x = 1$, $y=14-20\times1=-6$.
11. For $\begin{cases}y=(x + 3)^{2}-5\\y=11x\end{cases}$:
- Expand $(x + 3)^{2}-5=x^{2}+6x + 9 - 5=x^{2}+6x + 4$. Set $x^{2}+6x + 4=11x$, $x^{2}-5x + 4=0$. Factor: $(x - 1)(x - 4)=0$. So $x = 1$ or $x = 4$.
12. For $\begin{cases}y=2(x + 3)^{2}+3\\y=6x + 17\end{cases}$:
- Expand $2(x + 3)^{2}+3=2(x^{2}+6x + 9)+3=2x^{2}+12x+18 + 3=2x^{2}+12x + 21$. Set $2x^{2}+12x + 21=6x + 17$, $2x^{2}+6x + 4=0$, $x^{2}+3x + 2=0$, $(x + 1)(x + 2)=0$. So $x=-1$ or $x=-2$.
13. For $\begin{cases}y=(x - 5)^{2}+10\\y=-x + 17\end{cases}$:
- Expand $(x - 5)^{2}+10=x^{2}-10x + 25+10=x^{2}-10x + 35$. Set $x^{2}-10x + 35=-x + 17$, $x^{2}-9x + 18=0$, $(x - 3)(x - 6)=0$. So $x = 3$ or $x = 6$.
14. For $\begin{cases}y=-2x^{2}-8x\\y=8x + 30\end{cases}$:
- Set $-2x^{2}-8x=8x + 30$, $-2x^{2}-16x - 30=0$, $x^{2}+8x + 15=0$, $(x + 3)(x + 5)=0$. So $x=-3$ or $x=-5$.
15. For $\begin{cases}y=-2(x + 2)^{2}+2\\y=2\end{cases}$:
- Set $-2(x + 2)^{2}+2=2$, $-2(x + 2)^{2}=0$, $(x + 2)^{2}=0$, $x=-2$.
16. For $\begin{cases}y=-3x^{2}\\y=-3x\end{cases}$:
- Set $-3x^{2}=-3x$, $-3x^{2}+3x=0$, $-3x(x - 1)=0$. So $x = 0$ or $x = 1$.
17. For $\begin{cases}y = x^{2}+4x - 2\\y=-9x - 42\end{cases}$:
- Set $x^{2}+4x - 2=-9x - 42$, $x^{2}+13x + 40=0$, $(x + 5)(x + 8)=0$. So $x=-5$ or $x=-8$.
18. For $\begin{cases}y=(x - 2)^{2}\\y=-x + 4\end{cases}$:
- Expand $(x - 2)^{2}=x^{2}-4x + 4$. Set $x^{2}-4x + 4=-x + 4$, $x^{2}-3x=0$, $x(x - 3)=0$. So $x = 0$ or $x = 3$.
19. For $\begin{cases}y = x^{2}\\y=2x - 1\end{cases}$:
- Set $x^{2}=2x - 1$, $x^{2}-2x + 1=0$, $(x - 1)^{2}=0$, $x = 1$, $y=1$.
20. For $\begin{cases}y + 10x+12=x^{2}\\16x + y=-20\end{cases}$:
- Substitute $y=-20 - 16x$ into $y + 10x+12=x^{2}$, $-20 - 16x+10x+12=x^{2}$, $x^{2}+6x + 8=0$, $(x + 2)(x + 4)=0$. So $x=-2$ or $x=-4$.
21. For $\begin{cases}y = 2x^{2}+15\\y=6x + 15\end{cases}$:
- Set $2x^{2}+15=6x + 15$, $2x^{2}-6x=0$, $2x(x - 3)=0$. So $x = 0$ or $x = 3$.
22. For $\begin{cases}y+x^{2}=3x + 2\\y=7x + 6\end{cases}$:
- Substitute $y=7x + 6$ into $y+x^{2}=3x + 2$, $7x + 6+x^{2}=3x + 2$, $x^{2}+4x + 4=0$, $(x + 2)^{2}=0$, $x=-2$, $y=7\times(-2)+6=-8$.
23. For $\begin{cases}y + 5x+4=x^{2}\\y + 14=2x\end{cases}$:
- Substitute $y=2x - 14$ into $y + 5x+4=x^{2}$, $2x-14 + 5x+4=x^{2}$, $x^{2}-7x + 10=0$, $(x - 2)(x - 5)=0$. So $x = 2$ or $x = 5$.
24. For $\begin{cases}y = 6x^{2}-8\\6x + y=4\end{cases}$:
- Substitute $y=4 - 6x$ into $y = 6x^{2}-8$, $4 - 6x=6x^{2}-8$, $6x^{2}+6x - 12=0$, $x^{2}+x - 2=0$, $(x + 2)(x - 1)=0$. So $x=-2$ or $x = 1$.
25. For $\begin{cases}y=-\frac{1}{2}(x - 3)^{2}+4\\y=2\end{cases}$:
- Set $-\frac{1}{2}(x - 3)^{2}+4=2$, $-\frac{1}{2}(x - 3)^{2}=-2$, $(x - 3)^{2}=4$, $x - 3=\pm2$, $x = 1$ or $x = 5$.

Since you haven't provided the list of solutions to match these systems with, we can't give a final matching answer. But the general method to solve each system of equations (a quadratic and a linear equation) is as shown above. If you provide the list of solutions, we can match them accordingly.

Answer:

Step1: Set the two equations equal

Since in each system $y$ is expressed in two different ways, we set the right - hand sides equal to each other. For example, for the system

$$\begin{cases}y = x^{2}-5x - 3\\y=-7x - 3\end{cases}$$

, we have $x^{2}-5x - 3=-7x - 3$.

Step2: Rearrange to form a quadratic equation

For $x^{2}-5x - 3=-7x - 3$, we move all terms to one side: $x^{2}-5x + 7x-3 + 3 = 0$, which simplifies to $x^{2}+2x=0$. Factor out $x$: $x(x + 2)=0$. So $x = 0$ or $x=-2$.

Step3: Find the corresponding $y$ - values

When $x = 0$ in $y=-7x - 3$, $y=-3$. When $x=-2$ in $y=-7x - 3$, $y=-7\times(-2)-3=14 - 3 = 11$.
We repeat these steps for each system of equations:

1. For

$$\begin{cases}y = x^{2}-5x - 3\\y=-7x - 3\end{cases}$$

:

• Set $x^{2}-5x - 3=-7x - 3$, then $x^{2}+2x = 0$, $x(x + 2)=0$, $x = 0$ or $x=-2$.
• When $x = 0$, $y=-3$; when $x=-2$, $y = 11$.

1. For

$$\begin{cases}y = x^{2}+6x + 9\\y=3x + 19\end{cases}$$

:

• Set $x^{2}+6x + 9=3x + 19$, $x^{2}+3x - 10=0$. Factor: $(x + 5)(x - 2)=0$. So $x=-5$ or $x = 2$.
• When $x=-5$, $y=3\times(-5)+19 = 4$; when $x = 2$, $y=3\times2+19=25$.

1. For

$$\begin{cases}y=(x - 1)^{2}-5\\y=-4x - 4\end{cases}$$

:

• Expand $(x - 1)^{2}-5=x^{2}-2x+1 - 5=x^{2}-2x - 4$. Set $x^{2}-2x - 4=-4x - 4$, $x^{2}+2x=0$, $x(x + 2)=0$, $x = 0$ or $x=-2$.
• When $x = 0$, $y=-4$; when $x=-2$, $y=0$.

1. For

$$\begin{cases}y = x^{2}-x + 3\\y=3x + 3\end{cases}$$

:

• Set $x^{2}-x + 3=3x + 3$, $x^{2}-4x=0$, $x(x - 4)=0$, $x = 0$ or $x = 4$.
• When $x = 0$, $y=3$; when $x = 4$, $y=3\times4+3 = 15$.

1. For

$$\begin{cases}y=-3x^{2}+9x\\y=-3x + 9\end{cases}$$

:

• Set $-3x^{2}+9x=-3x + 9$, $-3x^{2}+12x - 9=0$, divide by $-3$ to get $x^{2}-4x + 3=0$. Factor: $(x - 1)(x - 3)=0$. So $x = 1$ or $x = 3$.
• When $x = 1$, $y=-3\times1+9 = 6$; when $x = 3$, $y=-3\times3+9 = 0$.

1. For

$$\begin{cases}y = 0.25x^{2}+2\\y=3\end{cases}$$

:

• Set $0.25x^{2}+2=3$, $0.25x^{2}=1$, $x^{2}=4$, $x=\pm2$.

1. For

$$\begin{cases}y=-(x + 2)^{2}\\y=-2x - 3\end{cases}$$

:

• Expand $-(x + 2)^{2}=-x^{2}-4x - 4$. Set $-x^{2}-4x - 4=-2x - 3$, $-x^{2}-2x - 1=0$, or $x^{2}+2x + 1=0$, $(x + 1)^{2}=0$, $x=-1$. Then $y=-2\times(-1)-3=-1$.

1. For

$$\begin{cases}y=-4x^{2}+9\\y=-4x + 1\end{cases}$$

:

• Set $-4x^{2}+9=-4x + 1$, $-4x^{2}+4x + 8=0$, divide by $-4$ to get $x^{2}-x - 2=0$. Factor: $(x - 2)(x+1)=0$. So $x = 2$ or $x=-1$.
• When $x = 2$, $y=-4\times2 + 1=-7$; when $x=-1$, $y=-4\times(-1)+1 = 5$.

1. For

$$\begin{cases}y = x^{2}+4x + 7\\y=4x + 8\end{cases}$$

:

• Set $x^{2}+4x + 7=4x + 8$, $x^{2}=1$, $x=\pm1$.

1. For

$$\begin{cases}y=-10x^{2}+4\\20x + y=14\end{cases}$$

:

• Substitute $y=14 - 20x$ into $y=-10x^{2}+4$, $14 - 20x=-10x^{2}+4$, $10x^{2}-20x + 10=0$, $x^{2}-2x + 1=0$, $(x - 1)^{2}=0$, $x = 1$, $y=14-20\times1=-6$.

1. For

$$\begin{cases}y=(x + 3)^{2}-5\\y=11x\end{cases}$$

:

• Expand $(x + 3)^{2}-5=x^{2}+6x + 9 - 5=x^{2}+6x + 4$. Set $x^{2}+6x + 4=11x$, $x^{2}-5x + 4=0$. Factor: $(x - 1)(x - 4)=0$. So $x = 1$ or $x = 4$.

1. For

$$\begin{cases}y=2(x + 3)^{2}+3\\y=6x + 17\end{cases}$$

:

• Expand $2(x + 3)^{2}+3=2(x^{2}+6x + 9)+3=2x^{2}+12x+18 + 3=2x^{2}+12x + 21$. Set $2x^{2}+12x + 21=6x + 17$, $2x^{2}+6x + 4=0$, $x^{2}+3x + 2=0$, $(x + 1)(x + 2)=0$. So $x=-1$ or $x=-2$.

1. For

$$\begin{cases}y=(x - 5)^{2}+10\\y=-x + 17\end{cases}$$

:

• Expand $(x - 5)^{2}+10=x^{2}-10x + 25+10=x^{2}-10x + 35$. Set $x^{2}-10x + 35=-x + 17$, $x^{2}-9x + 18=0$, $(x - 3)(x - 6)=0$. So $x = 3$ or $x = 6$.

1. For

$$\begin{cases}y=-2x^{2}-8x\\y=8x + 30\end{cases}$$

:

• Set $-2x^{2}-8x=8x + 30$, $-2x^{2}-16x - 30=0$, $x^{2}+8x + 15=0$, $(x + 3)(x + 5)=0$. So $x=-3$ or $x=-5$.

1. For

$$\begin{cases}y=-2(x + 2)^{2}+2\\y=2\end{cases}$$

:

• Set $-2(x + 2)^{2}+2=2$, $-2(x + 2)^{2}=0$, $(x + 2)^{2}=0$, $x=-2$.

1. For

$$\begin{cases}y=-3x^{2}\\y=-3x\end{cases}$$

:

• Set $-3x^{2}=-3x$, $-3x^{2}+3x=0$, $-3x(x - 1)=0$. So $x = 0$ or $x = 1$.

1. For

$$\begin{cases}y = x^{2}+4x - 2\\y=-9x - 42\end{cases}$$

:

• Set $x^{2}+4x - 2=-9x - 42$, $x^{2}+13x + 40=0$, $(x + 5)(x + 8)=0$. So $x=-5$ or $x=-8$.

1. For

$$\begin{cases}y=(x - 2)^{2}\\y=-x + 4\end{cases}$$

:

• Expand $(x - 2)^{2}=x^{2}-4x + 4$. Set $x^{2}-4x + 4=-x + 4$, $x^{2}-3x=0$, $x(x - 3)=0$. So $x = 0$ or $x = 3$.

1. For

$$\begin{cases}y = x^{2}\\y=2x - 1\end{cases}$$

:

• Set $x^{2}=2x - 1$, $x^{2}-2x + 1=0$, $(x - 1)^{2}=0$, $x = 1$, $y=1$.

1. For

$$\begin{cases}y + 10x+12=x^{2}\\16x + y=-20\end{cases}$$

:

• Substitute $y=-20 - 16x$ into $y + 10x+12=x^{2}$, $-20 - 16x+10x+12=x^{2}$, $x^{2}+6x + 8=0$, $(x + 2)(x + 4)=0$. So $x=-2$ or $x=-4$.

1. For

$$\begin{cases}y = 2x^{2}+15\\y=6x + 15\end{cases}$$

:

• Set $2x^{2}+15=6x + 15$, $2x^{2}-6x=0$, $2x(x - 3)=0$. So $x = 0$ or $x = 3$.

1. For

$$\begin{cases}y+x^{2}=3x + 2\\y=7x + 6\end{cases}$$

:

• Substitute $y=7x + 6$ into $y+x^{2}=3x + 2$, $7x + 6+x^{2}=3x + 2$, $x^{2}+4x + 4=0$, $(x + 2)^{2}=0$, $x=-2$, $y=7\times(-2)+6=-8$.

1. For

$$\begin{cases}y + 5x+4=x^{2}\\y + 14=2x\end{cases}$$

:

• Substitute $y=2x - 14$ into $y + 5x+4=x^{2}$, $2x-14 + 5x+4=x^{2}$, $x^{2}-7x + 10=0$, $(x - 2)(x - 5)=0$. So $x = 2$ or $x = 5$.

1. For

$$\begin{cases}y = 6x^{2}-8\\6x + y=4\end{cases}$$

:

• Substitute $y=4 - 6x$ into $y = 6x^{2}-8$, $4 - 6x=6x^{2}-8$, $6x^{2}+6x - 12=0$, $x^{2}+x - 2=0$, $(x + 2)(x - 1)=0$. So $x=-2$ or $x = 1$.

1. For

$$\begin{cases}y=-\frac{1}{2}(x - 3)^{2}+4\\y=2\end{cases}$$

:

• Set $-\frac{1}{2}(x - 3)^{2}+4=2$, $-\frac{1}{2}(x - 3)^{2}=-2$, $(x - 3)^{2}=4$, $x - 3=\pm2$, $x = 1$ or $x = 5$.

Since you haven't provided the list of solutions to match these systems with, we can't give a final matching answer. But the general method to solve each system of equations (a quadratic and a linear equation) is as shown above. If you provide the list of solutions, we can match them accordingly.