Given the system of equations. We have:
- For (5,0)
Substitute x = 5 and y = 0 in the equations:
![\begin{gathered} 5(5)+4(0)=-1 \\ 25+0=-1 \\ 25\\e-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/alz3ph3qvkzol3dodjmgo69otpgk9p67wa.png)
Answer: No
- For (-7, -8)
x = -7 and y = -8
![\begin{gathered} 5(-7)+4(-8)=-1 \\ -35-32=-1 \\ -67\\e-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/rraky371taqcdjdts7zakh8mp8ld8g6y4c.png)
Answer: No
- For (3, -4)
![\begin{gathered} 5(3)+4(-4)=-1 \\ 15-16=-1 \\ -1=-1 \\ \text{and} \\ 3(3)-2(-4)=-5 \\ 9+8=-5 \\ 17\\e-5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/r0m42yasjjncf7wlotd1tnqwryfzmrtjoi.png)
This is satisfied for the first equation but not for the second. Therefore it is not a solution to the system.
Answer: No
- For (-2, -2)
![\begin{gathered} 5(-2)+4(-2)=-1 \\ -10-8=-1 \\ -18\\e-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lwyi1g6ylofkme19a9bglld58y3zvhfzat.png)
Answer: No