Is (−2, 6) a solution to the system of these linear equations: x+2y=10 and 3x+y=0
? Why?
Responses
Yes, because the graphs don’t intersect at (−2, 6)
.
Yes, because the graphs don’t intersect at (−2, 6)
No, because the graphs intersect at (−2, 6)
.
No, because the graphs intersect at (−2, 6)
No, because the graphs don’t intersect at (−2, 6)
.
No, because the graphs don’t intersect at (−2, 6)
Yes, because the graphs intersect at (−2, 6)
.
No, because the point (−2, 6) does not satisfy both equations in the system.
When x = -2 and y = 6, the first equation x + 2y = 10 becomes -2 + 2(6) = 10, which is false.
Therefore, the point (−2, 6) is not a solution to the system of linear equations.
wrong
I apologize for the error. Let's check the point (−2, 6) in both equations of the system:
For the first equation x + 2y = 10:
-2 + 2(6) = -2 + 12 = 10. This equation holds true for the point (−2, 6).
For the second equation 3x + y = 0:
3(-2) + 6 = -6 + 6 = 0. This equation also holds true for the point (−2, 6).
Therefore, the point (−2, 6) is indeed a solution to the system of linear equations. Thank you for pointing out the mistake, and I appreciate your understanding.