What is the solution to the system of 3x+10y=-47 5x+7y=40 a. (1,-5) b. (1,5) c. (-1,-5) d. (-1,5) h e l p

First of all, we can solve this by elimination (adding the 2 equations and eliminating 1 variable).
3x+10y=-47
5x+7y=40
It looks like we can eliminate the x by making one 15x, and the other -15x (15 is the least common multiple [LCM] of 3 and 5).
5 (3x+10y=-47) --> 15x+50y=-235
-3 (5x+7y=40) --> -15x-21y=-120
15x-15x = 0, 50y-21y = 29y, -235-120 = -355
0 + 29y = -355
29y/29 = -355/29
y = 12.24
then plugged that into 3x+10y=-47
3x+122.4=-47
3x+122.4-122.4=-47-122.4
3x = -169.4
x = 56.47
Something is wrong... are you sure the 2nd equation wasn't equal to NEGATIVE 40??
Or the 1st one equal to POSITIVE 40??