The first thing you want to do is plug in x and y into both equations:
a(3) + b(4) = 4
b(3) + a(4) = 8
rearrange to line up a’s and b’s
3a + 4b = 4
4a + 3b = 8
now you want to choose a or b and multiply each equation by a number to make them have the same amount of a’s or b’s.
4(3a + 4b = 4) = 12a + 16b = 16
3(4a + 3b = 8) = 12a + 9b = 24
Now we subtract the bottom equation from the top and solve for b:
12a + 16b - (12a + 9b) = 16 - 24
7b = -8
b = -8/7
Now we plug back in for b to one of the original equations:
3a + 4(-8/7) = 4
3a + (-32/7) = 4
3a - (32/7) = 4
3a = 4 + (32/7)
3a = (28/7) + (32/7)
3a = 60/7
a = (60/7)/3 = 20/7.
Finally, plug a and b in together to double check using the second equation.
4a + 3b = 8
4(20/7) + 3(-8/7) = ?
(80/7) - (24/7) = ?
56/7 = 8.
I believe the first one is c
