Hi Jakob F
Did you notice that you have a given substitution that you can use?
If you write your equations directly as given you have an immediate substitution you can use to solve for B
Let M = muffins
Let B = Bananas
Let C = Cereal
Equation 1
2M + B + C = 10
Equation 2
M + 3B + C = 16.5
Equation 3
2M + C = 6
Equation 1 and Equation 3 both contain 2M + C which is 6
If you substitute 6 for 2M + C you can easily and quickly find B
B = 10 - 6 = 4
Plug in 4 for B in Equation 2 you will have another equation in M and C
M + 3B + C = 16.5
M + 3(4) + C = 16.5
M +12 + C = 16.5
M + C + 16.5 - 12
M + C = 4.5
Subtract M + C = 4.5 from 2M + C = 6 to solve for M
2M + C = 6
-(M + C = 4.5)
Gives
2M + C = 6
-M - C = -4.5
M = 1.5
You can use the information above to solve for C and check by plugging all the values back into the original equations. Of course it should agree with the elimination method used above as well.