Use A for third graders, B for fourth graders, and C for fifth graders.
33 = A + B + C
Since there are three variables, you need to come up with two other equations to substitute in.
C = 6 + A Since there are six more fifth graders than third graders.
A = B Since there is an equal amount of third and fourth graders.
Since A is in both equations, lets put each variable in terms of A.
B = A
C = 6 + A
So:
A+A+A+6=33
A+A+A=27 Since you subtract six from both sides.
3A=27 Since you combine all three A's.
A=9 Since you divide both sides by three.
Therefore, since A is the number of third graders, there are 9 third graders.
Linda · 7 years ago
Bob B
Use T for third graders, F for fourth graders and X for fifth graders and write what you know as equations:
T + F + X = 33 [There are 33 students in the band.]
X = T + 6 [6 more 5th graders than 3rd graders.]
T = F [Equal # of 3rd & 4th graders]
Since T equals F and X equals T + 6, substitute those into the first equation:
T + T + (T + 6) = 33
Combine into:
3T + 6 = 33
Subtract 6 from both sides:
3T = 27
Divide both sides by 3:
T = 9
There are 9 third graders.