Thread: setting up equations

If anyone can help me with this problem, that would be great. Thanks in advance.

I can't seem to set up this problem correctly.

Twice as many dimes as quarters
3 times as many nickles as quarters
how many dimes are there?
Total of 54 coins


This is what I came up with

Q+N+D=54
2d=Q
3n=Q
Q+3n+2d=54?

It's the other way round. Think of it this way.

D is bigger than Q. You need to multiply Q by 2 to get the equal amount of D.

So,

D = 2Q

Similarly,

N = 3Q

And

Q + N + D = 54

From there, put everything in Q.

Q + 3Q + 2Q = 54

Solve for Q. The rest will come easily.


Think in terms of "what should I do to X to make it equal to Y?"
It helped me a lot, maybe it can hel you too!