Bill has $2.00 in quarters and dimes. the number of quarters is 4 less than twice the number of dimes. find the number of coins of each type

Define your variables:
Q: number of quarters
D: number of dimes

Next, create your equations
Q = 2D-4
25Q+10D=200

Because one variable (Q) is already written in terms of the other variable, D, you can already start solving with substitution.
25(2D-4)+10D=200
50D-100+10D=200
60D=300
D=5

Substitute D back into the other equation to find Q. 
Q=2D-4
Q=2(5)-4
Q=6

Substitute D and Q into the second equation to check.
25Q+10D=200
25(6)+10(5)=200
150+50=200

This works, so the answer is right! Bill has 6 quarters and 5 dimes.

x-1 multiple 2x-1 = f(-x)