Mex has $2.65. He has only dimes and quarters. If Mex has more quarters than dimes, how many coins does he have in total?
We set up an equation first: 0.25q + 0.10d = 2.65
Multiplying by 100 and dividing by 5, we get 5q + 2d = 53
At this point, we can just use trial and error, noting that q can only be an odd number because when it's even, 3-0 = 3, which is not divisible by 2.
After testing out a few cases, the only pair that satisfies q > d is q=9 and d=4.
Let x be the number of dimes. y be the number of quarters.
y > x.
\(0.10 x + 0.25 y = 2.65\\ 2x + 5y = 53\)
You can find the total number of coins by a trial-and-error approach to find possible pairs of (x, y).