Algebraically you could represent it like this
2*5 + x*3 + y*4 = 25 Multiply 2*5
10 + 3x + 4y = 25 Subtract 10 from both sides.
3x + 4y = 15 Now try this. Let y = 2
3x + 4*2 = 15
3x + 8 = 15 Subtract 8 from both sides.
3x = 15 - 8
3x = 7 Ordinarily you would divide by 3 but the question doesn't let you.
7 is a neat number. You could use 4 + 3
So one answer is
2 fives.
3 Fours
1 Three
You could also do
2 fives
5 threes. Nothing says you must have a four.
You can also do.
3 fives
1 four
2 threes There is nothing to say you can't have another 5
