Question 75171
Let x=number of quarters
Then 30-x=number of dimes

(Lets deal in pennies to keep down confusion and simplify the problem)

Now we are told that:
25x+10(30-x)>520  get rid of parens
25x+300-10x>520  subtract 300 from both sides

25x-10x+300-300>520-300 collect like terms

15x>220  divide both sides by 15
x>14.667 This answer tells us that we have to round up to the nearest whole number because the number of quarters cannot be less than 14.667. So: 
x>=15  which means there's at least 15 quarters

If there's at least 15 quarters, then there's at most 15 dimes.

x>=15  multiply both sides by -1 (must reverse inequality sign)
-x<=-15  add 30 to both sides
30-x<=30-15 so
30-x<=15-------------------which means there's at least 15 dimes

CK
15*10+15*25>520 
150+375>520
525>520

now lets try 14 quarters and 16 dimes----this should not work
14*25+16*10>520
350+160>520
510 is not greater than 520


Hope this helps----ptaylor