Pennies, Nickels, Dimes, and Quarters Summation of Money Peter has only pennies, Norma only Nickels, Diane only dimes, and Quincy only quarters.  Peter and Norma have the same number of coins, and Diane and Quincy have the same number of coins.  What is the least number of coins they can all have if the sum of their money is \$4.87?   
So far i have $p=n$, and $d=q$ as far as number of coins, but where do i go from here?
 A: This is simple. First since we know that pennies = nickels and dimes = quaters  we shoudl now use varaibles. Let x represent the amount of pennies(also nickels) and y represent dimes (also quarters). So lets create an equation. 
xp + xn + yd + yq = 4.87
now convert the coins into their individual values in cents
x(.01) + x(.05) + y(.10) + y(.25) = 4.87
Factor out common variables and then add
x(.01 + .05)+ y(.10 + .25) = 4.87
x(.06) + y(.35) = 4.87
Now although it looks like we have come to a standstill there is somthing that can be done in order to help solve this question.....Simply solve for y but before this move all decimal places two to the right by multiplying all parts of equation by 100 to make things simpler.
6x + 35y = 487
35y = -6x + 487
y = -6x/35 + 487/35
As you can see we have now obtained a linear equation that can be graphed in a ti-83 graphing calulator(im assuming u have one). After it graphed simply look at the table and the chart you will notice that there are no visible integers that come in the y1 section. This is how you now that you donot have to solve for y but for x instead. 
IN linear equations it is standard to solve for a y value. However, in this question we randomly set up x and y so for teh second equaton we must switch the variables. In reality the options are.......
option 1 ) y = -6x/35 + 487/35 (we tried this it doesn't work)
option 2) 6y + 35x = 487 
y = -35x/6 + 487/6   ( Note: once again it is 6y for this option because we must solve for y this time with 6)
if this equation is graphed and you look at the y1 section it can be noticed that when x is 5 y1 is 52. This clearly works as both numbers are whole number and positive. It is also the least as the question stated. 
LOL but just to make sure plug values in so x = 5 and y = 52 and im too lazy too type but if done u should get 4.87.
Hope this helped!!!!!! :)
A: There is a lot detail supplied in the previous answers. Here is a more intuitive approach.
We agree that $6p+35d=487$ and we want to minimize the number of coins that gives us this total. In order to do this, we would want as many of our coins to be dimes and quarters as possible. Thus $d\le 13$.
Suppose $d=13$. But then $6p=32$ and this is impossible because $p$ needs to be an integer. 
Suppose $d=12$. But then $6p=67$ which is also impossible.
Suppose $d=11$. Then $6p=102$ which gives us $p=17$.
Thus the minimum number of coins is $11+11+17+17=56$.
