SOLUTION: Irving’s total cash was only $2.50. How many nickels and dimes would he have? Again, solve by the method of your choice.
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-> SOLUTION: Irving’s total cash was only $2.50. How many nickels and dimes would he have? Again, solve by the method of your choice.
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I can only help in a very general sense because you didn't bother to provide the entire problem. Your problem needs to have some sort of statement about how many coins Irving has. It could be that he has some specific number of coins or it might be that he has some number more dimes than nickels -- or some other relationship that would allow you to write a second equation.
Let's say that you know you have coins altogether.
Let represent the number of dimes Irving has, and let represent the number of nickels. Hence:
Since dimes are worth 10 cents each and nickels are worth 5 cents each and you can say that Irving's $2.50 is actually 250 cents, we can write:
Rearranging the first equation we get:
And substituting:
which, had you taken the trouble to provide a value for , we could easily solve for
You can put this solution on YOUR website! We assume that Irving only can have Nickels and dimes for his cash. Then if x is the number of Nickels and y is the number of dimes then it is clear that
5*x + 10*y = 250
so
y = -.5*x + 25
if we plot this equation we get:
any integer (x,y) on the line is nickels and dimes collection that equals $2,50. When x=0, no nickels all dimes, y=25. When y = 0, no dimes all nickels, x = 50.
We can't have 1 nickel because there is no 49 dimes combination that when added to 5 cent equals $2.50.
