If nickles + dimes =64 total coins but they also equip a price of $10.45

Answer 1

Let d be the number of dimes and n be the number of nickels. Then (because there are 64 coins), d + n = 64

Also, you can figure out the value of the money you have by multiplying the number of dimes by 0.10 and multiplying the number of nickels by 0.05. That means 0.10d + 0.05n = 10.45

This gives you 2 equations with the 2 unknowns (d and n). You can solve this in several ways, including substitution and elimination. I'll go with substitution because that's usually the first method taught.

If d + n = 64, then d = 64 - n (subtract "n" from both sides of the equation). Now since "d" is the same as "64 - n" you can plug into the second equation "64 - n" wherever you see "d". That gives you:

0.10(64 - n) + 0.05n = 10.45

6.4 - 0.10n + 0.05n = 10.45 (distributing to eliminate the parenthesis)

6.4 - 0.05n = 10.45 (combining like terms)

-0.05n = 4.05 (subtracting 6.4 from both sides)

n = -81 (dividing both sides by -0.05)

It is not possible to have negative 81 nickels so this problem does not have an answer. That makes sense if you think about what would happen if you had all dimes (that's the largest possible value of the money). If you had all dimes, you would have $6.40 - not even close to what they are telling you the value is.

This answer is telling you that if somehow you could have a negative 81 nickels, you would have to have 145 dimes because -81 + 145 = 64. That would give you -$4.05 worth of nickels but $14.50 worth of dimes, adding them together would give you $10.45. But, again, this doesn't make sense in the real word because there are no "negative" nickels.