Question

Joe has a collection of nickels and dimes that is worth $2.10. If the number of dimes was tripled and the number of nickels was increased by 24, the value of the coins would be $5.70. How many nickels and dimes does he have?

Answer 1

This is a two variable two equation type problem.

The first one is .05x + .10y = 2.10 where x is the number of nickels and y is the number of dimes. Ex. if you had 7 nickels and 8 dimes you would take .05 times 7 to find the value of the nickels (.35) and .10 * 8 to get the value of the dimes (.80) and then add them together to get the value of the coins.

2nd equation would be .05(x + 24) + .10(3y) = 5.70

so .05x + .10y = 2.10 -- multiply every term by 100 to get rid of decimals

5x + 10Y = 210

2nd equation .05(x + 24) + .10(3y) = 5.70
distribute
.05x + 1.20 + .30y = 5.70
I would multiply all terms by 100 to get
5x + 120 + 30y = 570
subtract 120 from each side
5x + 30y = 450

So you have
5x + 10y = 210
5x + 30Y = 450

subtract the bottom which means to change the signs on every term on the bottom

5x + 10Y = 210
-5x -30Y = -450

Add..
x's are gone leaving you
-20Y = -240
divide by -20
Y = 12
Substitue Y = 12 into any equation and solve for X
5x + 10Y = 210
5X + 10(12) = 210
5X + 120 = 210
5X = 90
X = 18

So 18 nickels and 12 dimes