Algebra word problem...?
Joe has a collection of nickels and dimes that is worth $8.55. If the number of dimes was tripled and the number of nickels was decreased by 2, the value of the coins would be $10.45. How many nickels and dimes does he have?
Update:any thoughts at all??
Comments
x = Number of Nickels, y - Number of Dimes
Nickels are worth .05 and dimes are worth .10
$0.05x + $0.10y = $8.55 (I hate working with decimals like this so I multiply by 100)
5x + 10y = 855
5(x - 2y) = 855
x -2y = 171
[x = 2y + 171]
$0.05(x - 2) + 3($0.10y) = $10.45
5(x - 2) + 30y = 1045
5x - 10 + 30y = 1045
5x - 30y = 1055
5(x + 6y) = 1055
x + 6y = 211
[x = 211 - 6y]
211 - 6y = 171 - 2y
211 - 4y = 171
40 = 4y
[y = 10] *
x = 171 - 2y
x = 171 - 2(10)
x = 171 - 20
[x = 151] *
$0.05x + $0.10y = $8.55
$0.05(151) + $ 0.10(10) = $8.55
$7.55 + $1.00 = $8.55
$8.55 = $8.55
check
$0.05(x - 2) + 3($0.10y) = $10.45
$0.05(151 - 2) + 3($0.10[10]) = $10.45
$0.05(149) + 3($1.00) = $10.45
$7.45 + $3.00 = $10.45
$10.45 = $10.45
Check