Question

Joe has a collection of nickels and dimes that is worth $9.60. If the number of nickels was tripled and the number of dimes was decreased by 31, the value of the coins would be $6.70. How many nickels and dimes does he have?

Answer 1

write a system of equations; the first one about the value of the money and the second about the number of coins.
.05x + .10y = 9.60 and 3x + y-31=6.70

Answer 2

ok thank you so much I forgot the money... i really need to finish my final, but also need to go to bed unfortunately the two don't mix well.

Answer 3

Do you have any questions?

Answer 4

.05n+.10d=9.60
-.05(3n+d-31=6.70)
3(.05n+.10d=9.60)

-.15n-.05d+1.55=-.335
.15n+.3d=28.8

.15n-.05d+1.55+.15n+.3d=-.335+28.8
.25d+1.55=28.465
.25d=26.915
d=107.66

3n+107.66-31=6.70
3n+76.66=6.70
3n=-69.96
n=-23.32
is this correct?

Answer 5

This can't be correct because the answers must be whole numbers and you cannot have a negative amount of coins...

Answer 6

I know what am I doing wrong?

Answer 7

give me a minute to work it out....

Answer 8

ok ty

Answer 9

ok. this is what I got:


.05n + .10d=9.60
and .05(3n)+.10(d-31)=6.70

then I solved the second eq in terms of d and got d=.98-1.5n

Then I substituted in the first and solved.
.05n + .10 (.98-1.5n)=9.60
-.10n=-.20
n= 2 nickels; then plug it back in the first one and I got 95 dimes...

Answer 10

thank you

Answer 11

yay!