Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes was tripled and the number of nickels was decreased by 49, the value of the coins would be $14.05. How many nickels and dimes does he have?

Question

anonymous · Answer

System of equations, where the number of dimes is x and the number of nickels is y.

.1X+.05y=6.1
3(.1x)+.05y-49(.05)=14.05

Solve

anonymous · Answer

how did u get the second equation?

anonymous · Answer

basically I just accounted for all of the changes. The new total is 14.05, and in order to get that, there are 3 times as many dimes, of which have the same .1 value, and the value for the nickels is essentially the same y value as before, again accounting for the .05 value, but taking into account that there are 49 less nickels each at a value of .05.

anonymous · Answer

i just tried it it doesnt makes sense..
this this the answer i got
.1X+.05y=6.1
3(.1x)+.05y-49(.05)=14.05

.3x+.05y-.0245=14.05
.3x+.05y-.0245+.0245=14.05+.0245
.3x+.05y-.05y=14.0745-.05y
.3x/.3=14.0745/.3-.05y/.3
x=46.914-.16666

anonymous · Answer

alright I'll work it out and see.

anonymous · Answer

1X+.05y=6.1
3(.1x)+.05y-49(.05)=14.05

.3x+.05y-2.45=14.05
.3x+.05y-.2.45-2.45=14.05+2.45
.3x+.05y-.05y=16.50-.05y
.3x/.3=16.5/.3-.05y/.3
x=55-.166667y 

ok i changed something i figured out whats wrong but i dont know where else im wrong

anonymous · Answer

You solved both equations incorrectly.

If you solve for y on the first equation I gave you, you get y=122-2x

Substitute that into the second equation to get

.3x +.05(122-2x)-2.45=14.05

Try solving from there.

Y

anonymous · Answer

now how did u get that...holy crap im more confused then when i wrote the question

anonymous · Answer

You should get x=52 and y=18

anonymous · Answer

You should get x=52 and y=18

anonymous · Answer

You should get x=52 and y=18

anonymous · Answer

Lol it's a system of equations... So you solve for one variable, in that instance  I solved for y first. Then I substituted it into Tyne second equation so I only have one variable in the equation. Then I solved for x. Once I had x, I could put it into the first equation, ie the y122-2x, in order to solve for the number of nickels.