SOLUTION: Maria bought 80 stamps at the post office in 33¢ and 25¢ denominations. If she paid $24 for the stamps, how many of each denomination did she buy? Here is what I tried: x=33

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Maria bought 80 stamps at the post office in 33¢ and 25¢ denominations. If she paid $24 for the stamps, how many of each denomination did she buy? Here is what I tried: x=33      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 899568: Maria bought 80 stamps at the post office in 33¢ and 25¢ denominations. If she paid $24 for the stamps, how many of each denomination did she buy?
Here is what I tried:
x=33¢ denomination
y=25¢ denomination
x+y=80
x=-y+80
33(x)+25(y)=24
33(-y+80)+25y=2400
-33y+(-2640)+25y=2400
Added 2640 to both sides
8y=50.4
Divide both sides by 8
y=6.30
x=-6.30+80
x=50.4
which Ido not think is right but I can not figure what I am doing wrong :(
Found 2 solutions by ewatrrr, richwmiller:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

x+y=80

 x=-y+80 

 33(x)+25(y)=24

 33(-y+80)+25y=2400  Nice Work...careful to the finish line 

 -33y+(2640)+25y=2400  

   -8y = -240

     y = 30, number of 25cent stamps.  50 33cent stamps 80-30

CHECKING our answer***

 30*25 + 50*33 = 2400

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x+y=80,
33*x+25*y=2400
x=80-y
33*(80-y)+25*y=2400
2640-33y+25*y=2400
-8*y=-240
y=30
x=80-y
x=50 y=30
check
50+30=80
33*x+25*y=2400
33*50+25*30=2400
1650+750=2400
2400=2400
ok