A stamp collector bought 295 stamps for $87.75. The purchase included 5 cent, 25 cent, and 40 cent stamps. How many of each type stamp was purchased?

Answer 1

As Steve already told your, there is no unique solution

https://www.jiskha.com/display.cgi?id=1509960580

picking up where he left off.... let's have a closer look
x+y+z = 295
5x+25y+40z = 8775
dividing the 2nd equation by 5:
x + 5y + 8z = 1755
subtract the first from that:
4y + 7z = 1460

y = (1460-7z)/4

Also x+y+z = 295
x + (1460-7z)/4 + z = 295
4x + 1460 - 7z + 4z = 1180
4x = 3z - 280
x = (3z - 280)/4

now pick a value of z which makes the right-side a whole number,
I would pick multiples of 4 for z, (since 1460 is divisible by 4 , the difference between 1460 and a multiple of 4 will be divisible by 4, the same is true for the x evaluation)
e.g. let z = 8
y = (1460-56)/4 = 351
AAHHH, but the total is only 295, we are already over that
so 1460-7z < 4(295)
-7z < -280
z > 40

let z = 44
y = (1460 - 308)/4 = 288 , we already have a total over 295
x = (3(44) - 280)/4 , which is negative,
so z = 44 doesn't work

try z = 160
y = 85
x = 50 , that works!

try z = 200
y = 15
x = 80 , that works!!

Have fun getting as many as you can

Answer 2

Thanks for the assistance Reiny. Much appreciated