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?

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

Thanks for the assistance Reiny. Much appreciated

To solve this problem, we can set up a system of equations. Let's represent the number of 5 cent stamps as 'x', the number of 25 cent stamps as 'y', and the number of 40 cent stamps as 'z'.

Based on the given information, we have three equations:

1. The total number of stamps is 295: x + y + z = 295
2. The total cost of the stamps is $87.75: 0.05x + 0.25y + 0.40z = 87.75

Now, we can use substitution or elimination to solve this system of equations. Let's use substitution:

From equation 1, we can express x in terms of y and z: x = 295 - y - z

Substituting this value of x into equation 2, we have:

0.05(295 - y - z) + 0.25y + 0.40z = 87.75

Expanding and simplifying, we get:

14.75 - 0.05y - 0.05z + 0.25y + 0.40z = 87.75
0.20y + 0.35z = 73

Now, we need to choose any two variables and solve for them. Let's choose y and z.

We can express y in terms of z by isolating y in the equation 0.20y + 0.35z = 73:

0.20y = 73 - 0.35z
y = (73 - 0.35z) / 0.20

We can substitute this expression for y in equation 1 to solve for x:

x + (73 - 0.35z) / 0.20 + z = 295

Multiplying through by 0.20 to clear the fraction, we get:

0.20x + (73 - 0.35z) + 0.20z = 59

Simplifying further:

0.20x + 73 - 0.15z + 0.20z = 59
0.20x + 73 + 0.05z = 59
0.20x + 0.05z = 59 - 73
0.20x + 0.05z = -14

Now, we have a linear equation in terms of x and z. To solve for x, we need to choose a value for z and substitute it back into the equation.

For example, let's choose z = 0:

0.20x + 0.05(0) = -14
0.20x = -14
x = -14 / 0.20
x = -70

Now, we can substitute the found values of x and z back into equation 1 to find y:

-70 + y + 0 = 295
y = 295 - (-70)
y = 365

So, the solution is x = -70, y = 365, and z = 0.

However, since we cannot have a negative number of stamps, the solution is not valid in this case. It's possible that there was an error or missing information in the given problem. Please double-check the question to ensure all information is provided correctly.