# How man stamps of each type did David buy? I've tried every way I can possibly find the variables. David paid $9.50 for some 15 cent, 25 cent, and 45 cent stamps. He bought 38 stamps in all....

How man stamps of each type did David buy?

I've tried every way I can possibly find the variables.

David paid $9.50 for some 15 cent, 25 cent, and 45 cent stamps. He bought 38 stamps in all. The number of 25 cent stamps was 8 more than twice the number of 45 cnt stamps.

My equations:

x=15

y=25

z=45

x+y+z=38

15x+25y+45z=950

8+2z=y

### 2 Answers | Add Yours

Your equations are mostly correct:

Let x= the **number** of 15 cent stamps (not .15)

Let y= the **number** of 25 cent stamps (not .25)

Let z= the **number** of 45 cent stamps (not .45)

Then:

`x+y+z=38` (Total number of stamps is 38)

`.15x+.25y+.45z=9.5` or `15x+25y+45z=950` (Total dollar amount)

`y=2z+8` (Number of 25 cent stamps is 8 more than twice the number of 45 cent stamps)

Then you have a system of three equations in three unknowns:

`x+y+z=38`

`15x+25y+45z=950`

`0x+y-2z=8`

You can use substitution (you know y=2z+8) or linear combinations or any number of matrix operations to solve this system. Using substitution we get:

`x+2z+8+z=38`

`15x+25(2z+8)+45z=950`

or

`x+3z=30`

`15x+95z=750`

Multiplying the first equation by -15 we get:

`-15x-45z=-450`

`15x+95z=750`

Adding these equations we get:

`50z=300==>z=6`

Then `x+18=30==>x=12`

and finally `y=2(6)+8=20`

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**The solution is 12 15 cent stamps, 20 25 cent stamps and 6 45 cent stamps**

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You had all of the constraints (the equations); you just didn't assign the variables correctly. You had a really nice start on the problem. Good luck!

Thank you so much! I had a whole bunch of problems and this was the only one I got stuck on. I couldn't figure out what I was doing wrong. Thank you for helping me!