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

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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`...

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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!

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