# Simple linear algebra problem

1. Oct 26, 2009

### schmolch

Hi All!

I hope i posted this in the correct forum. Technically its not homework, just a problem from a puzzle-book which i enjoyed very much until i got to this problem.

1. The problem statement, all variables and given/known data

Calvin Carpenter needed screws of three different sizes. He bought the same number of screws of each size at $.03,$.04, and \$.05 per screw respectively.
Wendy Woodall spent the same amount of money as Calvin, but divided her money equally among screws of the three sizes. In all, Wendy purchased six more screws than Calvin did. How much money did they each spend?

2. Relevant equations

3. The attempt at a solution

Calvin gets the same number of screws of each size:

3x+4x+5x=k (cents, x=number of screws, k=total price in cents)

Wendy gets screws for the same price each size:

3a+4b+5c=k
3a=4b=5c

I could mess around some more, but i just dont get it.
Im trying to understand this since more than a week and every day i feel dumber. If anyone could explain this to me i would be very gratefull.

schmolch

2. Oct 26, 2009

### lanedance

i think its in your definition of how wendy spends, but were close

first for Calvin as you wrote
3x+4x+5x= 12x = k

now wendy spends her money evenly but divides it equally, as there are three types of screws, each one gets k/3 spent on it

so if she buy a,b,c of respectively

then
3a = 4b = 5c = k/3

so now knowing k, give a,b,c & x

however there is one unused fact, wendy got 6 more screws than Calvin...