System of Linear Equations to Resolve Room Dimensions

[kyle1984]

Homework Statement

Twelve equal rectangular ceiling panels are placed in a four long by three wide grid. Each panel is 150mm longer than it is wide. Between the panels runs an edge/middle strip which totals 40.0m in total length. (This runs round the whole perimeter and between each panel). By developing a system of linear equations what are the dimensions of the room?

So I have been staring at this all day. I know I am missing something simple but translating the given detail into something I can develop and solve is just a block at the moment. I would love some guidance if possible.

Homework Equations

??

The Attempt at a Solution

Have made many but none are correct!

 

[HallsofIvy]

The phrase "The attempt at a solution" does NOT refer only to successfu ones. If you have made many attempts, you should be able to show one[b\]. Seeing your mistakes will help us know what hints will help you.

 

[kyle1984]

Ok well the best I have come up with so far is 4(4w+0.6)+5(3w)=40

Where w = width

 

[kyle1984]

Actually I think I have it now.

where w = width and l = length.

1. w=l-0.15
2. 16l+15w=40

Substituting into Eq 2

16l+15(l-0.15)=40
16l+15l-2.25=40
l=1.36m

Therefore:

w=1.36-0.15=1.21m

So overall room dimensions are 3w=3.63m and 4l=5.44m

That seems to work out and follow fairly sensible logic I think.

 

[verty]

So the strip runs between the tiles? I think you must calculate half the sum of the length of any side that touches another tile. Set this to 40 and you'll be able to find the solution.