Here's the problem:

Let L={(x,y) | 2x+3y=6} and L'={a(3,0) + b(0,2) | a+b=1}

Prove that L=L'

What I've done:

I've picked a few examples, so I believe that this is true. I know that I have to prove that L<L' and L'<L to conclude that L=L'.

I just can't seem to get how to go about doing it.

I think i did the second part:

if (m,n) is in L', then (m,n)=a(3,0)+b(0,2) where a+b=1

so, m=3a and n=2b

but a+b=1, so b=1-a

so (m,n)=(3a,2b)=(3a,2(1-a))

so then I tried to prove that this point is in L:

2x+3y=2(3a)+3(2(1-a))=6a+6-6a=6

I think this proves the second half, L'=L. Does it?

For the first part, I have:

(p,q) is in L, so 2p+3q=6

and I didn't know what to do so I tried to work backwards from the conclusion to get an idea

(p,q)=(p,0)+(0+q)=(p/3)(3,0)+(q/2)(0,2)

does p/3 + q/2 =1? but I can't figure out how to get any further.

Any help or hints or ideas or anything would be greatly appreciated!