Proof of Linear Algebra Solution Using Systems of Equations

[nickw00tz]

Hi, first post here. I need help with a proof from linear algebra.
It states:
suppose that (x,y)=(r,s) is a solution of:

system of equations #1
ax+by=p
cx+dy=q

and that (x,y)=(u,v) is a solution of:

system of equations #2
ax+by=0
cx+dy=0

prove that (x,y)=(r+u , s+v) is a solution for system of equations #1

. The attempt at a solution

ar+bs=p
cr+dy=q

au+bv=0
cu+dv=0

au+bv=cu+dv

i then tried solving for a,b,c and d and plugging them back into the first system of equations, however after doing so my equations become very long and confusing. I tried working it backwards but i still seem to get stuck. There has to be a simpler way of solving it but i can't seem to figure it out.

 

[micromass]

Just plug in (r+u,s+v) in your equations and see if it works out.

 

[nickw00tz]

Just plug in (r+u,s+v) in your equations and see if it works out.

I attempted to do that and working backwards from there but i get stuck here:

a(r+u)+b(s+v)=p
c(r+u)+d(s+v)=q

ar+au+bs+bv=p
cr+cu+ds+sv=q

 

[micromass]

I attempted to do that and working backwards from there but i get stuck here:

a(r+u)+b(s+v)=p
c(r+u)+d(s+v)=q

ar+au+bs+bv=p
cr+cu+ds+sv=q

Yes, and now use that au+bv=0 and cu+dv=0

 

[Mark44]

i then tried solving for a,b,c and d and plugging them back into the first system of equations

You should not be trying to solve for these numbers. The variables in your problem are x and y. Everything else is a constant.

 

[nickw00tz]

Yes, and now use that au+bv=0 and cu+dv=0

Oh i see now, thank you!