# Too many equations. Can I eliminate one this way?

• Mar 2nd 2011, 04:32 AM
salohcin
Too many equations. Can I eliminate one this way?
Hello all,

I have a big long system of equations I am trying to solve and I have more equations than unknowns... let me tell you how I got there. Lets say I am trying to solve for x, y, w/x and z/y, and right now I have 4 equations (we are fine right now, but x and y don't mean much in the context I am working in). Two of these four equations are...

y = a x + b z
x = c y + d w

I divide both sides of the top equation by y and both sides of the bottom equation by x. Now I have

1 = a (x/y) + b (z/y)
1 = c (y/x) + d (w/x)

Okay so now I am just trying to solve for x/y w/x and z/y (but I have four equations). There is no reason why I cant just set a (x/y) + b (z/y) equal to c (y/x) + d (w/x) to eliminate this extra equation right?

Cheers!

Nick
• Mar 2nd 2011, 06:38 AM
tonio
Quote:

Originally Posted by salohcin
Hello all,

I have a big long system of equations I am trying to solve and I have more equations than unknowns... let me tell you how I got there. Lets say I am trying to solve for x, y, w/x and z/y, and right now I have 4 equations (we are fine right now, but x and y don't mean much in the context I am working in). Two of these four equations are...

y = a x + b z
x = c y + d w

I divide both sides of the top equation by y and both sides of the bottom equation by x. Now I have

1 = a (x/y) + b (z/y)
1 = c (y/x) + d (w/x)

Okay so now I am just trying to solve for x/y w/x and z/y (but I have four equations). There is no reason why I cant just set a (x/y) + b (z/y) equal to c (y/x) + d (w/x) to eliminate this extra equation right?

Cheers!

Nick

Certainly you have 4 equations but two of them are directly derived from the other two, and you still have 3 unknowns:

putting $\displaystyle{A:=\frac{x}{y}\,,\,\,B:=\frac{z}{y}\ ,,\,\,C:=\frac{w}{x}}$, the last two equations you have are

$\displaystyle{aA+bB=1\,,\,\,\frac{c}{A}+dC=1}$ ....you're missing one (independent) equation if you want to solve this.

Tonio