# How to solve a system of linear equations when there are less rows than variables?

• Sep 22nd 2009, 01:56 PM
PeterTheBohemian
How to solve a system of linear equations when there are less rows than variables?
I took a year off from Math before deciding to major in it, so I'm a little rusty.

I've come to a point in a problem where I have 2 linear equations:

-2a+3b+9c=0
and
-2a+5b+4c=0

I'm unsure of how to solve these 2 equations. I can cancel out one of the variables, but after than, I don't know how to eliminate the next one so I can isolate the value of the 3rd variable. The Cramster expert answer says a = -33, b=-10, and c = -4.

Any help is greatly appreciated.

P.S. I know this looks like a linear algebra question, but this problem came from my multivariable calculus HW where those equations are the equation of the line orthogonal to a plane and the equation of the point on that plane.
• Sep 22nd 2009, 04:56 PM
Dark Sun
Peter, subtract the two equations to get:

$-2b+5c=0\Rightarrow 5c=2b$

Then, all points along this line will satisfy the system. Select $c=-4$.

Which gives us: $2b=5(-4)=-20\Rightarrow b=-10$

Then, plug these two results into our first equation, to get:

$0=-2a+3b+9c=-2a+3(-10)+9(-4)=-2a-30-36\Rightarrow 66=-2a\Rightarrow -33=a$

To give us a final result of $(-33,-10,-4)$.

This is not the only correct answer for this system. Hope this helps, thanks!
• Sep 23rd 2009, 07:40 AM
PeterTheBohemian
Quote:

Originally Posted by Dark Sun
Peter, subtract the two equations to get:

$-2b+5c=0\Rightarrow 5c=2b$

Then, all points along this line will satisfy the system. Select $c=-4$.

Which gives us: $2b=5(-4)=-20\Rightarrow b=-10$

Then, plug these two results into our first equation, to get:

$0=-2a+3b+9c=-2a+3(-10)+9(-4)=-2a-30-36\Rightarrow 66=-2a\Rightarrow -33=a$

To give us a final result of $(-33,-10,-4)$.

This is not the only correct answer for this system. Hope this helps, thanks!

Sorry, but I'm confused as to how you came up with c = -4. Why or how did you come to select -4 as the value of c?
• Sep 23rd 2009, 06:02 PM
Dark Sun
There are more than one solution. You could also select $c=0$.

I selected -4 so that the answer would agree with the solution that you had already known.

Any point along the line $5c=2b$ would return a valid solution.

Do you understand now?
• Sep 24th 2009, 01:10 AM
PeterTheBohemian
Yup, thanks a lot man.