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

1. ## 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.

2. Peter, subtract the two equations to get:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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?

4. There are more than one solution. You could also select $\displaystyle c=0$.

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

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

Do you understand now?

5. Yup, thanks a lot man.