Basic linear

• Apr 1st 2010, 04:14 AM
Henryt999
Basic linear
Hi!
Iīm having some issues solving this equation system.

$x_1+x_2+2x_3 = 3$
$x_1-x_2+8x_3 = -1$
$2x_1+3x_2+x_3 = 8$

Nomatter what I do I cant eliminate the 2 terms to get one of the unknows alone. No problem eliminating 1 term thought if you ad the first 2 terms x_2 is eliminated... great and now lets eliminate one more term....emm emmm canīt do it! whats wrong...?
Any hints?
• Apr 1st 2010, 06:18 AM
craig
Quote:

Originally Posted by Henryt999
Hi!
Iīm having some issues solving this equation system.

$x_1+x_2+2x_3 = 3$
$x_1-x_2+8x_3 = -1$
$2x_1+3x_2+x_3 = 8$

Nomatter what I do I cant eliminate the 2 terms to get one of the unknows alone. No problem eliminating 1 term thought if you ad the first 2 terms x_2 is eliminated... great and now lets eliminate one more term....emm emmm canīt do it! whats wrong...?
Any hints?

There are infinitely many solutions to this set of equations, you cannot get simple numeric values for $x_1, x_2, x_3$.

The way I would do this is let one of the variables, (I'll use $x_3$), and let this equal any constant $\lambda$, so we have $x_3 = \lambda$

Now try and solve your equations in terms of lamba( $\lambda$).

You already know that $x_3 = \lambda$, your other 2 solutions will have the form $x_n = a + b\lambda$ where $a,b \in \mathbb{R}$.

Hope this helps.
• Apr 1st 2010, 06:47 AM
HallsofIvy
You can add the first two equations to get $2x_1+ 10x_3= 2$ which is the same as $x_1+ 5x_3= 1$. If you multiply the first equation by 3 and subtract the third equation from it you also get $x_1+ 5x_3= 1$.

Since those are the same equation, you are right- you cannot continue and get a single equation in one variable. And, as craig says, that means there are an infinite number of solutions.

What you can do is this: solve $x_1+ 5x_3= 1$ for $x_1$ in terms of $x_3$: $x_1= 1- 5x_3$. Now put that back into the first equation, $x_1+ x_2+ 2x_3= 1- 5x_3+ x_2+ 2x_3= 1- 3x_3+ x_2= 3$ so $x_2= 2+ 3x_3$.

We can now write the solution set as $(x_1, x_2, x_3)= (1- 5x_3, 2+ 3x_3, x_3)$. We can also write it as $(1, 2, 0)+ t(-5, 3, 1)$ where I have taken $x_3$ to be the parameter t. The solution set is a plane which, because the set of equations is not homogeneous, does not contain the origin.