# Thread: Basic linear

1. ## Basic linear

Hi!
I´m 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 can´t do it! whats wrong...?
Any hints?
Thanks in advance

2. Originally Posted by Henryt999
Hi!
I´m 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 can´t do it! whats wrong...?
Any hints?
Thanks in advance
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.

3. 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.