Find the general solution of each of the following equations (presented, as in the text

Step 1 of 3

a)Because there are variables in this one equation,specifying two determines the third.So let’s set two of them equal to arbitrary variable names.

Let \(x_{2}=s\)

 and \(x_{3}=t\)

Then we see that \(x_{1}-x_{2}+3 x_{3}=4=x_{1}=4+2 s-3 t\).

Thus the solution set is all vectors of the form

\(X=\left(x_{1}, x_{2}, X-3\right)=(4+2 s-3 t, s, t)=(4,0,0)+s(2,1,0)+t(-3,0,1)\)

b)

In the rest of these we proceed exactly in (a).

So Let \(x_{2}=r\)

\(x_{3}=s\) 

and \(x_{4}=t\).

Then

\(x=(-r+s-2 t, r, s, t)=r(-1,1,0,0)+s(1,0,1,0)+t(-2,0,0,1)\)