> with(linalg);
Using the four equations given, we get this matrix: > A:=matrix(4,4, [[3,2,-1,-15], [5,3,2,0], [3,1,3,11], [-6,-4,2,30]]);
A := [ 3 2 -1 -15 ]
[ 5 3 2 0 ]
[ 3 1 3 11 ]
[ -6 -4 2 30 ]
> B:=mulrow(A, 1, 1/3);
B := [ 1 2/3 -1/3 -5 ]
[ 5 3 2 0 ]
[ 3 1 3 11 ]
[ -6 -4 2 30 ]
> C:=addrow(B, 1,2, -5);
C := [ 1 2/3 -1/3 -5 ]
[ 0 -1/3 11/3 25 ]
[ 3 1 3 11 ]
[ -6 -4 2 30 ]
> D:=addrow(C, 1, 3, -3);
D := [ 1 2/3 -1/3 -5 ]
[ 0 -1/3 11/3 25 ]
[ 0 -1 4 26 ]
[ -6 -4 2 30 ]
> A:=addrow(D, 1,4,6);
A := [ 1 2/3 -1/3 -5 ]
[ 0 -1/3 11/3 25 ]
[ 0 -1 4 26 ]
[ 0 0 0 0 ]
> B:=mulrow(A, 2, -3);
B := [ 1 2/3 -1/3 -5 ]
[ 0 1 -11 75 ]
[ 0 -1 4 26 ]
[ 0 0 0 0 ]
> C:=addrow(B, 2, 3, 1);
C := [ 1 2/3 -1/3 -5 ]
[ 0 1 -11 75 ]
[ 0 0 -7 -49 ]
[ 0 0 0 0 ]
> D:=mulrow(C, 3, -1/7);
D := [ 1 2/3 -1/3 -5 ]
[ 0 1 -11 75 ]
[ 0 0 1 7 ]
[ 0 0 0 0 ]
> A:=addrow(D, 3, 2, 11);
A := [ 1 2/3 -1/3 -5 ]
[ 0 1 0 2 ]
[ 0 0 1 7 ]
[ 0 0 0 0 ]
> B:=addrow(A, 3, 1, 1/3);
B := [ 1 2/3 0 8/3 ]
[ 0 1 0 2 ]
[ 0 0 1 7 ]
[ 0 0 0 0 ]
> C:=addrow(B, 2, 1, -2/3);
C := [ 1 0 0 -4 ]
[ 0 1 0 2 ]
[ 0 0 1 7 ]
[ 0 0 0 0 ] 
Therefore: x1= -4 
           x2= 2 
           x3= 7
Using the given equations, we get the following matrix > A:=matrix(5,5,[[0,10,-4,1,1],[1,4,-1,1,2],[3,2,1,2,5],[-2,-8,2,-2,-4],[1,-6,3,0,1]]);
A := [ 0 10 -4 1 1 ]
[ 1 4 -1 1 2 ]
[ 3 2 1 2 5 ]
[ -2 -8 2 -2 -4 ]
[ 1 -6 3 0 1 ]
> B:=swaprow(A, 1,2);
B := [ 1 4 -1 1 2 ]
[ 0 10 -4 1 1 ]
[ 3 2 1 2 5 ]
[ -2 -8 2 -2 -4 ]
[ 1 -6 3 0 1 ]
> C:=addrow(B, 1,3, -3);
C := [ 1 4 -1 1 2 ]
[ 0 10 -4 1 1 ]
[ 0 -10 4 -1 -1 ]
[ -2 -8 2 -2 -4 ]
[ 1 -6 3 0 1 ]
> D:=addrow(C, 1,4,2);
D := [ 1 4 -1 1 2 ]
[ 0 10 -4 1 1 ]
[ 0 -10 4 -1 -1 ]
[ 0 0 0 0 0 ]
[ 1 -6 3 0 1 ]
> A:=addrow(D, 1,5,-1);
A := [ 1 4 -1 1 2 ]
[ 0 10 -4 1 1 ]
[ 0 -10 4 -1 -1 ]
[ 0 0 0 0 0 ]
[ 0 -10 4 -1 -1 ]
> B:=mulrow(A, 2, 1/10);
B := [ 1 4 -1 1 2 ]
[ 0 1 -2/5 1/10 1/10 ]
[ 0 -10 4 -1 -1 ]
[ 0 0 0 0 0 ]
[ 0 -10 4 -1 -1 ]
> C:=addrow(B, 2, 3, 10);
C := [ 1 4 -1 1 2 ]
[ 0 1 -2/5 1/10 1/10 ]
[ 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
[ 0 -10 4 -1 -1 ]
> D:=addrow(C, 2, 5, 10);
D := [ 1 4 -1 1 2 ]
[ 0 1 -2/5 1/10 1/10 ]
[ 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
> A:=addrow(D, 2,1,-4);
A := [ 1 0 3/5 3/5 8/5 ]
[ 0 1 -2/5 1/10 1/10 ]
[ 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
Since we get rows that are completely zeroes, we know that z and w are constants. Therefore: 
 


		x= 8/5 - 3/5z - 3/5w 
		y=1/10 + 2/5z - 1/10w
Number 17: using the given equations gives us this matrix > A:=matrix(3, 4,[ [1,2,-3,4],[3,-1,5,2],[4,1,a2-14,a+2]]);
A := [ 1 2 -3 4 ]
[ 3 -1 5 2 ]
[ 4 1 a2-14 a+2 ]
> B:=addrow(A,1,2,-3);
B := [ 1 2 -3 4 ]
[ 0 -7 14 -10 ]
[ 4 1 a2-14 a+2 ]
> C:=addrow(B, 1,3,-4);
C := [ 1 2 -3 4 ]
[ 0 -7 14 -10 ]
[ 0 -7 -2+a2 -14+a ]
> D:=mulrow(C,2,-1/7);
D := [ 1 2 -3 4 ]
[ 0 1 -2 10/7 ]
[ 0 -7 -2+a2 -14+a ]
> A:=addrow(D, 2,3,7);
A := [ 1 2 -3 4 ]
[ 0 1 -2 10/7 ]
[ 0 0 -16+a2 -4+a ]

This gives us three equations:
					x + 2y - 3z = 4 
					     y - 2z = 10/7 
					  (a2-16)z = -4 + a
If a is equal to 4, then the last equation becomes 0=0, and there will be infinitely many solutions.

If a=-4, the equation will be 0=4 which is not true, so there will be no solutions.

All other values for a will give only one solution.