There is obviously a trivial solution x=0, y=0,
in which t can be any number.  So for any other
case we will assume there are other soluitions.

For

4x+tx-20y = 0 

Factor x out of the first two terms on the left

x(4+t) - 20y = 0

(4+t)x - 20y = 0

For 

x-4y+ty = 0 

Factor -y out of the last two terms on the left

x-y(4-t) = 0

x-(4-t)y = 0

So the system of equations is

(4+t)x -    20y = 0
     x - (4-t)y = 0

Since both constant terms on the right are 0,
that means the determinant of coefficient
matrix must be 0, so 

det(A) = 0



Evaluate the determinant on the left and we have

-(4+t)(4-t) - (-20)(1) = 0

-(16-t²) + 20 = 0
 
     -(16-t²) = -20

        16-t² = 20
          -t² = 4
           t² = -4
            t = ±√-4
            t = ±2i

Oh, oh!!!!!  Something is wrong with the problem!!!!!
There are no real values of t!!!!!

You must have copied a sign or number wrong, for t 
cannot be a real number as the problem is given 
here, except for the one trivial solution x=0 and 
y=0, in which case t can be any number.

If you will state the problem correctly in the 
thank-you note form, we can help you.  Be sure to 
check the signs and numbers very carefully.  

Edwin