x - y =  1
x² - xy - y² = -5 

The other tutor solved the first equation for x. 
Just to be different I'll solve it for y:

                          x - y = 1
                             -y = 1 - x
                              y = -1 + x
                              y = x - 1

Substitute (x - 1) for y in the second equation,
but first factor -y out of the second and third terms
on the left:

                   x² - xy - y² = -5
                  x² - y(x + y) = -5
      x² - (x - 1)[x + (x - 1)] = -5  
        x² - (x - 1)(x + x - 1) = -5 
           x² - (x - 1)(2x - 1) = -5
            x² - (2x² - 3x + 1) = -5
              x² - 2x² + 3x - 1 = -5
                   -x² + 3x + 4 =  0
                    x² - 3x - 4 =  0
                 (x - 4)(x + 1) =  0
                x - 4 = 0;  x + 1 = 0  
                    x = 4       x = -1

To find the y-value for x = 4, substitute 4 for x in

                
                              y = x - 1     
                              y = 4 - 1
                              y = 3      

So one solution is (x,y) = (4,3)

To find the y-value for x = -1, substitute -1 for x in

                
                              y = x - 1     
                              y = -1 - 1
                              y = -2      

So the other solution is (x,y) = (-1,-2)

Edwin