Question 1009252
xy=12  (equation 1)
x+y=7  (equation 2)

divide equation 1, both sides, by -X --> -y= -12/x  --> Y = 12/x  (mult by -1)

Substitute Y = 12X into equation 2  for Y 

a) X + 12/X =7          (substituting 12X for Y into equation 2)
b) X + 12/X - 7 = 0     (subtract 7 from both sides)
C) X(X +12/X -7) = x(0)  (Multiply both sides by X)
   X^2 + 12 - 7x  = 0    (distribute)

d) (x-3)(X-4) = 0        (factor)
e) Set x-3 =0  and x-4 =0
f)  Therefore there are 2 solutions for X : x=3 and x=4

g) Take the first value of X=3 and substitute for X in Equation 2:

h) 3 +y =7 --> Y = 4  Therefore there is an intersection at (3,4)
i)  Take the second value of X=4 and substitute into Equation 2
j) 4 + Y =7 ---> Y = 3  Therefore there is an intersection at (4,3)

k)  Substitute the 2 solutions back into both equations to make sure there are no extraneous equations.

l) 3+4 =7 : 4+3 =7  (Equation 2)
m) 3*4 =7 : 4*3 =7   (equation 1)

Therefore the solutions are  (3,4) and (4,3)