Solve the system of equations algebraically x^2+y^2=100 x-y=2

Given:-

(x^2) + (y^2) = 100
..........(1)

x - y =
2...........(2)

Squaring (2) on both sides we
get

(x-y)^2 = (2)^2

or, (x^2)
+ (y^2) - 2xy = 4

Putting the value of (x^2) + (y^2) from
(1) in the above equation we get

100 - 2xy =
4

or, 2xy =
96...........(3)

Now, (1) + (3)
 gives

(x^2) + (y^2) + 2xy = 100 +
96

or, (x+y)^2 = 196

or,
(x+y)^2 = (14)^2

or, (x+y) =
14.........(4)

or, (x+y) =
-14..........(5)

Now, (2) + (4)
gives

2x = 16

or, x =
8 

Thus, y =
6...........(6)

Also, (2) + (5)
gives

2x = -12

or, x =
-6

Thus, y =
-8..........(7)

Hence the two values of (x,y) are (8,6)
& (-6,-8)