Question

Solve the two following simultaneous equations and find the possible values of x and y.
`x^2 +y^2-2xy = 64`


`x^2 - y^2 = 40`

Answer 1

Explanation:

If u assume x=y=1, then x+y=2 and x^2+y^2=2

And now, xy=1.

Therefore x=y=1.

Let me know, if anyone come with other answers too.

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Answer 2

Explanation:

(*) x^2 + y^2 - 2xy = 64

<=> (x - y)^2 = 64

<=> x - y = 8 or x - y = -8

(**) x^2 - y^2 = 40

<=> (x - y)(x + y) = 40

Case 1: x - y = 8

(**) <=> 8(x + y) = 40

<=> x + y = 5

we have: x+y = 5 and x-y = 8

--> x = (5+8)/2 = 6.5

--> y = x - 8 = 6.5 - 8 = -1.5

so (x,y) = (6.5 , -1.5)

Case 2: x - y = -8

(**) <=> -8(x + y) = 40

<=> x + y = -5

we have: x + y = -5 and x - y = -8

--> x = [ (-5) + (-8) ]/2 = -6.5

--> y = -5 - x = -5 + 6.5 = 1.5

so (x,y) = (-6.5 , 1.5)

Answer: (x,y) = (6.5 , -1.5) and (-6.5 , 1.5)