Question

find x and y if (x^2-3x, y^2-y)= (-2,6)

50 points answer be quick​

Answer 1

Hii

you are welcome in my ans

x^2 -3x +2=0

(x-1)(x-2)=0

x = 1 and 2

y^2 - y - 6=0

y ^2 - 2y + 1/4 - 25/2 =0

(x - 1/2)^2 - 25/2 =0

{x - (1/2 - 5/sqrt2) }{x - (1/2 +5/sqrt2)}=0

x = 1/2 - 5/sqrt2

and

x = 1/2 +5/sqrt2

_/\_☹️

Hope it may helps you

Answer 2

Answer:

ok,

x² - 3x = -2

x² - 3x + 2 = 0

x² - 2x - x + 2 = 0

x(x - 2) -1( x - 2) = 0

(x -2) or (x - 1) = 0

( x = 2) or (x = 1)

y² - y = 6

y² - y - 6 = 0

y² - 3y + 2y - 6 = 0

y( y - 3) + 2(y - 3) = 0

(y - 3) or (y + 2) = 0

(y = 3) or (y = -2)

There is two solutions for Value of x and y

(x , y) = (2,3) or (x , y) = (1, -2)