Question

how to solve this equation "if x^2-3xy-2x=4 and y^2-xy+x+y=2
find the value of x-y.

Answer 1

x² - 3xy - 2x = 4

y =  x²  - 2x - 4

           3x

y² - xy + x + y =2

y² - x(y - 1) + y = 2

x  =  y²  + y - 2  

            y - 1

x  =  y²  -y + 2y - 2  

            y - 1

x  =  y(y - 1)  + 2(y - 1)  

            y - 1

x  =  (y + 2 )(y - 1)  

           ( y - 1)

x =  y + 2

x - y =   y + 2    -    x²  - 2x - 4  

                                   3x

=   3x(y + 2)  -   x²  + 2x + 4

                       3x

=    3xy  +  6x  -   x² +  2x  +  4  

                        3x

If I made mistakes in claculation pls do it. I am tired of doing it.

Answer 2

Answer:

x^2 - 3xy - 2x = 4

x^2-2xy-xy-2x-4=0

x^2 -2xy + y^2 - y^2 - xy -2x - 4 = 0

(x - y)^2 - ( y^2 + xy + 2x + 4) = 0

(x - y)^2 = (y^2 + xy + 2x +4)   ........equation 1

now;

second equation;

y^2 + xy + x + y = 2

y^2 + xy + x = 2 - y  ...... equation 2

(x-y)^2 = (y^2 + xy +  x) + x + 4

(x-y)^2 = (2 - y) + x + 4 ........... substituting from equation 2

(x-y)^2 = 2 + (x - y) + 4

(x-y)^2 = (x - y) + 6  .....................equation 3

let (x-y) be A

therefore;

A^2 - A - 6 = 0 ....modified equation 3

now solving by splitting the middle term factor

A^2 - (3A - 2A) - 6 = 0

A^2 - 3A + 2A - 6 = 0

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

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

therefore;

A = -2 or 3

or

(x - y) = -2 or 3

Step-by-step explanation: