Help with A-maths problem?

Question

The line x – y = 3 intersects the curve x^2 - 3xy + y^2 + 19 = 0 at P and Q.
Find the coordinates of P and Q.

Answer 1

x-y=3 => x= y+3
plugged it to the second equatione

(y+3)^2-3(y+3)y+y^2+19=0
(y+3)(y+3)-3y(y+3)+y^2+19= 0
y^2+3y+3y+9 -3y^2-9y +y^2+19= 0
y^2+6y+9 -3y^2-9y+y^2+19=0
combine like term
-y^2-3y+28=0
time negative 1 to both sides
y^2+3y-28=0
(y+7)(y-4)=0

y+7=0 or y -4 = 0
y= -7, and y = 4

put it back to x= y+3 to find x values

y= -7 so x = -7+3 = -4

y= 4 so x = 4+3= 7

so P(-4,- 7) Q(7,4)

Answer 2

x – y = 3 >>>>>>>>>>>>>>>equation 1
x² - 3xy + y² + 19 = 0 >>>>>>>>>>>>>>>equation 2

From equation 1, x = 3 + y >>>>>>>>>>>>>>>equation 3
Plug it into equation 2

(3 + y)² - 3(3 + y)y + y² + 19 = 0
(3 + y)² - 3y(3 + y) + y² + 19 = 0
9 + 6y + y² - 9y - 3y² + y² + 19 = 0
- y² - 3y + 28 = 0
(- y + 7)(y - 4) = 0 [By using cross-multiplication]

- y + 7 = 0
- y = 7
y = -7

y - 4 = 0
y = 4

Substitute y = -7 and y = 4 into equation 3
x = 3 + y
x = 3 + (-7)
x = -4

x = 3 + y
x = 3 + 4
x = 7

Thus x = -4 , y = -7 or x = 7 , y = 4
So coordinates of P(-4 , -7) and Q(7 , 4)

Answer 3

x - y = 3 ==> y = x - 3

Substitute this expression for y into the second equation.

x^2 - 3xy + y^2 + 19 = 0
x^2 - 3x(x - 3) + (x - 3)^2 + 19 = 0

Expand the polynomial products and combine like terms, and then solve as you would any quadratic equation. This should give you two values for x. Use them in the expression for y to get the corresponding values of y, and match them up into ordered pairs. Make sure you put each y-value with the x-value from which you got it.

Answer 4

its easy.
put y=x-3 in the equation of the curve .u will get a quadratic eqn. in x.solve it

u will get 2 values for x. then find out y for each x value uget.
thus u will get co-ordinates of P & Q.