How do I solve this simultaneous equation?

x = 11 - y
Now substitute x in the second equation:
(11 - y)y = 30
11y - y^2 = 30
y^2 - 11y + 30 = 0
(y-5)(y-6) = 0
hence, y = 5 or y = 6
if y= 5 then x = 6, if y = 6 then x = 5.

x + y = 11 (solve by using substitution)
xy = 30

x + y = 11
x = 11 - y

xy = 30
(11 - y)y = 30
11y - y^2 = 30
y^2 - 11y + 30 = 0
y^2 - 5y - 6y + 30 = 0
(y^2 - 5y) - (6y + 30) = 0
y(y - 5) - 6(y - 5) = 0
(y - 5)(y - 6) = 0

y - 5 = 0
y = 5

y - 6 = 0
y = 6

x + y = 11
x + 5 = 11
x = 11 - 5
x = 6

x + y = 11
x + 6 = 11
x = 11 - 6
x = 5

∴ (x = 6 , y = 5) , (x = 5 , y = 6)

x + y = 11 : x = 11 -y
xy = 30

sub eqn 1 into 2

(11-y)y = 30
y^2 -11 y + 30 = 0
(y -5)(y-6) = 0
y = 5 and x = 6
or
y = 6 and x = 5

from the first equation

x=11-y

substitute this value of x in the 2nd eqn

(11-y)y =30

-y^2 + 11y = 30

or

y^2 - 11y + 30 = 0

use quadratic formula to find y and then substitute its value/s into the 1st eqn to find x

you can do it from here

x+y=11------1
xy=30------2

x=11-y------- 3
sub 3 into 2

y(11-y)=30
11y-y^2=30
y^2-11y+30=0
(y-5)(y-6)=0
y=5 or y=6

sub them back into 1

x+5=11 so x=6
x+6=11 so x=5

so x=6 when y=5 and x=5 when y=6

(i thinnk its right)

x + 30/x = 11
x² + 30 = 11x
x² - 11x + 30 = 0
(x - 6)(x - 5) = 0
x = 6 , x = 5
y = 5 , y = 6

(6,5) , (5,6)

its very easy, just think of the factors of 30, the first one that pops into my head is 5X6. That equals 30. 5+6 also equals 11.