Let x + y = a and xy = b. Then the given equations read a+b+1 = 12 > a+b = 11 and ab = 30. So, b = 30/a and therefore a + 30/a = 11 > a^2 - 11a = 30 > a = 6, b= 5 or a = 5, b =6. If a= 6, b= 5, x+y = 6 and y = 5/x > x^2 - 6x + 5 = 0 > (x,y) = (5,1), (1,5). If a= 5, b=6, x+y = 5 and y = 6/x > x^2 - 5x + 6 = 0 > (x,y) = (3,2), (2,3).
a+b=11 and ab=30 (a-b)^2=(a+b)^2-4ab=121-120=1 a-b=+/-1 Now solve 2 sets of equations a+b=11 and a-b=1 gives a=6 and b=5 a+b=11 and a-b =-1 gives a=5 and b=6 Use similar method to determine x and y.
Let x + y = a and xy = b. Then the given equations read a+b+1 = 12 > a+b = 11 and ab = 30. So, b = 30/a and therefore a + 30/a = 11 > a^2 - 11a = 30 > a = 6, b= 5 or a = 5, b =6. If a= 6, b= 5, x+y = 6 and y = 5/x > x^2 - 6x + 5 = 0 > (x,y) = (5,1), (1,5). If a= 5, b=6, x+y = 5 and y = 6/x > x^2 - 5x + 6 = 0 > (x,y) = (3,2), (2,3).
More reasonable is to choose linear dependence, substitute into nonlinear, anyways you receive the same equation
a+b=11 and ab=30
(a-b)^2=(a+b)^2-4ab=121-120=1
a-b=+/-1
Now solve 2 sets of equations
a+b=11 and a-b=1 gives a=6 and b=5
a+b=11 and a-b =-1 gives a=5 and b=6
Use similar method to determine x and y.
X.Y=3or2
x=2, y=3
Factorize instead of using formulas.
xy + (x + y) + 1 = 12 => xy + (x + y) = 11
xy(x + y) = 30 => x + y = 30/(xy)
xy + 30/(xy) = 11
(xy)² - 11xy + 30 = 0
xy = (11 ± 1)/2
xy = 6 or xy = 5
xy = 6 => y = 6/x
6 + x + 6/x = 11
x² - 5x + 6 = 0
x = (5 ± 1)/2
*x = 3 => y = 2*
*x = 2 => y = 3*
xy = 5 => y = 5/x
5 + x + 5/x = 11
x² - 6x + 5 = 0
x = (6 ± 4)/2
*x = 5 => y = 1*
*x = 1 => y = 5*