Solving non linear simultaneous equations

a)
x-3y=31
xy+12=0
b)
xy=30
3x+y=21
c)
2x+3y=1
4x^2-9y^2=-17
d)
x-3y=1
x^2-2xy-y^2=7
3 days ago

x - 3y = 31, x = 3y + 31
xy + 12 = 0
(3y + 31)y + 12 = 0
3y^2 + 31y + 12 = 0
y = (-31 ± √(961 - 144))/6
y = (-31 ± √817)/6
x = (-31 ± √817)/2 + 31 = (31 ± √817)/2

xy = 30, x = 30/y
3x + y = 21
90/y + y = 21
90 + y^2 = 21y
y^2 - 21y + 90 = 0
y^2 - 6y - 15y + 90 = 0
y(y - 6) - 15(y - 6) = 0
(y - 6)(y - 15) = 0
=> y - 6 = 0, y = 6
=> y - 15 = 0, y = 15
y = 6: x = 5
y = 15: x = 2

2x + 3y = 1, 2x = 1 - 3y, x = (1 - 3y)/2
4x^2 - 9y^2 = -17
(1 - 3y)^2 - 9y^2 = -17
1 - 6y + 9y^2 - 9y^2 = -17
1 - 6y = -17
6y = 18
y = 3
x = -4

x - 3y = 1
x = 1 + 3y
x^2 - 2xy - y^2 = 7
(1 + 3y)^2 - 2(1 + 3y)y - y^2 = 7
1 + 6y + 9y^2 - 2y - 6y^2 - y^2 = 7
2y^2 + 4y - 6 = 0
y^2 + 2y - 3 = 0
y^2 - y + 3y - 3 = 0
y(y - 1) + 3(y - 1) = 0
(y - 1)(y + 3) = 0
=> y - 1 = 0, y = 1
=> y + 3 = 0, y = -3
y = 1: x = 4
y = -3: x = -8

All of these can be solved by substitution from the linear equation to the quadratic equation, then applying the quadratic formula or factorisation to find two roots, then solving the other variable from the linear equation.

The substitution in C can be x=(1-3y)/2, but you can substitute 2x=1-3y as well, and get away without fractions entirely. In others, the choice for the substitution is quite easy.

ex:
2x+3y=1
2x=1-3y

4x^2-9y^2=-17
(1-3y)^2-9y^2=-17
1-6y+9y^2-9y^2=-17
1-6y=-17 (this time we got lucky and the quadratic term canceled out)
-6y=-18
y=3

2x+3y=1
2x+18=1
2x=-17
x=-17/2

a)
Solve the first equation for x:
x - 3y = 31
x = 3y + 31

Now plug that in for for x in the second equation and solve for y:
(3y + 31)*y + 12 = 0
3y^2 + 31y + 12 = 0