set of equations (system of equations)?

Question

please solve each system for both x & y. and are there mulitiple answers for each?

Question # 1. 18y-6x+12=0 & 18x-6y^2-156=0

Question # 2. 12xy^2+410+2y=0 & 6x^2(2y)+640+2y

Thank you

Answer 1

The easiest way to do this would be by substitution; solve one equation for a variable, and plug it into the other equation.

Q1: 18y-6x+12=0 -> 3y-x+2=0 -> x=3y+2

Plug x into the second equation:

18x-6y^2-156=0 -> 18(3y+2)-6y^2-156=0 -> 54y+36-6y^2-156=0 ->
-> 6y^2-54y+120=0 -> y^2-9y+20=0 -> (y-5)(y-4)=0 -> y=4 and y=5

Thus, x=14 when y=4 and x=17 when y=5.

Q2: I will assume you meant to set the second equation equal to zero. So:

6x^2(2y)+640+2y=0 -> 12yx^2+640+2y=0 -> y(12x^2+2)+640=0 ->
-> y=(-640)/(12x^2+2)=(-320)/(6x^2+1)

Plug this into the first equation:

12xy^2+410+2y=0 -> 12x(-320/(6x^2+1))^2+410+2(-320/(6x^2+1))=0 ->
-> 1228800x/(36x^4+12x^2+1)+410+(-640/(6x^2+1))=0 ->
-> 1228800x+410(6x^2+1)^2+(-640)(6x^2+1)=0 ->
-> 1228800x+410(36x^4+12x^2+1)-3840x^2-640=0 ->
-> 1228800x+24760x^4+4920x^2+410-3840x^2-640=0 ->
-> 34760x^4+1080x^2+1228800x-230=0 ->
-> 3476x^4+108x^2+122880x-23=0 -> ...

Answer 2

Question # 1.

18y-6x+12=0
18x-6y^2-156=0

Reducing the equations by dividing by 6.

3y - x + 2 = 0
3x - y^2 - 26 = 0

Solve the first equation for x.
x = 3y + 2

Plugging into the second equation.
3(3y + 2) - y^2 - 26 = 0
9y + 6 - y^2 - 26 = 0
9y - y^2 - 20 = 0
y^2 - 9y + 20 = 0
(y - 4)(y - 5) = 0
y = 4,5

x = 3y + 2 = 14, 17

Solutions are P(x,y) = (14,4) and (17,5)