Question

How do I solve the system?
x^2 + xy = 1
xy + y^2 = 3

Answer 1

you can maybe solve one of them for either x or y, and put all that in for the variable in the other equation, to reduce to single var

Answer 2

Solve for `xy` in either equation. I'm going to pick on the second equation


xy + y^2 = 3
xy + y^2-y^2 = 3-y^2
xy = 3-y^2


then plug this into the other equation


x^2 + xy = 1
x^2 + 3-y^2 = 1
x^2-y^2 + 3 = 1
x^2-y^2 + 3-3 = 1-3
x^2 - y^2 = -2


This is as far as you can go really. We'd need another equation to figure out what x and y are. In this case, there are infinitely many solutions. Each solution (x,y) lies along a hyperola

Answer 3

plug in (1/2 , 3/2)

Answer 4

x.x+xy=x(x+y)=1
xy+y.y=y(x+y)=3
divide first eqn. by second,
x/y=1/3
or y=3x
now solve 1st eqn by substitution and solve for y, then find x

Answer 5

@arnavguddu has the right idea. There will be 2 solutions in the form (x,y)

Answer 6

yeah that is good, y = 1/x - x solving the first equation for the y, and using that, you can get to the end too

Answer 7

the process you did @jim_thompson5910 looks like the same as adding (-1)times the first equation to the other eq

does that linear system work for those nonlinear things too

Answer 8

linear sytem solving rules

Answer 9

constant multioplication and adding equations and changing order, linear algebra