Question

Find the solution of system of equations `x+y=2` and `x^2+2y^2=11`?

Answer 1

We have two solution sets `(3,-1)` and `(-1/3,7/3)`

Explanation:

For solving system of two equations in two variables, what is required is first eliminating one variable to get the value of another and then substituting second in either to get value of first variable.

First equation gives us `y=2-x`, putting this in `x^2+2y^2=11` we get

`x^2+2(2-x)^2=11` or

`x^2+2(4-4x+x^2)=11` or `3x^2-8x-3=0`

This can be factorized as

`3x^2-9x+x-3=0`

or `3x(x-3)+1(x-3)=0` or `(3x+1)(x-3)=0`

i.e. `x=3` or `-1/3`. Putting these values in `y=2-x`, we get

`y=-1` or `7/3`

Hence we have two solution sets `(3,-1)` and `(-1/3,7/3)`

graph{(x+y-2)(x^2+2y^2-11)((x-3)^2+(y+1)^2-0.01)((x+1/3)^2+(y-7/3)^2-0.01)=0 [-4.42, 5.58, -1.78, 3.22]}