Please tell me how to solve this simultaneous equation?

Question

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

I know you have to do this:

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

but when I try to solve it I always get a quadratic that just can't factorise, help!

Answer 1

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

The aim is to get one equation to contain like terms (ie. only x or only y). ok, so we know that y= 2-x, so sub that in to the 2nd equation like so...
x^2 + 2x(2-x)= 3
now multiply out the bracket part like so...
2x(2- x) ------> 4x - 2x^2
so the 2nd equation can be written as...
x^2 + 4x - 2x^2= 3
this can be simplified to...
4x- x^2= 3
then... 4x- x^2 -3=0
then... x^2 -4x + 3=0 (I've just changed the signs around so that the x^2 term is positive for simplicity)

now we can factorise
x^2- 4x+3=0
(x- 1)(x- 3)=0
so x= 1 & 3

now we can find y...
we know y= 2-x
so y= 2-1= 1
& y= 2- 3= -1

so the pairs of answers are
when x= 1 y=1
when x= 3 y=-1

sorted!

Answer 2

x ² + 2 x y = 3
x ² + 2 x (2 - x) = 3
x ² + 4 x - 2 x ² = 3
x ² - 4 x + 3 = 0
(x - 1) (x - 3) = 0
x = 1 , x = 3
y = 1 , y = - 1
(1 , 1) and (3,-1) are solutions.

Answer 3

No, you can't do it that way:

y = 2- x

Plug y in the second equation:

x^2 + 2x (2-x) = 3
x^2 -2x^2 + 4x - 3 = 0
-x^2 + 4x - 3 = 0

And continue working on it.
Hint: use the property of delta.

Answer 4

Actually you want to substitute the first equation into the second equation:

x^2 + 2x*(2-x) = 3

It should be pretty easy to solve from there.

Answer 5

y=2-x
y+x=2
(y+χ)²=4
y²+2xy+x²=4 but 2xy+x²=3
y²+3=4
y²=1
y=1 or y=-1

if y=1 then 1=2-x i.e x=1
if y=-1 then -1=2-x i.e x=3
then you have 2 solutions (x,y)=(1,1) and (x,y)=(3,-1)