how do i solve for x and y x+y=2 and 3x^2+4xy+y^2=6

Answer 1

You have to isolate one variable in the first equation first because its easier, then substitute it into the second one:

x=2-y

Now plug this in for x in the second equation:

3(2-y)^2+4(2-y)y+y^2=6

Now solve for y, then plug your answer back into:
x=2-y

3(2-y)(2-y)+8y-4y^2+y^2=6
3(4-4y+y^2)+8y-3y^2=6
12-12y+3y^2+8y-3y^2=6
-4y+12=6
-4y=6-12
y=-1.5

Now plug it in:
x=2-(-1.5)
x=3.5

Hope that helped! :)

Answer 2

my bad, the 1.5 should be positive! so y=1.5 and x=0.5

Answer 3

x + y = 2 Subtract x to both sides

x + y - x = 2 - x

y = 2 - x

3 x ^ 2 + 4 x y + y ^ 2 = 6

3 x ^ 2 + 4 x ( 2 - x ) + ( 2 - x ) ^ 2 = 6

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Remark :

( a - b ) ^ 2 = a ^ 2 - 2 a b + b ^ 2

So :

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

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3 x ^ 2 + 4 x * 2 + 4 x * ( - x ) + 2 ^ 2 - 2 * 2 * x + x ^ 2 = 6

3 x ^ 2 + 8 x - 4 x ^ 2 + 4 - 4 x + x ^ 2 = 6

3 x ^ 2 - 4 x ^ 2 + x ^ 2 + 8 x + 4 - 4 x = 6

- x ^ 2 + x ^ 2 + 8 x - 4 x + 4 = 6

4 x + 4 = 6 Subtract 4 to both sides

4 x + 4 - 4 = 6 - 4

4 x = 2 Divide both sides by 4

4 x / 4 = 2 / 4

x = 2 / ( 2 * 2 )

x = 1 / 2

y = 2 - x

y = 2 - 1 / 2

y = 4 / 2 - 1 / 2

y = 3 / 2

The solutions are :

x = 1 / 2 , y = 3 / 2