Another Maths Q

y - x = 2 can be rewritten as y = 2 + x therefore you can substitute that into the equation and solve for x.

2x^2 + 3x( 2 + x ) + ( 2 + x )^2 = 8
2x^2 + 6x + 3x^2 + 4 + 4x + x^2 = 8
6x^2 + 10x + 4 = 8
6x^2 + 10x - 4 = 0
( 6x - 2 )( x + 2 ) = 0
x = 1/3, -2
 
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Substitute y into X
So it's like
2X^2+3x(2+X)+(2+x)^2=8
2x^2 +6x+3x^2+4+4x+x^2=8
6x^2+10x-4=0
Probably made a mistake somewhere as it's this time of night, then just factorise or quadratic equation.
Night night.
 
Garee said:
y - x = 2 can be rewritten as y = 2 + x therefore you can substitute that into the equation and solve for x.
Click to expand...

And 2x^2 + 3xy + y^2 = (2x+y)(x+y), which simplifies it a bit.

So (2x+2+x)(x+2+x) = 8
(3x+2)(2x+2)=8
6x^2+10x+4=8
6x^2+10x=4

Then you do something clever... I used trial and error, answer is x = -2

EDIT: Below is the actual way to do it, it has been too long. :D
 
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y - x = 2
2x^2 + 3xy + y^2 = 8

Rearrange y - x = 2 to get y = x + 2, then substitute it in and solve as a quadratic:

y = (x+2)

2x^2 + 3x(x+2) + (x+2)^2 = 8

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

6x^2 + 10x - 4 = 0

3x^2 + 5x - 2 = 0

(3x-1)(x+2) = 0

x = 1/3 or -2
 
y - x = 2
2x^2 + 3xy + y^2 = 8

x = y-2

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

6y^2 -14y = 0
2y(3y - 7) = 0

y = 0 | y = 7/3

x = -2 | x = 1/3 (sub y into original equation)
 
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