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Can someone help me to solve this two variable quadratic?

Here is the quadratic:
xy+2x+2y+4=0-xy + 2x + 2y + 4 = 0

I have tried to factorise it but I can only group it to this:
x(y2)+2(y+2)=0 -x(y - 2) + 2(y + 2) = 0
This is close but I cannot seem to finish it off.

Could someone please help?
(edited 6 years ago)
Original post by milan_tom
Here is the quadratic:
-xy + 2x + 2y + 4 = 0

I have tried to factorise it but I can only group it to this:
-x (y - 2) + 2 (y + 2) = 0
This is close but I cannot seem to finish it off.

Could someone please help?


Get all the terms in y over to one side of the equation:

xy2y=2x+4 xy - 2y = 2x + 4

Can you take it from there?
Reply 2
Original post by Gregorius
Get all the terms in y over to one side of the equation:

xy2y=2x+4 xy - 2y = 2x + 4

Can you take it from there?


Let me try:
xy2y=2x+4[br]y(x2)=2x+4[br]y=2x+4x2[br]y=2(x+2)x2 xy - 2y = 2x + 4[br]y(x-2) = 2x+4[br]y = \dfrac{2x+4}{x-2}[br]y = \dfrac{2(x+2)}{x-2}

How can I simplify it from here?
(edited 6 years ago)
Original post by milan_tom
Let me try:
xy - 2y = 2x 4
y(x-2) = 2x 4
y = (2x+4)/x-2
y = (2(x+4))/x-2

How can I simplify it from here?


Your final line should be y=(2(x+2))/x-2, you didn’t factor out the 4.
(edited 6 years ago)
Original post by milan_tom
Let me try:
xy - 2y = 2x + 4
y(x-2) = 2x+4
y = (2x+4)/x-2
y = (2(x+4))/x-2

How can I simplify it from here?


That's as far as you need to take it (modulo correcting the factoring out pointed out by @Y11_Maths) So a final answer of

y=2(x+2)x2 \displaystyle y = \frac{2(x+2)}{x-2}

should do nicely.
Reply 5
Original post by Gregorius
That's as far as you need to take it (modulo correcting the factoring out pointed out by @Y11_Maths) So a final answer of

y=2(x+2)x2 \displaystyle y = \frac{2(x+2)}{x-2}

should do nicely.


I need to solve for x and y.
Reply 6
Original post by Y11_Maths
Your final line should be y=(2(x+2))/x-2, you didn’t factor out the 4.


Yes, sorry, that was a typo.
Original post by milan_tom
I need to solve for x and y.


You have! Equations like this will not have simple x = a y = b type solutions. To see this, this of the analogous, but simpler, equation xy = 1. The solution to it is y = 1/x (clearly!)
Reply 8
Original post by Gregorius
You have! Equations like this will not have simple x = a y = b type solutions. To see this, this of the analogous, but simpler, equation xy = 1. The solution to it is y = 1/x (clearly!)


Thanks!
Original post by milan_tom
I need to solve for x and y.


To get numerical values for x and y you need another equation.

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