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Can someone help me to solve this two variable quadratic? watch

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    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?
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    (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:

     xy - 2y = 2x + 4

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

     xy - 2y = 2x + 4

    Can you take it from there?
    Let me try:
     xy - 2y = 2x + 4

y(x-2) = 2x+4

y = \dfrac{2x+4}{x-2}

y = \dfrac{2(x+2)}{x-2}

    How can I simplify it from here?
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    (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.
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    (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

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

    should do nicely.
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    (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

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

    should do nicely.
    I need to solve for x and y.
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    (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.
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    (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!)
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    (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!
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    (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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