help solve simultaneous equation












2












$begingroup$


I need to solve the system



$$6x+y=3tag 1 $$



$$x^2+y^2=16tag 2$$



So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.



I get $$x^2+(3-6x)^2 = 16$$



which is $$x^2 + 36x^2-36x+9-16=0$$



which is $$37x^2-36x-7=0$$.



This i need to solve by completing the square.






systems-of-equations completing-the-square







share|cite|improve this question











$endgroup$












  • $begingroup$
    If completing the square confuses you, you could use quadratic formula if you are allowed.
    $endgroup$
    – randomgirl
    Jan 12 '17 at 23:46










  • $begingroup$
    Ive been told to do it by completing the square. thanks though :)
    $endgroup$
    – user406484
    Jan 12 '17 at 23:50
















2












$begingroup$


I need to solve the system



$$6x+y=3tag 1 $$



$$x^2+y^2=16tag 2$$



So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.



I get $$x^2+(3-6x)^2 = 16$$



which is $$x^2 + 36x^2-36x+9-16=0$$



which is $$37x^2-36x-7=0$$.



This i need to solve by completing the square.






systems-of-equations completing-the-square







share|cite|improve this question











$endgroup$












  • $begingroup$
    If completing the square confuses you, you could use quadratic formula if you are allowed.
    $endgroup$
    – randomgirl
    Jan 12 '17 at 23:46










  • $begingroup$
    Ive been told to do it by completing the square. thanks though :)
    $endgroup$
    – user406484
    Jan 12 '17 at 23:50














2












2








2





$begingroup$


I need to solve the system



$$6x+y=3tag 1 $$



$$x^2+y^2=16tag 2$$



So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.



I get $$x^2+(3-6x)^2 = 16$$



which is $$x^2 + 36x^2-36x+9-16=0$$



which is $$37x^2-36x-7=0$$.



This i need to solve by completing the square.






systems-of-equations completing-the-square







share|cite|improve this question











$endgroup$




I need to solve the system



$$6x+y=3tag 1 $$



$$x^2+y^2=16tag 2$$



So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.



I get $$x^2+(3-6x)^2 = 16$$



which is $$x^2 + 36x^2-36x+9-16=0$$



which is $$37x^2-36x-7=0$$.



This i need to solve by completing the square.






systems-of-equations completing-the-square




systems-of-equations completing-the-square






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 '17 at 23:51









Namaste

1




1










asked Jan 12 '17 at 23:45









user406484user406484

111




111












  • $begingroup$
    If completing the square confuses you, you could use quadratic formula if you are allowed.
    $endgroup$
    – randomgirl
    Jan 12 '17 at 23:46










  • $begingroup$
    Ive been told to do it by completing the square. thanks though :)
    $endgroup$
    – user406484
    Jan 12 '17 at 23:50


















  • $begingroup$
    If completing the square confuses you, you could use quadratic formula if you are allowed.
    $endgroup$
    – randomgirl
    Jan 12 '17 at 23:46










  • $begingroup$
    Ive been told to do it by completing the square. thanks though :)
    $endgroup$
    – user406484
    Jan 12 '17 at 23:50
















$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46




$begingroup$
If completing the square confuses you, you could use quadratic formula if you are allowed.
$endgroup$
– randomgirl
Jan 12 '17 at 23:46












$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50




$begingroup$
Ive been told to do it by completing the square. thanks though :)
$endgroup$
– user406484
Jan 12 '17 at 23:50










2 Answers
2






active

oldest

votes


















0












$begingroup$

To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
$$37^2x^2-36times37x-7times37=0 .$$
To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
$$(37x-18)^2=7times37+18^2 .$$
I'm sure you can finish it from here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. When i was taught this i was told you turn it into this format
    $endgroup$
    – user406484
    Jan 13 '17 at 0:00










  • $begingroup$
    Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
    $endgroup$
    – user406484
    Jan 13 '17 at 0:06










  • $begingroup$
    The formula you gave in your comment is missing a square but I assume that's just a typo.
    $endgroup$
    – David
    Jan 13 '17 at 0:09










  • $begingroup$
    Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
    $endgroup$
    – David
    Jan 13 '17 at 0:10



















0












$begingroup$

Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.



There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are



Given $ax^2+bx+c=0$,




  1. Note the values of $4a=148$ and $b^2=1296$

  2. Subtract $c$ from both sides: $37x^2-36x=7$

  3. Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$

  4. Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$

  5. The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$


From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.






share|cite|improve this answer











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    2 Answers
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    0












    $begingroup$

    To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
    $$37^2x^2-36times37x-7times37=0 .$$
    To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
    $$(37x-18)^2=7times37+18^2 .$$
    I'm sure you can finish it from here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format
      $endgroup$
      – user406484
      Jan 13 '17 at 0:00










    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
      $endgroup$
      – user406484
      Jan 13 '17 at 0:06










    • $begingroup$
      The formula you gave in your comment is missing a square but I assume that's just a typo.
      $endgroup$
      – David
      Jan 13 '17 at 0:09










    • $begingroup$
      Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
      $endgroup$
      – David
      Jan 13 '17 at 0:10
















    0












    $begingroup$

    To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
    $$37^2x^2-36times37x-7times37=0 .$$
    To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
    $$(37x-18)^2=7times37+18^2 .$$
    I'm sure you can finish it from here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format
      $endgroup$
      – user406484
      Jan 13 '17 at 0:00










    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
      $endgroup$
      – user406484
      Jan 13 '17 at 0:06










    • $begingroup$
      The formula you gave in your comment is missing a square but I assume that's just a typo.
      $endgroup$
      – David
      Jan 13 '17 at 0:09










    • $begingroup$
      Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
      $endgroup$
      – David
      Jan 13 '17 at 0:10














    0












    0








    0





    $begingroup$

    To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
    $$37^2x^2-36times37x-7times37=0 .$$
    To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
    $$(37x-18)^2=7times37+18^2 .$$
    I'm sure you can finish it from here.






    share|cite|improve this answer









    $endgroup$



    To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get
    $$37^2x^2-36times37x-7times37=0 .$$
    To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives
    $$(37x-18)^2=7times37+18^2 .$$
    I'm sure you can finish it from here.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 12 '17 at 23:55









    DavidDavid

    69.4k667130




    69.4k667130












    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format
      $endgroup$
      – user406484
      Jan 13 '17 at 0:00










    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
      $endgroup$
      – user406484
      Jan 13 '17 at 0:06










    • $begingroup$
      The formula you gave in your comment is missing a square but I assume that's just a typo.
      $endgroup$
      – David
      Jan 13 '17 at 0:09










    • $begingroup$
      Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
      $endgroup$
      – David
      Jan 13 '17 at 0:10


















    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format
      $endgroup$
      – user406484
      Jan 13 '17 at 0:00










    • $begingroup$
      Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
      $endgroup$
      – user406484
      Jan 13 '17 at 0:06










    • $begingroup$
      The formula you gave in your comment is missing a square but I assume that's just a typo.
      $endgroup$
      – David
      Jan 13 '17 at 0:09










    • $begingroup$
      Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
      $endgroup$
      – David
      Jan 13 '17 at 0:10
















    $begingroup$
    Thank you. When i was taught this i was told you turn it into this format
    $endgroup$
    – user406484
    Jan 13 '17 at 0:00




    $begingroup$
    Thank you. When i was taught this i was told you turn it into this format
    $endgroup$
    – user406484
    Jan 13 '17 at 0:00












    $begingroup$
    Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
    $endgroup$
    – user406484
    Jan 13 '17 at 0:06




    $begingroup$
    Thank you. When i was taught this i was told you turn it into this format x^2+ax+b=0 -> (x+(1/2a))-(1/2a)^2)+b=0 then go from there. in this case it would be (37x-18)^2-324-7=0 When i follow on from this it never comes out right. can you explain please?
    $endgroup$
    – user406484
    Jan 13 '17 at 0:06












    $begingroup$
    The formula you gave in your comment is missing a square but I assume that's just a typo.
    $endgroup$
    – David
    Jan 13 '17 at 0:09




    $begingroup$
    The formula you gave in your comment is missing a square but I assume that's just a typo.
    $endgroup$
    – David
    Jan 13 '17 at 0:09












    $begingroup$
    Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
    $endgroup$
    – David
    Jan 13 '17 at 0:10




    $begingroup$
    Apart from that it should work. However you will have a lot of fractions in the working so it would be easy to make an arithmetic error. Doing it my way avoids most of the fractions and gives the same result in the end.
    $endgroup$
    – David
    Jan 13 '17 at 0:10











    0












    $begingroup$

    Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.



    There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are



    Given $ax^2+bx+c=0$,




    1. Note the values of $4a=148$ and $b^2=1296$

    2. Subtract $c$ from both sides: $37x^2-36x=7$

    3. Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$

    4. Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$

    5. The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$


    From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.



      There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are



      Given $ax^2+bx+c=0$,




      1. Note the values of $4a=148$ and $b^2=1296$

      2. Subtract $c$ from both sides: $37x^2-36x=7$

      3. Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$

      4. Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$

      5. The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$


      From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.



        There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are



        Given $ax^2+bx+c=0$,




        1. Note the values of $4a=148$ and $b^2=1296$

        2. Subtract $c$ from both sides: $37x^2-36x=7$

        3. Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$

        4. Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$

        5. The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$


        From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.






        share|cite|improve this answer











        $endgroup$



        Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.



        There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are



        Given $ax^2+bx+c=0$,




        1. Note the values of $4a=148$ and $b^2=1296$

        2. Subtract $c$ from both sides: $37x^2-36x=7$

        3. Multiply both sides by $4a=148$ from step $1$: $5476x^2-5238x=1036$

        4. Add $b^2=1296$ from step $1$ to both sides: $5476x^2-5328x+1296=2332$

        5. The left side now equals the square $(2ax+b)^2$ of the derivative of $ax^2+bx+c$: $(74x-36)^2=2332$


        From there the solution $x=dfrac{18pmsqrt{583}}{37}$ is straightforward.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 13 '17 at 4:26

























        answered Jan 13 '17 at 1:59









        John Wayland BalesJohn Wayland Bales

        14.6k21238




        14.6k21238






























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