How to solve these simultaneous equation?












0












$begingroup$


I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows



begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}



Just need the values and how to to them.. not case sensitive just in case
Thanks






algebra-precalculus systems-of-equations







share|cite|improve this question











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  • $begingroup$
    Where did you “stumble” upon this system of equations?
    $endgroup$
    – amd
    Jan 15 at 20:44
















0












$begingroup$


I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows



begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}



Just need the values and how to to them.. not case sensitive just in case
Thanks






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where did you “stumble” upon this system of equations?
    $endgroup$
    – amd
    Jan 15 at 20:44














0












0








0





$begingroup$


I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows



begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}



Just need the values and how to to them.. not case sensitive just in case
Thanks






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$




I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows



begin{array}{r l}
bc+3d+15a-db-15c-3b=60 & (1) \
d-15c=0 & (2) \
3a-b= -6 & (3) \
b-d-c+a=0 & (4)
end{array}



Just need the values and how to to them.. not case sensitive just in case
Thanks






algebra-precalculus systems-of-equations




algebra-precalculus systems-of-equations






share|cite|improve this question















share|cite|improve this question













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edited Jan 15 at 15:57









Dylan

12.9k31026




12.9k31026










asked Jan 15 at 14:33









user10276651user10276651

1




1












  • $begingroup$
    Where did you “stumble” upon this system of equations?
    $endgroup$
    – amd
    Jan 15 at 20:44


















  • $begingroup$
    Where did you “stumble” upon this system of equations?
    $endgroup$
    – amd
    Jan 15 at 20:44
















$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44




$begingroup$
Where did you “stumble” upon this system of equations?
$endgroup$
– amd
Jan 15 at 20:44










2 Answers
2






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2












$begingroup$

From $(2)$,
$$d=15c ldots(5)$$
Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
$$3a+6-15c-c+a=0$$
$$4a-16c+6=0$$
$$a = 4c-frac32 ldots(6)$$
So$$begin{align}
b&=3left(4c-frac32right)+6\
&=12c+frac32 ldots(7)
end{align}$$
Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.



Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.






share|cite|improve this answer









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    0












    $begingroup$

    Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      2












      $begingroup$

      From $(2)$,
      $$d=15c ldots(5)$$
      Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
      $$3a+6-15c-c+a=0$$
      $$4a-16c+6=0$$
      $$a = 4c-frac32 ldots(6)$$
      So$$begin{align}
      b&=3left(4c-frac32right)+6\
      &=12c+frac32 ldots(7)
      end{align}$$
      Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.



      Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        From $(2)$,
        $$d=15c ldots(5)$$
        Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
        $$3a+6-15c-c+a=0$$
        $$4a-16c+6=0$$
        $$a = 4c-frac32 ldots(6)$$
        So$$begin{align}
        b&=3left(4c-frac32right)+6\
        &=12c+frac32 ldots(7)
        end{align}$$
        Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.



        Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          From $(2)$,
          $$d=15c ldots(5)$$
          Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
          $$3a+6-15c-c+a=0$$
          $$4a-16c+6=0$$
          $$a = 4c-frac32 ldots(6)$$
          So$$begin{align}
          b&=3left(4c-frac32right)+6\
          &=12c+frac32 ldots(7)
          end{align}$$
          Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.



          Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.






          share|cite|improve this answer









          $endgroup$



          From $(2)$,
          $$d=15c ldots(5)$$
          Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$,
          $$3a+6-15c-c+a=0$$
          $$4a-16c+6=0$$
          $$a = 4c-frac32 ldots(6)$$
          So$$begin{align}
          b&=3left(4c-frac32right)+6\
          &=12c+frac32 ldots(7)
          end{align}$$
          Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.



          Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 15 at 15:47









          Faiq IrfanFaiq Irfan

          813317




          813317























              0












              $begingroup$

              Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 14:42









                  Robert IsraelRobert Israel

                  324k23214468




                  324k23214468






























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