convert non linear equation to linear form (Y=mX+c)












0












$begingroup$


I'm hoping someone can help.
I have been given 3 equations that are non linear and I have to reduce them down to linear if possible where x & y are variables and a & b are unknown constants.
But I'm having trouble reducing them down especially the 1st and 3rd. Does anyone have any ideas or solutions.
begin{eqnarray}
y&=&ln(ax^2+bx)\
y&=& frac{a}{bx+c}\
y&=&frac{1}{(x-a)(x-b)}\
end{eqnarray}



Any help would be greatly appreciated






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Use completing the square to remove the double x's
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:32










  • $begingroup$
    but your equations are not linear and can not convert into the form $$y=mx+c$$ the last two equations can convert into the form $$Ax+By+C=0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '16 at 14:32












  • $begingroup$
    So first take e to the power for the first and reciprocate the 2nd and 3rd. Then substitutions of $Y=e^y$ and $Y=1/y$ makes them linear.
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:33












  • $begingroup$
    ok so for the second one would it be: (a/y)=bx+c where you would plot 1/y on y axis and x on x axis? b=gradient and c=intercept?
    $endgroup$
    – S1981
    Dec 15 '16 at 14:44


















0












$begingroup$


I'm hoping someone can help.
I have been given 3 equations that are non linear and I have to reduce them down to linear if possible where x & y are variables and a & b are unknown constants.
But I'm having trouble reducing them down especially the 1st and 3rd. Does anyone have any ideas or solutions.
begin{eqnarray}
y&=&ln(ax^2+bx)\
y&=& frac{a}{bx+c}\
y&=&frac{1}{(x-a)(x-b)}\
end{eqnarray}



Any help would be greatly appreciated






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Use completing the square to remove the double x's
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:32










  • $begingroup$
    but your equations are not linear and can not convert into the form $$y=mx+c$$ the last two equations can convert into the form $$Ax+By+C=0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '16 at 14:32












  • $begingroup$
    So first take e to the power for the first and reciprocate the 2nd and 3rd. Then substitutions of $Y=e^y$ and $Y=1/y$ makes them linear.
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:33












  • $begingroup$
    ok so for the second one would it be: (a/y)=bx+c where you would plot 1/y on y axis and x on x axis? b=gradient and c=intercept?
    $endgroup$
    – S1981
    Dec 15 '16 at 14:44
















0












0








0





$begingroup$


I'm hoping someone can help.
I have been given 3 equations that are non linear and I have to reduce them down to linear if possible where x & y are variables and a & b are unknown constants.
But I'm having trouble reducing them down especially the 1st and 3rd. Does anyone have any ideas or solutions.
begin{eqnarray}
y&=&ln(ax^2+bx)\
y&=& frac{a}{bx+c}\
y&=&frac{1}{(x-a)(x-b)}\
end{eqnarray}



Any help would be greatly appreciated






algebra-precalculus







share|cite|improve this question











$endgroup$




I'm hoping someone can help.
I have been given 3 equations that are non linear and I have to reduce them down to linear if possible where x & y are variables and a & b are unknown constants.
But I'm having trouble reducing them down especially the 1st and 3rd. Does anyone have any ideas or solutions.
begin{eqnarray}
y&=&ln(ax^2+bx)\
y&=& frac{a}{bx+c}\
y&=&frac{1}{(x-a)(x-b)}\
end{eqnarray}



Any help would be greatly appreciated






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '16 at 14:38









Kitter Catter

1,179917




1,179917










asked Dec 15 '16 at 14:29









S1981S1981

11




11












  • $begingroup$
    Hint: Use completing the square to remove the double x's
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:32










  • $begingroup$
    but your equations are not linear and can not convert into the form $$y=mx+c$$ the last two equations can convert into the form $$Ax+By+C=0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '16 at 14:32












  • $begingroup$
    So first take e to the power for the first and reciprocate the 2nd and 3rd. Then substitutions of $Y=e^y$ and $Y=1/y$ makes them linear.
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:33












  • $begingroup$
    ok so for the second one would it be: (a/y)=bx+c where you would plot 1/y on y axis and x on x axis? b=gradient and c=intercept?
    $endgroup$
    – S1981
    Dec 15 '16 at 14:44




















  • $begingroup$
    Hint: Use completing the square to remove the double x's
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:32










  • $begingroup$
    but your equations are not linear and can not convert into the form $$y=mx+c$$ the last two equations can convert into the form $$Ax+By+C=0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 15 '16 at 14:32












  • $begingroup$
    So first take e to the power for the first and reciprocate the 2nd and 3rd. Then substitutions of $Y=e^y$ and $Y=1/y$ makes them linear.
    $endgroup$
    – jugglingmike
    Dec 15 '16 at 14:33












  • $begingroup$
    ok so for the second one would it be: (a/y)=bx+c where you would plot 1/y on y axis and x on x axis? b=gradient and c=intercept?
    $endgroup$
    – S1981
    Dec 15 '16 at 14:44


















$begingroup$
Hint: Use completing the square to remove the double x's
$endgroup$
– jugglingmike
Dec 15 '16 at 14:32




$begingroup$
Hint: Use completing the square to remove the double x's
$endgroup$
– jugglingmike
Dec 15 '16 at 14:32












$begingroup$
but your equations are not linear and can not convert into the form $$y=mx+c$$ the last two equations can convert into the form $$Ax+By+C=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32






$begingroup$
but your equations are not linear and can not convert into the form $$y=mx+c$$ the last two equations can convert into the form $$Ax+By+C=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32














$begingroup$
So first take e to the power for the first and reciprocate the 2nd and 3rd. Then substitutions of $Y=e^y$ and $Y=1/y$ makes them linear.
$endgroup$
– jugglingmike
Dec 15 '16 at 14:33






$begingroup$
So first take e to the power for the first and reciprocate the 2nd and 3rd. Then substitutions of $Y=e^y$ and $Y=1/y$ makes them linear.
$endgroup$
– jugglingmike
Dec 15 '16 at 14:33














$begingroup$
ok so for the second one would it be: (a/y)=bx+c where you would plot 1/y on y axis and x on x axis? b=gradient and c=intercept?
$endgroup$
– S1981
Dec 15 '16 at 14:44






$begingroup$
ok so for the second one would it be: (a/y)=bx+c where you would plot 1/y on y axis and x on x axis? b=gradient and c=intercept?
$endgroup$
– S1981
Dec 15 '16 at 14:44












1 Answer
1






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oldest

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0












$begingroup$

Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
$$begin{align}
y&=log_{10}(ax^2+bx+c)\
10^y&=ax^2+bx+c\
&=aleft(x^2+frac baxright)+c\
&=aleft(x^2+frac bax+left(frac b{2a}right)^2-left(frac b{2a}right)^2right)+c\
&=aleft(x^2+frac bax+left(frac b{2a}right)^2right)-aleft(frac b{2a}right)^2+c\
10^y&=aleft(x+frac b{2a}right)^2+c-frac {b^2}{4a}
end{align}$$
Letting $Y=10^y$ and $X=left(x+frac b{2a}right)^2$, we have $Y=aX+c-frac {b^2}{4a}$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac {b^2}{4a}$).






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    0












    $begingroup$

    Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
    $$begin{align}
    y&=log_{10}(ax^2+bx+c)\
    10^y&=ax^2+bx+c\
    &=aleft(x^2+frac baxright)+c\
    &=aleft(x^2+frac bax+left(frac b{2a}right)^2-left(frac b{2a}right)^2right)+c\
    &=aleft(x^2+frac bax+left(frac b{2a}right)^2right)-aleft(frac b{2a}right)^2+c\
    10^y&=aleft(x+frac b{2a}right)^2+c-frac {b^2}{4a}
    end{align}$$
    Letting $Y=10^y$ and $X=left(x+frac b{2a}right)^2$, we have $Y=aX+c-frac {b^2}{4a}$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac {b^2}{4a}$).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
      $$begin{align}
      y&=log_{10}(ax^2+bx+c)\
      10^y&=ax^2+bx+c\
      &=aleft(x^2+frac baxright)+c\
      &=aleft(x^2+frac bax+left(frac b{2a}right)^2-left(frac b{2a}right)^2right)+c\
      &=aleft(x^2+frac bax+left(frac b{2a}right)^2right)-aleft(frac b{2a}right)^2+c\
      10^y&=aleft(x+frac b{2a}right)^2+c-frac {b^2}{4a}
      end{align}$$
      Letting $Y=10^y$ and $X=left(x+frac b{2a}right)^2$, we have $Y=aX+c-frac {b^2}{4a}$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac {b^2}{4a}$).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
        $$begin{align}
        y&=log_{10}(ax^2+bx+c)\
        10^y&=ax^2+bx+c\
        &=aleft(x^2+frac baxright)+c\
        &=aleft(x^2+frac bax+left(frac b{2a}right)^2-left(frac b{2a}right)^2right)+c\
        &=aleft(x^2+frac bax+left(frac b{2a}right)^2right)-aleft(frac b{2a}right)^2+c\
        10^y&=aleft(x+frac b{2a}right)^2+c-frac {b^2}{4a}
        end{align}$$
        Letting $Y=10^y$ and $X=left(x+frac b{2a}right)^2$, we have $Y=aX+c-frac {b^2}{4a}$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac {b^2}{4a}$).






        share|cite|improve this answer









        $endgroup$



        Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
        $$begin{align}
        y&=log_{10}(ax^2+bx+c)\
        10^y&=ax^2+bx+c\
        &=aleft(x^2+frac baxright)+c\
        &=aleft(x^2+frac bax+left(frac b{2a}right)^2-left(frac b{2a}right)^2right)+c\
        &=aleft(x^2+frac bax+left(frac b{2a}right)^2right)-aleft(frac b{2a}right)^2+c\
        10^y&=aleft(x+frac b{2a}right)^2+c-frac {b^2}{4a}
        end{align}$$
        Letting $Y=10^y$ and $X=left(x+frac b{2a}right)^2$, we have $Y=aX+c-frac {b^2}{4a}$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac {b^2}{4a}$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '16 at 1:14









        W. ZhuW. Zhu

        685314




        685314






























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