convert non linear equation to linear form (Y=mX+c)
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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.
begineqnarray
y&=&ln(ax^2+bx)\
y&=& fracabx+c\
y&=&frac1(x-a)(x-b)\
endeqnarray
Any help would be greatly appreciated
algebra-precalculus
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
asked Dec 15 '16 at 14:29
S1981
11
add a comment |Â
up vote
0
down vote
favorite
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.
begineqnarray
y&=&ln(ax^2+bx)\
y&=& fracabx+c\
y&=&frac1(x-a)(x-b)\
endeqnarray
Any help would be greatly appreciated
algebra-precalculus
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
asked Dec 15 '16 at 14:29
S1981
11
Hint: Use completing the square to remove the double x's
– jugglingmike
Dec 15 '16 at 14:32
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 $$Ax+By+C=0$$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32
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.
– jugglingmike
Dec 15 '16 at 14:33
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?
– S1981
Dec 15 '16 at 14:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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.
begineqnarray
y&=&ln(ax^2+bx)\
y&=& fracabx+c\
y&=&frac1(x-a)(x-b)\
endeqnarray
Any help would be greatly appreciated
algebra-precalculus
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
asked Dec 15 '16 at 14:29
S1981
11
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.
begineqnarray
y&=&ln(ax^2+bx)\
y&=& fracabx+c\
y&=&frac1(x-a)(x-b)\
endeqnarray
Any help would be greatly appreciated
algebra-precalculus
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
asked Dec 15 '16 at 14:29
S1981
11
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
edited Dec 15 '16 at 14:38
Kitter Catter
1,177917
1,177917
asked Dec 15 '16 at 14:29
S1981
11
asked Dec 15 '16 at 14:29
S1981
11
asked Dec 15 '16 at 14:29
S1981
11
11
Hint: Use completing the square to remove the double x's
– jugglingmike
Dec 15 '16 at 14:32
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 $$Ax+By+C=0$$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32
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.
– jugglingmike
Dec 15 '16 at 14:33
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?
– S1981
Dec 15 '16 at 14:44
add a comment |Â
Hint: Use completing the square to remove the double x's
– jugglingmike
Dec 15 '16 at 14:32
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 $$Ax+By+C=0$$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32
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.
– jugglingmike
Dec 15 '16 at 14:33
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?
– S1981
Dec 15 '16 at 14:44
Hint: Use completing the square to remove the double x's
– jugglingmike
Dec 15 '16 at 14:32
Hint: Use completing the square to remove the double x's
– jugglingmike
Dec 15 '16 at 14:32
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 $$Ax+By+C=0$$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32
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 $$Ax+By+C=0$$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32
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.
– jugglingmike
Dec 15 '16 at 14:33
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.
– jugglingmike
Dec 15 '16 at 14:33
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?
– S1981
Dec 15 '16 at 14:44
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?
– S1981
Dec 15 '16 at 14:44
add a comment |Â
1 Answer
1
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up vote
0
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Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
$$beginalign
y&=log_10(ax^2+bx+c)\
10^y&=ax^2+bx+c\
&=aleft(x^2+frac baxright)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2-left(frac b2aright)^2right)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2right)-aleft(frac b2aright)^2+c\
10^y&=aleft(x+frac b2aright)^2+c-frac b^24a
endalign$$
Letting $Y=10^y$ and $X=left(x+frac b2aright)^2$, we have $Y=aX+c-frac b^24a$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac b^24a$).
answered Dec 16 '16 at 1:14
W. Zhu
644212
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
$$beginalign
y&=log_10(ax^2+bx+c)\
10^y&=ax^2+bx+c\
&=aleft(x^2+frac baxright)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2-left(frac b2aright)^2right)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2right)-aleft(frac b2aright)^2+c\
10^y&=aleft(x+frac b2aright)^2+c-frac b^24a
endalign$$
Letting $Y=10^y$ and $X=left(x+frac b2aright)^2$, we have $Y=aX+c-frac b^24a$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac b^24a$).
answered Dec 16 '16 at 1:14
W. Zhu
644212
add a comment |Â
up vote
0
down vote
Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
$$beginalign
y&=log_10(ax^2+bx+c)\
10^y&=ax^2+bx+c\
&=aleft(x^2+frac baxright)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2-left(frac b2aright)^2right)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2right)-aleft(frac b2aright)^2+c\
10^y&=aleft(x+frac b2aright)^2+c-frac b^24a
endalign$$
Letting $Y=10^y$ and $X=left(x+frac b2aright)^2$, we have $Y=aX+c-frac b^24a$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac b^24a$).
answered Dec 16 '16 at 1:14
W. Zhu
644212
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
$$beginalign
y&=log_10(ax^2+bx+c)\
10^y&=ax^2+bx+c\
&=aleft(x^2+frac baxright)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2-left(frac b2aright)^2right)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2right)-aleft(frac b2aright)^2+c\
10^y&=aleft(x+frac b2aright)^2+c-frac b^24a
endalign$$
Letting $Y=10^y$ and $X=left(x+frac b2aright)^2$, we have $Y=aX+c-frac b^24a$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac b^24a$).
answered Dec 16 '16 at 1:14
W. Zhu
644212
Take powers or logarithms or complete the square or do whatever you need to isolate $x$ and $y$. I will give you an example.
$$beginalign
y&=log_10(ax^2+bx+c)\
10^y&=ax^2+bx+c\
&=aleft(x^2+frac baxright)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2-left(frac b2aright)^2right)+c\
&=aleft(x^2+frac bax+left(frac b2aright)^2right)-aleft(frac b2aright)^2+c\
10^y&=aleft(x+frac b2aright)^2+c-frac b^24a
endalign$$
Letting $Y=10^y$ and $X=left(x+frac b2aright)^2$, we have $Y=aX+c-frac b^24a$ (of the form $Y=MX+C$, where $M=a$ and $C=c-frac b^24a$).
answered Dec 16 '16 at 1:14
W. Zhu
644212
answered Dec 16 '16 at 1:14
W. Zhu
644212
answered Dec 16 '16 at 1:14
W. Zhu
644212
answered Dec 16 '16 at 1:14
W. Zhu
644212
644212
add a comment |Â
add a comment |Â
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Hint: Use completing the square to remove the double x's
– jugglingmike
Dec 15 '16 at 14:32
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 $$Ax+By+C=0$$
– Dr. Sonnhard Graubner
Dec 15 '16 at 14:32
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.
– jugglingmike
Dec 15 '16 at 14:33
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?
– S1981
Dec 15 '16 at 14:44