Simpler proof that the coefficients must be proportional for an equation involving a ratio of two linear...
$begingroup$
I'm an adult learner, but I have recently been doing some math olympiad preparation problems to bone up my algebra--it has been a lot of fun!
Recently I came across a problem that involved something like
$$frac{ax+b}{cx+d}=frac{ex+f}{gx+h} $$
The problem only involved a specific case, and I was able to solve it a couple different ways, but I started to get curious about the possibility that if this expression is true for all x (where defined) that
$$frac{a}{e}=frac{b}{f}=frac{c}{g}=frac{d}{h}$$
I was able to persuade myself of that, because if $x=0$ we immediately have $frac{b}{d}=frac{f}{h}$ and so $frac{b}{f}=frac{d}{h}$. And if we consider $lim_{x to infty}$ we have $frac{a}{c}=frac{e}{g}$ and so $frac{a}{e}=frac{c}{g}$. Also, the zeros of both sides, when defined, demand $ax+b=ex+f$ and so $frac{a}{b}=frac{e}{f}$ and $frac{a}{e}=frac{b}{f}$ (and where the zeros aren't defined, we have constant ratios, so we're done).
I have a hunch, though, that I am somehow radically overcomplicating this--does anyone know off hand if there is a (potentially much) simpler way?
algebra-precalculus
edited Dec 28 '18 at 19:39
Larry
2,53531131
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
$endgroup$
add a comment |
$begingroup$
I'm an adult learner, but I have recently been doing some math olympiad preparation problems to bone up my algebra--it has been a lot of fun!
Recently I came across a problem that involved something like
$$frac{ax+b}{cx+d}=frac{ex+f}{gx+h} $$
The problem only involved a specific case, and I was able to solve it a couple different ways, but I started to get curious about the possibility that if this expression is true for all x (where defined) that
$$frac{a}{e}=frac{b}{f}=frac{c}{g}=frac{d}{h}$$
I was able to persuade myself of that, because if $x=0$ we immediately have $frac{b}{d}=frac{f}{h}$ and so $frac{b}{f}=frac{d}{h}$. And if we consider $lim_{x to infty}$ we have $frac{a}{c}=frac{e}{g}$ and so $frac{a}{e}=frac{c}{g}$. Also, the zeros of both sides, when defined, demand $ax+b=ex+f$ and so $frac{a}{b}=frac{e}{f}$ and $frac{a}{e}=frac{b}{f}$ (and where the zeros aren't defined, we have constant ratios, so we're done).
I have a hunch, though, that I am somehow radically overcomplicating this--does anyone know off hand if there is a (potentially much) simpler way?
algebra-precalculus
edited Dec 28 '18 at 19:39
Larry
2,53531131
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
$endgroup$
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– lhf
Dec 28 '18 at 16:05
$begingroup$
Thank you for the reference! It has been interesting reading, but the connection is a bit above my level. Can you elaborate slightly?
$endgroup$
– mcgosc
Dec 29 '18 at 14:54
add a comment |
$begingroup$
I'm an adult learner, but I have recently been doing some math olympiad preparation problems to bone up my algebra--it has been a lot of fun!
Recently I came across a problem that involved something like
$$frac{ax+b}{cx+d}=frac{ex+f}{gx+h} $$
The problem only involved a specific case, and I was able to solve it a couple different ways, but I started to get curious about the possibility that if this expression is true for all x (where defined) that
$$frac{a}{e}=frac{b}{f}=frac{c}{g}=frac{d}{h}$$
I was able to persuade myself of that, because if $x=0$ we immediately have $frac{b}{d}=frac{f}{h}$ and so $frac{b}{f}=frac{d}{h}$. And if we consider $lim_{x to infty}$ we have $frac{a}{c}=frac{e}{g}$ and so $frac{a}{e}=frac{c}{g}$. Also, the zeros of both sides, when defined, demand $ax+b=ex+f$ and so $frac{a}{b}=frac{e}{f}$ and $frac{a}{e}=frac{b}{f}$ (and where the zeros aren't defined, we have constant ratios, so we're done).
I have a hunch, though, that I am somehow radically overcomplicating this--does anyone know off hand if there is a (potentially much) simpler way?
algebra-precalculus
edited Dec 28 '18 at 19:39
Larry
2,53531131
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
$endgroup$
I'm an adult learner, but I have recently been doing some math olympiad preparation problems to bone up my algebra--it has been a lot of fun!
Recently I came across a problem that involved something like
$$frac{ax+b}{cx+d}=frac{ex+f}{gx+h} $$
The problem only involved a specific case, and I was able to solve it a couple different ways, but I started to get curious about the possibility that if this expression is true for all x (where defined) that
$$frac{a}{e}=frac{b}{f}=frac{c}{g}=frac{d}{h}$$
I was able to persuade myself of that, because if $x=0$ we immediately have $frac{b}{d}=frac{f}{h}$ and so $frac{b}{f}=frac{d}{h}$. And if we consider $lim_{x to infty}$ we have $frac{a}{c}=frac{e}{g}$ and so $frac{a}{e}=frac{c}{g}$. Also, the zeros of both sides, when defined, demand $ax+b=ex+f$ and so $frac{a}{b}=frac{e}{f}$ and $frac{a}{e}=frac{b}{f}$ (and where the zeros aren't defined, we have constant ratios, so we're done).
I have a hunch, though, that I am somehow radically overcomplicating this--does anyone know off hand if there is a (potentially much) simpler way?
algebra-precalculus
algebra-precalculus
edited Dec 28 '18 at 19:39
Larry
2,53531131
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
edited Dec 28 '18 at 19:39
Larry
2,53531131
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
edited Dec 28 '18 at 19:39
Larry
2,53531131
edited Dec 28 '18 at 19:39
Larry
2,53531131
edited Dec 28 '18 at 19:39
Larry
2,53531131
2,53531131
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
asked Dec 28 '18 at 15:41
mcgoscmcgosc
112
112
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– lhf
Dec 28 '18 at 16:05
$begingroup$
Thank you for the reference! It has been interesting reading, but the connection is a bit above my level. Can you elaborate slightly?
$endgroup$
– mcgosc
Dec 29 '18 at 14:54
add a comment |
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– lhf
Dec 28 '18 at 16:05
$begingroup$
Thank you for the reference! It has been interesting reading, but the connection is a bit above my level. Can you elaborate slightly?
$endgroup$
– mcgosc
Dec 29 '18 at 14:54
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– lhf
Dec 28 '18 at 16:05
$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– lhf
Dec 28 '18 at 16:05
$begingroup$
Thank you for the reference! It has been interesting reading, but the connection is a bit above my level. Can you elaborate slightly?
$endgroup$
– mcgosc
Dec 29 '18 at 14:54
$begingroup$
Thank you for the reference! It has been interesting reading, but the connection is a bit above my level. Can you elaborate slightly?
$endgroup$
– mcgosc
Dec 29 '18 at 14:54
add a comment |
1 Answer
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$begingroup$
good question and welcome to the platform.
I haven't fully worked it out, but this might push you in the right direction..
Given,
$$frac{ax+b}{cx+d} = frac{ex+f}{gx+h}$$
multiply out to get
$$agx^2 + ahx + bgx + bh = cex^2 + cfx + dex + df.$$
For these to have any chance of being equal, we must have $agx^2 = cex^2$. This implies $ag = ce rightarrow frac{a}{e} = frac{c}{g}$. Similarly, we must have $bh = df rightarrow frac{b}{f} = frac{d}{h}$.
Then you have to equate $frac{a}{e} = frac{c}{g}$ and $frac{b}{f} = frac{d}{h}$ some way.
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
$endgroup$
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
add a comment |
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$begingroup$
good question and welcome to the platform.
I haven't fully worked it out, but this might push you in the right direction..
Given,
$$frac{ax+b}{cx+d} = frac{ex+f}{gx+h}$$
multiply out to get
$$agx^2 + ahx + bgx + bh = cex^2 + cfx + dex + df.$$
For these to have any chance of being equal, we must have $agx^2 = cex^2$. This implies $ag = ce rightarrow frac{a}{e} = frac{c}{g}$. Similarly, we must have $bh = df rightarrow frac{b}{f} = frac{d}{h}$.
Then you have to equate $frac{a}{e} = frac{c}{g}$ and $frac{b}{f} = frac{d}{h}$ some way.
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
$endgroup$
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
add a comment |
$begingroup$
good question and welcome to the platform.
I haven't fully worked it out, but this might push you in the right direction..
Given,
$$frac{ax+b}{cx+d} = frac{ex+f}{gx+h}$$
multiply out to get
$$agx^2 + ahx + bgx + bh = cex^2 + cfx + dex + df.$$
For these to have any chance of being equal, we must have $agx^2 = cex^2$. This implies $ag = ce rightarrow frac{a}{e} = frac{c}{g}$. Similarly, we must have $bh = df rightarrow frac{b}{f} = frac{d}{h}$.
Then you have to equate $frac{a}{e} = frac{c}{g}$ and $frac{b}{f} = frac{d}{h}$ some way.
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
$endgroup$
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
add a comment |
$begingroup$
good question and welcome to the platform.
I haven't fully worked it out, but this might push you in the right direction..
Given,
$$frac{ax+b}{cx+d} = frac{ex+f}{gx+h}$$
multiply out to get
$$agx^2 + ahx + bgx + bh = cex^2 + cfx + dex + df.$$
For these to have any chance of being equal, we must have $agx^2 = cex^2$. This implies $ag = ce rightarrow frac{a}{e} = frac{c}{g}$. Similarly, we must have $bh = df rightarrow frac{b}{f} = frac{d}{h}$.
Then you have to equate $frac{a}{e} = frac{c}{g}$ and $frac{b}{f} = frac{d}{h}$ some way.
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
$endgroup$
good question and welcome to the platform.
I haven't fully worked it out, but this might push you in the right direction..
Given,
$$frac{ax+b}{cx+d} = frac{ex+f}{gx+h}$$
multiply out to get
$$agx^2 + ahx + bgx + bh = cex^2 + cfx + dex + df.$$
For these to have any chance of being equal, we must have $agx^2 = cex^2$. This implies $ag = ce rightarrow frac{a}{e} = frac{c}{g}$. Similarly, we must have $bh = df rightarrow frac{b}{f} = frac{d}{h}$.
Then you have to equate $frac{a}{e} = frac{c}{g}$ and $frac{b}{f} = frac{d}{h}$ some way.
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
answered Dec 28 '18 at 19:03
T. FoT. Fo
496311
496311
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
add a comment |
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
$begingroup$
Thank you for the warm welcome!
$endgroup$
– mcgosc
Dec 29 '18 at 14:53
add a comment |
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$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– lhf
Dec 28 '18 at 16:05
$begingroup$
Thank you for the reference! It has been interesting reading, but the connection is a bit above my level. Can you elaborate slightly?
$endgroup$
– mcgosc
Dec 29 '18 at 14:54