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Easiest way to solve this system of equations
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up vote
15
down vote
favorite
I have these two equations:
$$x=fracab(1+k)b+ka\ y=fracab(1+k)a+kb$$
where $a,b$ are constants and $k$ is a parameter to be eliminated.
A relation between $x,y$ is to be found. What is the best way to do it? Cross multiplying and solving is a bit too hectic. Is there a way we can maybe exploit the symmetry of the situation? Thanks!!
algebra-precalculus systems-of-equations
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 3:25
tatan
5,02442053
 |Â
show 1 more comment
up vote
15
down vote
favorite
I have these two equations:
$$x=fracab(1+k)b+ka\ y=fracab(1+k)a+kb$$
where $a,b$ are constants and $k$ is a parameter to be eliminated.
A relation between $x,y$ is to be found. What is the best way to do it? Cross multiplying and solving is a bit too hectic. Is there a way we can maybe exploit the symmetry of the situation? Thanks!!
algebra-precalculus systems-of-equations
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 3:25
tatan
5,02442053
5
Don't know if this helps but $frac1x + frac1y = frac1a + frac1b$
– iamwhoiam
Jul 16 at 3:37
1
@iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer.
– Mason
Jul 16 at 3:40
I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^-1$ and $y^-1$.
– Cornman
Jul 16 at 3:42
@iamwhoiam This was what I was looking for!! Thanks a lot!
– tatan
Jul 16 at 3:43
1
@iamwhoiam I think you should post it as an answer!
– tatan
Jul 16 at 3:48
 |Â
show 1 more comment
up vote
15
down vote
favorite
up vote
15
down vote
favorite
I have these two equations:
$$x=fracab(1+k)b+ka\ y=fracab(1+k)a+kb$$
where $a,b$ are constants and $k$ is a parameter to be eliminated.
A relation between $x,y$ is to be found. What is the best way to do it? Cross multiplying and solving is a bit too hectic. Is there a way we can maybe exploit the symmetry of the situation? Thanks!!
algebra-precalculus systems-of-equations
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 3:25
tatan
5,02442053
I have these two equations:
$$x=fracab(1+k)b+ka\ y=fracab(1+k)a+kb$$
where $a,b$ are constants and $k$ is a parameter to be eliminated.
A relation between $x,y$ is to be found. What is the best way to do it? Cross multiplying and solving is a bit too hectic. Is there a way we can maybe exploit the symmetry of the situation? Thanks!!
algebra-precalculus systems-of-equations
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 3:25
tatan
5,02442053
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
edited Jul 16 at 8:35
Rodrigo de Azevedo
12.5k41751
12.5k41751
asked Jul 16 at 3:25
tatan
5,02442053
asked Jul 16 at 3:25
tatan
5,02442053
asked Jul 16 at 3:25
tatan
5,02442053
5,02442053
5
Don't know if this helps but $frac1x + frac1y = frac1a + frac1b$
– iamwhoiam
Jul 16 at 3:37
1
@iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer.
– Mason
Jul 16 at 3:40
I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^-1$ and $y^-1$.
– Cornman
Jul 16 at 3:42
@iamwhoiam This was what I was looking for!! Thanks a lot!
– tatan
Jul 16 at 3:43
1
@iamwhoiam I think you should post it as an answer!
– tatan
Jul 16 at 3:48
 |Â
show 1 more comment
5
Don't know if this helps but $frac1x + frac1y = frac1a + frac1b$
– iamwhoiam
Jul 16 at 3:37
1
@iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer.
– Mason
Jul 16 at 3:40
I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^-1$ and $y^-1$.
– Cornman
Jul 16 at 3:42
@iamwhoiam This was what I was looking for!! Thanks a lot!
– tatan
Jul 16 at 3:43
1
@iamwhoiam I think you should post it as an answer!
– tatan
Jul 16 at 3:48
5
5
Don't know if this helps but $frac1x + frac1y = frac1a + frac1b$
– iamwhoiam
Jul 16 at 3:37
Don't know if this helps but $frac1x + frac1y = frac1a + frac1b$
– iamwhoiam
Jul 16 at 3:37
1
1
@iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer.
– Mason
Jul 16 at 3:40
@iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer.
– Mason
Jul 16 at 3:40
I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^-1$ and $y^-1$.
– Cornman
Jul 16 at 3:42
I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^-1$ and $y^-1$.
– Cornman
Jul 16 at 3:42
@iamwhoiam This was what I was looking for!! Thanks a lot!
– tatan
Jul 16 at 3:43
@iamwhoiam This was what I was looking for!! Thanks a lot!
– tatan
Jul 16 at 3:43
1
1
@iamwhoiam I think you should post it as an answer!
– tatan
Jul 16 at 3:48
@iamwhoiam I think you should post it as an answer!
– tatan
Jul 16 at 3:48
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
44
down vote
accepted
Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $frac1x$ and $frac1y$. With this approach, we observe that $$frac1x + frac1y = frac1a + frac1b$$
answered Jul 16 at 3:55
iamwhoiam
1,016612
1
Nicely done (+1).
– dxiv
Jul 16 at 4:00
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
7
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
2
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
1
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
 |Â
show 1 more comment
up vote
16
down vote
Direct elimination doesn't look so hectic in this case:
$$(b+ka)x=ab(1+k) iff ka(x-b)=b(a-x)iff k = - fracb(x-a)a(x-b)$$
Doing the same for the second equation then equating eliminates $,k,$.
answered Jul 16 at 3:41
dxiv
54.3k64797
1
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
add a comment |Â
up vote
4
down vote
Alternatively, note that $$frac xy=fraca+kbb+kaimplies (bx-ay)=k(by-ax)implies k=fracbx-ayby-ax$$ and equate with @dxiv's answer.
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
44
down vote
accepted
Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $frac1x$ and $frac1y$. With this approach, we observe that $$frac1x + frac1y = frac1a + frac1b$$
answered Jul 16 at 3:55
iamwhoiam
1,016612
1
Nicely done (+1).
– dxiv
Jul 16 at 4:00
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
7
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
2
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
1
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
 |Â
show 1 more comment
up vote
44
down vote
accepted
Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $frac1x$ and $frac1y$. With this approach, we observe that $$frac1x + frac1y = frac1a + frac1b$$
answered Jul 16 at 3:55
iamwhoiam
1,016612
1
Nicely done (+1).
– dxiv
Jul 16 at 4:00
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
7
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
2
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
1
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
 |Â
show 1 more comment
up vote
44
down vote
accepted
up vote
44
down vote
accepted
Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $frac1x$ and $frac1y$. With this approach, we observe that $$frac1x + frac1y = frac1a + frac1b$$
answered Jul 16 at 3:55
iamwhoiam
1,016612
Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $frac1x$ and $frac1y$. With this approach, we observe that $$frac1x + frac1y = frac1a + frac1b$$
answered Jul 16 at 3:55
iamwhoiam
1,016612
answered Jul 16 at 3:55
iamwhoiam
1,016612
answered Jul 16 at 3:55
iamwhoiam
1,016612
answered Jul 16 at 3:55
iamwhoiam
1,016612
1,016612
1
Nicely done (+1).
– dxiv
Jul 16 at 4:00
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
7
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
2
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
1
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
 |Â
show 1 more comment
1
Nicely done (+1).
– dxiv
Jul 16 at 4:00
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
7
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
2
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
1
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
1
1
Nicely done (+1).
– dxiv
Jul 16 at 4:00
Nicely done (+1).
– dxiv
Jul 16 at 4:00
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
What step am I missing to get to $frac1a + frac1b $?$frac1x + frac1y = fraca + ak + b + kbab(1 + k) iff fraca (1 + k) + b (1 + k)ab(1 + k) iff frac(a + b) (1 + k)ab(1 + k) iff frac(a + b)ab iff ???$
– Phil Patterson
Jul 16 at 20:43
7
7
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
@PhilPatterson $frac(a+b)ab$ <-> $fracaab + fracbab$ <-> $frac1b + frac1a$ <-> $frac1a+frac1b$
– pizzapants184
Jul 16 at 21:18
2
2
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
@pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me!
– Phil Patterson
Jul 17 at 4:28
1
1
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW
– Phil Patterson
Jul 19 at 17:37
 |Â
show 1 more comment
up vote
16
down vote
Direct elimination doesn't look so hectic in this case:
$$(b+ka)x=ab(1+k) iff ka(x-b)=b(a-x)iff k = - fracb(x-a)a(x-b)$$
Doing the same for the second equation then equating eliminates $,k,$.
answered Jul 16 at 3:41
dxiv
54.3k64797
1
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
add a comment |Â
up vote
16
down vote
Direct elimination doesn't look so hectic in this case:
$$(b+ka)x=ab(1+k) iff ka(x-b)=b(a-x)iff k = - fracb(x-a)a(x-b)$$
Doing the same for the second equation then equating eliminates $,k,$.
answered Jul 16 at 3:41
dxiv
54.3k64797
1
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
add a comment |Â
up vote
16
down vote
up vote
16
down vote
Direct elimination doesn't look so hectic in this case:
$$(b+ka)x=ab(1+k) iff ka(x-b)=b(a-x)iff k = - fracb(x-a)a(x-b)$$
Doing the same for the second equation then equating eliminates $,k,$.
answered Jul 16 at 3:41
dxiv
54.3k64797
Direct elimination doesn't look so hectic in this case:
$$(b+ka)x=ab(1+k) iff ka(x-b)=b(a-x)iff k = - fracb(x-a)a(x-b)$$
Doing the same for the second equation then equating eliminates $,k,$.
answered Jul 16 at 3:41
dxiv
54.3k64797
edited Jul 16 at 3:45
answered Jul 16 at 3:41
dxiv
54.3k64797
answered Jul 16 at 3:41
dxiv
54.3k64797
answered Jul 16 at 3:41
dxiv
54.3k64797
54.3k64797
1
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
add a comment |Â
1
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
1
1
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-)
– tatan
Jul 16 at 3:44
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
@tatan I agree, and will upvote that once posted as an answer ;-)
– dxiv
Jul 16 at 3:47
add a comment |Â
up vote
4
down vote
Alternatively, note that $$frac xy=fraca+kbb+kaimplies (bx-ay)=k(by-ax)implies k=fracbx-ayby-ax$$ and equate with @dxiv's answer.
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
add a comment |Â
up vote
4
down vote
Alternatively, note that $$frac xy=fraca+kbb+kaimplies (bx-ay)=k(by-ax)implies k=fracbx-ayby-ax$$ and equate with @dxiv's answer.
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Alternatively, note that $$frac xy=fraca+kbb+kaimplies (bx-ay)=k(by-ax)implies k=fracbx-ayby-ax$$ and equate with @dxiv's answer.
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
Alternatively, note that $$frac xy=fraca+kbb+kaimplies (bx-ay)=k(by-ax)implies k=fracbx-ayby-ax$$ and equate with @dxiv's answer.
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
answered Jul 16 at 8:24
TheSimpliFire
9,70261951
9,70261951
add a comment |Â
add a comment |Â
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5
Don't know if this helps but $frac1x + frac1y = frac1a + frac1b$
– iamwhoiam
Jul 16 at 3:37
1
@iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer.
– Mason
Jul 16 at 3:40
I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^-1$ and $y^-1$.
– Cornman
Jul 16 at 3:42
@iamwhoiam This was what I was looking for!! Thanks a lot!
– tatan
Jul 16 at 3:43
1
@iamwhoiam I think you should post it as an answer!
– tatan
Jul 16 at 3:48