Mixing
Simultaneous equations with two unknowns
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
The question I've been given is
$$beginarrayct&v\hline3&38\12&200endarray$$
Modelling equation is
$$v=ksqrtt-a$$
Calculate $a$ and $k$.
I tried to solve like:
$$38 = k sqrt12-a$$
$$200 = k sqrt3-a$$
$$162 = k sqrt12-a - k sqrt3-a$$
$$frac162k = sqrt12-a - sqrt3-a$$
$$(162/k)^2 = 12-a - 3 + a$$
$$(162/k)^2 = 9$$
$$162/k = 3$$
$$162/3 = k$$
$$k = 54$$
WHICH IS GIVEN AS WRONG. $k$ should be $65.45$
systems-of-equations radicals
edited Jul 23 at 11:47
Harry Peter
5,45311438
asked Jul 22 at 4:43
user578719
51
add a comment |Â
up vote
0
down vote
favorite
The question I've been given is
$$beginarrayct&v\hline3&38\12&200endarray$$
Modelling equation is
$$v=ksqrtt-a$$
Calculate $a$ and $k$.
I tried to solve like:
$$38 = k sqrt12-a$$
$$200 = k sqrt3-a$$
$$162 = k sqrt12-a - k sqrt3-a$$
$$frac162k = sqrt12-a - sqrt3-a$$
$$(162/k)^2 = 12-a - 3 + a$$
$$(162/k)^2 = 9$$
$$162/k = 3$$
$$162/3 = k$$
$$k = 54$$
WHICH IS GIVEN AS WRONG. $k$ should be $65.45$
systems-of-equations radicals
edited Jul 23 at 11:47
Harry Peter
5,45311438
asked Jul 22 at 4:43
user578719
51
mathjax resources to learn how to typeset maths.
– Siong Thye Goh
Jul 22 at 4:54
You swapped $3$ and $12$ !
– Yves Daoust
Jul 23 at 11:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question I've been given is
$$beginarrayct&v\hline3&38\12&200endarray$$
Modelling equation is
$$v=ksqrtt-a$$
Calculate $a$ and $k$.
I tried to solve like:
$$38 = k sqrt12-a$$
$$200 = k sqrt3-a$$
$$162 = k sqrt12-a - k sqrt3-a$$
$$frac162k = sqrt12-a - sqrt3-a$$
$$(162/k)^2 = 12-a - 3 + a$$
$$(162/k)^2 = 9$$
$$162/k = 3$$
$$162/3 = k$$
$$k = 54$$
WHICH IS GIVEN AS WRONG. $k$ should be $65.45$
systems-of-equations radicals
edited Jul 23 at 11:47
Harry Peter
5,45311438
asked Jul 22 at 4:43
user578719
51
The question I've been given is
$$beginarrayct&v\hline3&38\12&200endarray$$
Modelling equation is
$$v=ksqrtt-a$$
Calculate $a$ and $k$.
I tried to solve like:
$$38 = k sqrt12-a$$
$$200 = k sqrt3-a$$
$$162 = k sqrt12-a - k sqrt3-a$$
$$frac162k = sqrt12-a - sqrt3-a$$
$$(162/k)^2 = 12-a - 3 + a$$
$$(162/k)^2 = 9$$
$$162/k = 3$$
$$162/3 = k$$
$$k = 54$$
WHICH IS GIVEN AS WRONG. $k$ should be $65.45$
systems-of-equations radicals
edited Jul 23 at 11:47
Harry Peter
5,45311438
asked Jul 22 at 4:43
user578719
51
edited Jul 23 at 11:47
Harry Peter
5,45311438
edited Jul 23 at 11:47
Harry Peter
5,45311438
edited Jul 23 at 11:47
Harry Peter
5,45311438
5,45311438
asked Jul 22 at 4:43
user578719
51
asked Jul 22 at 4:43
user578719
51
asked Jul 22 at 4:43
user578719
51
51
mathjax resources to learn how to typeset maths.
– Siong Thye Goh
Jul 22 at 4:54
You swapped $3$ and $12$ !
– Yves Daoust
Jul 23 at 11:50
add a comment |Â
mathjax resources to learn how to typeset maths.
– Siong Thye Goh
Jul 22 at 4:54
You swapped $3$ and $12$ !
– Yves Daoust
Jul 23 at 11:50
mathjax resources to learn how to typeset maths.
– Siong Thye Goh
Jul 22 at 4:54
mathjax resources to learn how to typeset maths.
– Siong Thye Goh
Jul 22 at 4:54
You swapped $3$ and $12$ !
– Yves Daoust
Jul 23 at 11:50
You swapped $3$ and $12$ !
– Yves Daoust
Jul 23 at 11:50
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
0
down vote
accepted
It seems to me that in general this question would be easier if you were to square both sides:
$$
v^2=k^2(t-a)
$$
With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have
$$
1444 = 3k^2-ak^2 \
40000 = 12k^2-ak^2
$$
Subtracting the former from the latter gives us
$$
9k^2 = 38556 \
k^2 = 4284
k = sqrt4284 approx 65.452
$$
Then substituting $k^2 = 4284$ back into either of the above yields
$$
a = 12-frac400004284 = 3-frac14444284 approx 2.663
$$
Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.
answered Jul 22 at 5:00
Brian Tung
25.2k32453
Thanks, that got it
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
First mistake:
When time is $3$, the volume is $38$.
When time is $12$, the volume is $200$.
You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.
Second mistake:
- In general, $(sqrta - sqrtb)^2 ne a - b$.
Guide:
- You might like to divide the equations rather than subtract the equations.
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
Using your data, you have in fact $$38 = k sqrt3-atag 1$$
$$200 = k sqrt12-atag 2$$ Make the ratio
$$frac38200=fracsqrt3-a sqrt12-a $$ Square both sides to get
$$frac36110000=frac3-a12-aimplies a=frac28521071$$ Plug in $(1)$ to get $k=6 sqrt119approx 65.4523$.
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
Rewrite the model as
$$k^2t-k^2a=v^2$$
and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.
$$begincases3k^2-k^2a&=38^2,\12k^2-k^2a&=200^2.endcases$$
Then solving the system
$$k^2= 4284,$$ $$k^2a=11408.$$
The rest is easy.
answered Jul 23 at 12:01
Yves Daoust
111k665204
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It seems to me that in general this question would be easier if you were to square both sides:
$$
v^2=k^2(t-a)
$$
With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have
$$
1444 = 3k^2-ak^2 \
40000 = 12k^2-ak^2
$$
Subtracting the former from the latter gives us
$$
9k^2 = 38556 \
k^2 = 4284
k = sqrt4284 approx 65.452
$$
Then substituting $k^2 = 4284$ back into either of the above yields
$$
a = 12-frac400004284 = 3-frac14444284 approx 2.663
$$
Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.
answered Jul 22 at 5:00
Brian Tung
25.2k32453
Thanks, that got it
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
accepted
It seems to me that in general this question would be easier if you were to square both sides:
$$
v^2=k^2(t-a)
$$
With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have
$$
1444 = 3k^2-ak^2 \
40000 = 12k^2-ak^2
$$
Subtracting the former from the latter gives us
$$
9k^2 = 38556 \
k^2 = 4284
k = sqrt4284 approx 65.452
$$
Then substituting $k^2 = 4284$ back into either of the above yields
$$
a = 12-frac400004284 = 3-frac14444284 approx 2.663
$$
Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.
answered Jul 22 at 5:00
Brian Tung
25.2k32453
Thanks, that got it
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It seems to me that in general this question would be easier if you were to square both sides:
$$
v^2=k^2(t-a)
$$
With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have
$$
1444 = 3k^2-ak^2 \
40000 = 12k^2-ak^2
$$
Subtracting the former from the latter gives us
$$
9k^2 = 38556 \
k^2 = 4284
k = sqrt4284 approx 65.452
$$
Then substituting $k^2 = 4284$ back into either of the above yields
$$
a = 12-frac400004284 = 3-frac14444284 approx 2.663
$$
Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.
answered Jul 22 at 5:00
Brian Tung
25.2k32453
It seems to me that in general this question would be easier if you were to square both sides:
$$
v^2=k^2(t-a)
$$
With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have
$$
1444 = 3k^2-ak^2 \
40000 = 12k^2-ak^2
$$
Subtracting the former from the latter gives us
$$
9k^2 = 38556 \
k^2 = 4284
k = sqrt4284 approx 65.452
$$
Then substituting $k^2 = 4284$ back into either of the above yields
$$
a = 12-frac400004284 = 3-frac14444284 approx 2.663
$$
Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.
answered Jul 22 at 5:00
Brian Tung
25.2k32453
answered Jul 22 at 5:00
Brian Tung
25.2k32453
answered Jul 22 at 5:00
Brian Tung
25.2k32453
answered Jul 22 at 5:00
Brian Tung
25.2k32453
25.2k32453
Thanks, that got it
– user578719
Jul 22 at 5:59
add a comment |Â
Thanks, that got it
– user578719
Jul 22 at 5:59
Thanks, that got it
– user578719
Jul 22 at 5:59
Thanks, that got it
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
First mistake:
When time is $3$, the volume is $38$.
When time is $12$, the volume is $200$.
You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.
Second mistake:
- In general, $(sqrta - sqrtb)^2 ne a - b$.
Guide:
- You might like to divide the equations rather than subtract the equations.
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
First mistake:
When time is $3$, the volume is $38$.
When time is $12$, the volume is $200$.
You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.
Second mistake:
- In general, $(sqrta - sqrtb)^2 ne a - b$.
Guide:
- You might like to divide the equations rather than subtract the equations.
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First mistake:
When time is $3$, the volume is $38$.
When time is $12$, the volume is $200$.
You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.
Second mistake:
- In general, $(sqrta - sqrtb)^2 ne a - b$.
Guide:
- You might like to divide the equations rather than subtract the equations.
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
First mistake:
When time is $3$, the volume is $38$.
When time is $12$, the volume is $200$.
You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.
Second mistake:
- In general, $(sqrta - sqrtb)^2 ne a - b$.
Guide:
- You might like to divide the equations rather than subtract the equations.
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
answered Jul 22 at 4:47
Siong Thye Goh
77.6k134795
77.6k134795
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
add a comment |Â
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that
– user578719
Jul 22 at 5:59
add a comment |Â
up vote
0
down vote
Using your data, you have in fact $$38 = k sqrt3-atag 1$$
$$200 = k sqrt12-atag 2$$ Make the ratio
$$frac38200=fracsqrt3-a sqrt12-a $$ Square both sides to get
$$frac36110000=frac3-a12-aimplies a=frac28521071$$ Plug in $(1)$ to get $k=6 sqrt119approx 65.4523$.
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
Using your data, you have in fact $$38 = k sqrt3-atag 1$$
$$200 = k sqrt12-atag 2$$ Make the ratio
$$frac38200=fracsqrt3-a sqrt12-a $$ Square both sides to get
$$frac36110000=frac3-a12-aimplies a=frac28521071$$ Plug in $(1)$ to get $k=6 sqrt119approx 65.4523$.
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using your data, you have in fact $$38 = k sqrt3-atag 1$$
$$200 = k sqrt12-atag 2$$ Make the ratio
$$frac38200=fracsqrt3-a sqrt12-a $$ Square both sides to get
$$frac36110000=frac3-a12-aimplies a=frac28521071$$ Plug in $(1)$ to get $k=6 sqrt119approx 65.4523$.
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
Using your data, you have in fact $$38 = k sqrt3-atag 1$$
$$200 = k sqrt12-atag 2$$ Make the ratio
$$frac38200=fracsqrt3-a sqrt12-a $$ Square both sides to get
$$frac36110000=frac3-a12-aimplies a=frac28521071$$ Plug in $(1)$ to get $k=6 sqrt119approx 65.4523$.
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
answered Jul 22 at 6:07
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
up vote
0
down vote
Rewrite the model as
$$k^2t-k^2a=v^2$$
and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.
$$begincases3k^2-k^2a&=38^2,\12k^2-k^2a&=200^2.endcases$$
Then solving the system
$$k^2= 4284,$$ $$k^2a=11408.$$
The rest is easy.
answered Jul 23 at 12:01
Yves Daoust
111k665204
add a comment |Â
up vote
0
down vote
Rewrite the model as
$$k^2t-k^2a=v^2$$
and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.
$$begincases3k^2-k^2a&=38^2,\12k^2-k^2a&=200^2.endcases$$
Then solving the system
$$k^2= 4284,$$ $$k^2a=11408.$$
The rest is easy.
answered Jul 23 at 12:01
Yves Daoust
111k665204
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Rewrite the model as
$$k^2t-k^2a=v^2$$
and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.
$$begincases3k^2-k^2a&=38^2,\12k^2-k^2a&=200^2.endcases$$
Then solving the system
$$k^2= 4284,$$ $$k^2a=11408.$$
The rest is easy.
answered Jul 23 at 12:01
Yves Daoust
111k665204
Rewrite the model as
$$k^2t-k^2a=v^2$$
and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.
$$begincases3k^2-k^2a&=38^2,\12k^2-k^2a&=200^2.endcases$$
Then solving the system
$$k^2= 4284,$$ $$k^2a=11408.$$
The rest is easy.
answered Jul 23 at 12:01
Yves Daoust
111k665204
answered Jul 23 at 12:01
Yves Daoust
111k665204
answered Jul 23 at 12:01
Yves Daoust
111k665204
answered Jul 23 at 12:01
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859098%2fsimultaneous-equations-with-two-unknowns%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
mathjax resources to learn how to typeset maths.
– Siong Thye Goh
Jul 22 at 4:54
You swapped $3$ and $12$ !
– Yves Daoust
Jul 23 at 11:50