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Science Forum Index » Math - Symbolic Forum » how to solve this system of 3 equations?
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| Bo |
 Posted: Thu Feb 15, 2007 2:58 pm |
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Guest
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I have a system of equations that is driving me in circles
eq 1: (a-10)/c + c/b = pi/4;
eq 2: c^2 = b(a-10);
eq 3: ac = 10000;
I think the form for solving c should be a quadratic, based on a similar
example(below), but I cannot seem to get there.
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a quadratic
also--but I don't know what the steps are to come up with the equation for
c.
Can anyone help?
Thanks,
Bo |
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| Nasser Abbasi |
| Posted: Thu Feb 15, 2007 4:05 pm |
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"Bo" <bo@cephus.com> wrote in message
news:3b096$45d4ad64$41a26d4d$26244@API-DIGITAL.COM...
Quote:
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a quadratic
also--but I don't know what the steps are to come up with the equation for
c.
Can anyone help?
Thanks,
Bo
I think the equation you showed for 'c' is not the right equation? becuase
its roots are:
restart;
eq:=c^2+800/(Pi*c)-800000/Pi=0.;
solve(eq,c)
0.001000000000, 504.6260044, -504.6270044
while the solution for 'c' from your initial 3 equation is
eq1:= 100/c-a/c-c/b=-Pi/4.:
eq2:= c^2=b*(a-100):
eq3:= a*c=100000:
{c = 393.1174912, b = 1001.065471, a = 254.3768778},
{c = -647.7654001, b = -1649.521046, a = -154.3768778}
which does not match. the 'c' values are not the same as the roots of the
above 'c' equation.
Nasser |
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| Nasser Abbasi |
| Posted: Thu Feb 15, 2007 4:09 pm |
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"Nasser Abbasi" <nma@12000.org> wrote in message
news:q23Bh.28931$KW3.1047@newsfe17.phx...
Quote:
"Bo" <bo@cephus.com> wrote in message
news:3b096$45d4ad64$41a26d4d$26244@API-DIGITAL.COM...
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a
quadratic also--but I don't know what the steps are to come up with the
equation for c.
Can anyone help?
Thanks,
Bo
Quote: I think the equation you showed for 'c' is not the right equation? becuase
its roots are:
restart;
eq:=c^2+800/(Pi*c)-800000/Pi=0.;
solve(eq,c)
0.001000000000, 504.6260044, -504.6270044
while the solution for 'c' from your initial 3 equation is
eq1:= 100/c-a/c-c/b=-Pi/4.:
eq2:= c^2=b*(a-100):
eq3:= a*c=100000:
{c = 393.1174912, b = 1001.065471, a = 254.3768778},
{c = -647.7654001, b = -1649.521046, a = -154.3768778}
which does not match. the 'c' values are not the same as the roots of the
above 'c' equation.
Nasser
oh, sorry, it is the right quation. I mangled it up, here is its solution,
and it does match the solution obtain from the 3 set of equations:
restart;
eq:=c^2+800/Pi*c-800000/Pi=0.;
solve(eq,c);
393.1174912, -647.7654001
Nasser |
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| A N Niel |
| Posted: Thu Feb 15, 2007 4:11 pm |
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In article <3b096$45d4ad64$41a26d4d$26244@API-DIGITAL.COM>, Bo
<bo@cephus.com> wrote:
Quote: I have a system of equations that is driving me in circles
eq 1: (a-10)/c + c/b = pi/4;
eq 2: c^2 = b(a-10);
eq 3: ac = 10000;
I think the form for solving c should be a quadratic, based on a similar
example(below), but I cannot seem to get there.
a = 10000/c, b = -(1/10)*c^3/(-1000+c), where -80000+80*c+pi*c^2=0 |
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| John O'Flaherty |
| Posted: Thu Feb 15, 2007 4:33 pm |
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On Feb 15, 12:58 pm, "Bo" <b...@cephus.com> wrote:
Quote: I have a system of equations that is driving me in circles
eq 1: (a-10)/c + c/b = pi/4;
eq 2: c^2 = b(a-10);
eq 3: ac = 10000;
I think the form for solving c should be a quadratic, based on a similar
example(below), but I cannot seem to get there.
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a quadratic
also--but I don't know what the steps are to come up with the equation for c.
Solve equation 3 for a, then plug that into equation 2 and 1.
Now you have equations in c and b.
1a) (10000/c-10)/c + c/b = pi/4
1b) c^2 = b (10000/c) - 10/b
Get some of the c's out of the denominator by multiplying in extra
c's...
1c) 1e4 - 10c + 1/b * c^3 = pi/4 * c^2
2c) c^3 = 1e4b - 10bc
Now substitute the value for c^3 in 2c) back into 1c), getting
pi/4 * c^2 +20c - 2e4 = 0
That's your quadratic. The solutions are c~= 147.35, -172.82
--
John |
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| Chip Eastham |
| Posted: Thu Feb 15, 2007 6:12 pm |
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On Feb 15, 1:58 pm, "Bo" <b...@cephus.com> wrote:
Quote: I have a system of equations that is driving me in circles
eq 1: (a-10)/c + c/b = pi/4;
eq 2: c^2 = b(a-10);
eq 3: ac = 10000;
I think the form for solving c should be a quadratic, based on a similar
example(below), but I cannot seem to get there.
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a quadratic
also--but I don't know what the steps are to come up with the equation for
c.
Can anyone help?
Thanks,
Bo
Equation 1 contains indicated divisions by both
b and c, so I'll assume neither is allowed to
be zero.
Then equation 2 (upon division by bc) yields:
c/b = (a-10)/c
which you'll recognize as the two terms in the
lefthand side of equation 1:
(a-10)/c + c/b = pi/4
Therefore c/b = (a-10)/c = pi/8, and equation 3:
ac = 100000
combine to give us (multiplying both sides:
a(a-10) = 12500*pi
Complete the square:
a^2 - 10a + 25 = 12500*pi + 25
a = 5 +/- sqrt(12500*pi + 25)
and values for b,c are easily derived from:
c = 100000/a
b = 8c/pi
regards, chip |
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| Bo |
| Posted: Mon Feb 19, 2007 4:06 pm |
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"John O'Flaherty" <quiasmox@yahoo.com> wrote in message
news:1171571596.037769.193560@a75g2000cwd.googlegroups.com...
Quote: On Feb 15, 12:58 pm, "Bo" <b...@cephus.com> wrote:
I have a system of equations that is driving me in circles
eq 1: (a-10)/c + c/b = pi/4;
eq 2: c^2 = b(a-10);
eq 3: ac = 10000;
I think the form for solving c should be a quadratic, based on a similar
example(below), but I cannot seem to get there.
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a
quadratic
also--but I don't know what the steps are to come up with the equation
for c.
Solve equation 3 for a, then plug that into equation 2 and 1.
Now you have equations in c and b.
1a) (10000/c-10)/c + c/b = pi/4
1b) c^2 = b (10000/c) - 10/b
Get some of the c's out of the denominator by multiplying in extra
c's...
1c) 1e4 - 10c + 1/b * c^3 = pi/4 * c^2
2c) c^3 = 1e4b - 10bc
Now substitute the value for c^3 in 2c) back into 1c), getting
pi/4 * c^2 +20c - 2e4 = 0
That's your quadratic. The solutions are c~= 147.35, -172.82
--
John
Cool!
THanks John! |
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| osquiozo |
| Posted: Wed Feb 21, 2007 9:41 pm |
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Guest
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On Feb 15, 12:58 pm, "Bo" <b...@cephus.com> wrote:
Quote: I have a system of equations that is driving me in circles
eq 1: (a-10)/c + c/b = pi/4;
eq 2: c^2 = b(a-10);
eq 3: ac = 10000;
I think the form for solving c should be a quadratic, based on a similar
example(below), but I cannot seem to get there.
The example set is:
100/c - a/c - c/b = -pi/4;
c^2 = b(a-100);
ac = 100000
and the text shows the equation for c as:
c^2 + 800/pi *c - 800000/pi = 0;
But it does not show how that was derived.
Since my equations are so simliar, I'm thinking my c should be a quadratic
also--but I don't know what the steps are to come up with the equation for
c.
Can anyone help?
Thanks,
Bo
Hi Bo
You just can try in Mathematica:
In[3]:=
NSolve[{(a - 10)/c + c/b == Pi/4, c^2 == b*(a - 10),
a*c == 10000}, {a, b, c}]
and it should give you the following result:
{{a -> -57.8649, b -> -440.073, c -> -172.816}, {a -> 67.8649,
b -> 375.228, c -> 147.352}}
where this system have two sets of solutions.
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