proble in using "solve function

i am using solve function to equate two equations but the result are not the one that i want and also have problem with the format of the result . the exp. is
[L D] = solve(L*(D-d)==0.1386,L/(D-d)==3.6,'Real',true)
==============
the result is in fraction form or some times show empty matrix

1 Kommentar

L = (9/250)*sqrt(385), D = (1/100)*sqrt(385)+d
L = -(9/250)*sqrt(385), D = -(1/100)*sqrt(385)+d

Antworten (2)

Mischa Kim
Mischa Kim am 20 Jan. 2014
Bearbeitet: Mischa Kim am 20 Jan. 2014

0 Stimmen

Have you assigned a numeric value for d ?
syms L D
d = 1;
[L D] = solve(L*(D-d)==0.1386,L/(D-d)==3.6,'Real',true)
L =
385^(1/2)/100 + 1
1 - 385^(1/2)/100
D =
(9*385^(1/2))/250
-(9*385^(1/2))/250
double(L(1))
ans =
1.196214168703486

1 Kommentar

vikas
vikas am 20 Jan. 2014
i have assigned value for d=0.1382 and i am getting the results same as yours but i want to get results in the format "double(L(1))" without using this code
Walter Roberson
Walter Roberson am 20 Jan. 2014

0 Stimmen

3 Kommentare

replace
syms L D
with
L = sym('L', 'd');
D = sym('D', 'd');
vikas
vikas am 22 Jan. 2014
when i used * L=sym('L','positive'); D=sym('D','positive'); solve(L*(D-d)-Aw,L/(D-d)-3.6,'Real',true) *
i got result
ans =
D: [1x1 sym]
L: [1x1 sym]
i have to select only one value of L and D which are dimensions
S = solve(L*(D-d)-Aw,L/(D-d)-3.6,'Real',true);
S.D
S.L
and if appropriate,
double(S.D)
double(S.L)

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am 20 Jan. 2014

Geschlossen:

am 20 Aug. 2021

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