How to calculate the equation with letter and variable

Question

HONG CHENG 2017-5-4

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https://ww2.mathworks.cn/matlabcentral/answers/338609-how-to-calculate-the-equation-with-letter-and-variable

编辑： Stephen23 2017-10-17

采纳的回答： Star Strider

在 MATLAB Online 中打开

Dear everyone,

I want to calculate an equation with letter and variable.

Now I can get the variable value presented with the letters I used

but I don't know how to change the letters as real numbers.

this is the example

%declear syms
syms x0 y0 x1 y1 x2 y2 a b p q d positive;
[x0 y0] = solve('(x-p)^2+(y-q)^2=d^2','(p-a)*(y-b)=(x-a)*(q-b)')

then I can get the result

but the problem is how to enter the real number value of letters like

a = 1;
b = 2;
p = 3;
q = 4;
d = 5;

then I can get the numerical value of x, y ???

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HONG CHENG 2017-5-4

编辑：HONG CHENG 2017-5-4

在 MATLAB Online 中打开

Maybe we can use the subs() function to do this

f=subs(solve('(x-p)^2+(2-q)^2=d^2') ,{a,b,p,q,d},{1,2,3,4,5})

but here we just can use one equation.

if anyone have other or better method, please show your idea

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Answer 1

Star Strider 2017-5-4

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The substitution will occur automatically. The problem is that you must put ‘x’, ‘x0’ and ‘y0’ in the equations you want to solve for them.

This will do the substitutions:

syms x0 y0 x1 y1 x2 y2 a b p q d positive
a = 1;
b = 2;
p = 3;
q = 4;
d = 5;
Eq1 = (x-p)^2+(y-q)^2==d^2;
Eq2 = (p-a)*(y-b)==(x-a)*(q-b);
[x0 y0] = solve(Eq1, Eq2);

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HONG CHENG 2017-5-4

在 MATLAB Online 中打开

And without the real number we get this

clear
clc
%declear syms
syms x0 y0 x1 y1 x2 y2 a b p q d positive;
%%the coodinates of S(x0,y0)
% a = 1;
% b = 2;
% p = 3;
% q = 4;
% d = 5;
Eq1 = (x0-p)^2+(y0-q)^2==d^2;
Eq2 = (p-a)*(y0-b)==(x0-a)*(q-b);
[x0 y0] = solve(Eq1, Eq2);
% y0 = round(y0)
% x0 = round(x0)
x0
x0 =
   -(a*q - b*p - a*(q^3/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + b*d*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (a^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*b*q^2)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + (b^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - d*q*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (p^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*a*p*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2)) + p*(q^3/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + b*d*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (a^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*b*q^2)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + (b^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - d*q*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (p^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*a*p*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2)))/(b - q)
   -(a*q - b*p - a*(q^3/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - b*d*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (a^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*b*q^2)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + (b^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + d*q*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (p^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*a*p*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2)) + p*(q^3/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - b*d*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (a^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*b*q^2)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + (b^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + d*q*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (p^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*a*p*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2)))/(b - q)
y0
y0 =
   q^3/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + b*d*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (a^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*b*q^2)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + (b^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - d*q*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (p^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*a*p*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2)
   q^3/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - b*d*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (a^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*b*q^2)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + (b^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) + d*q*(1/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2))^(1/2) + (p^2*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2) - (2*a*p*q)/(a^2 - 2*a*p + b^2 - 2*b*q + p^2 + q^2)

Walter Roberson 2017-5-7

HONG CHENG comments on Star Strider's Answer:

kind and big god in MATLAB

Karan Gill 2017-5-9

See my answer below for why you only get one solution.

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Answer 2

Karan Gill 2017-5-9

2
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/338609-how-to-calculate-the-equation-with-letter-and-variable#answer_266285

编辑：Stephen23 2017-10-17

在 MATLAB Online 中打开

Use subs to substitute values, as shown below. You only get one solution because in the other solution, "x" is negative, which is not allowed due to the assumption that it is positive.

BUT if don't substitute values before solving, then you get two solutions because the second solution can be positive under certain conditions. "solve" also issues a warning stating that conditions apply to the solutions. If you use the "ReturnConditions" option, then you get these conditions. Applying these conditions will let you find correct values. See the doc: https://www.mathworks.com/help/symbolic/solve-an-algebraic-equation.html.

syms x y a b p q d positive
eqn1 = (x-p)^2+(y-q)^2 == d^2;
eqn2 = (p-a)*(y-b) == (x-a)*(q-b);
vars = [a b p q d];
vals = sym([1 2 3 4 5]);
eqn1 = subs(eqn1,vars,vals);
eqn2 = subs(eqn2,vars,vals);
[xSol ySol] = solve(eqn1, eqn2)
xSol =
(5*2^(1/2))/2 + 3
ySol =
(5*2^(1/2))/2 + 4

Lastly, do not redeclare symbolic variables as doubles because you are overwriting them. So don't do this.

syms a
a = 1

Just do

a = 1

Or use "subs" to substitute for "a" in an expression

f = a^2
subs(f,a,2)

Karan (Symbolic doc)

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Karan Gill 2017-5-9

Glad to hear that! If my answer was helpful, could you accept it so that others can find the reason.

HONG CHENG 2017-5-10

编辑：HONG CHENG 2017-5-10

Sorry, I will reopen another question. Can you answer in this link - a new question ?

Because I think Another Answer does a lot help for me too.

I will accept the answer in this new link. Thanks a lot !!!1

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