How can I let Matlab to put different numbers in different variables?

I have to complete my code that should solve an equation with 7 variables to get a specific number.

my equation is like this:

5*x1 + 6*x2 + 7*x3 + 8*x4 + 9*x5 + 10*x6 + 11*x7 = 77 (to get more answers, I will put range like = from 75 to 78).

I will get many solutions, but I am confused how to code it.

I started with initiating x1,2,3,4,5,6,7 and each time I make loop for changing one variable and keep other constants. Then, I do the same thing with the second variable.

This way is not efficient. Can you suggest a coding way that is more efficient?

Thank you

[Walter Roberson]

Are there constraints on the values of the variables? For example are they restricted to positive integers? If they are restricted to integers then you would have a Diophantine Equation and there are techniques for solving those.

[Ali Alnemer]

No constraints at all

Answer 1

[Star Strider]

You have 1 equation in 7 unknowns, so you will get an infinity of solutions. Solving it seems to be pointless.

To code it, I would use an anonymous function:

f = @(x) 5*x(1) + 6*x(2) + 7*x(3) + 8*x(4) + 9*x(5) + 10*x(6) + 11*x(7) - 77;

Your x-values are now one vector: [x(1),x(2),...,x(7)]. Given a random starting vector, you could use the fsolve function to ‘solve’ it for one set of variables. Whether that is an efficient use of your time is for you to decide.

[Ali Alnemer]

Later, I will have six other equations.

But for now, I search for a code for a this equation that gives me infinity solutions

[Star Strider]

If all seven are linear equations, you can simply use the mldivide,\ (link) function to solve them as a matrix equation.

One function that will provide an infinity of solutions is fsolve (link).

You can use the function I wrote as the objective function.

[Ali Alnemer]

I am a beginner in Matlab. Can you please explain how to specify

f = @(x) 5*x(1) + 6*x(2) + 7*x(3) + 8*x(4) + 9*x(5) + 10*x(6) + 11*x(7) - 77; with vector matrix?

Also, can you help me with loop method ? That we start with a given values of x1,x2,x3,x4..... and ask Matlab to change the values and break when it comes up with results that equal to range from 75-78.

it would be great if you could help. Thank you

[Star Strider]

I will do my best.

First, to pass a vector of values to the function, create the vector and then use it as a function argument as you would with any other function:

f = @(x) 5*x(1) + 6*x(2) + 7*x(3) + 8*x(4) + 9*x(5) + 10*x(6) + 11*x(7) - 77;

xv = [1 2 3 4 5 6 7];
q = f(xv)

q =
   175

Second, use the function with the Optimization Toolbox fsolve function to calculate a suitable vector of estimates for a solution:

[xs,fval] = fsolve(f, xv)

xs =
     -0.83824     -0.20588      0.42647       1.0588       1.6912       2.3235       2.9559

fval =
   7.7575e-09

The ‘fval’ output is the function value when the fsolve converges. Depending on the initial estimate vector (I used the same ‘xv’ here), the solution will be different.

Note — If all seven of your equations are linear, fsolve would be ‘overkill’. Just use the matrix left division operator ‘\’ to solve them. I will help you with that when you have defined the rest of your equations.

[Ali Alnemer]

Very nice. Actually, the above equation was only example that is similar to what I have. I will give 3 of my equations and I will see how you proceed with it, and I hope I can follow you with the rest of equations.

first equation: 42.4*x1 + 83*x2 + 100.1*x3 + 100*x4 + 0*x5 + 62.6*x6 + 118*x7 = 77

second equation: 46.4*x1 + 77.2*x2 + 84.9*x3 + 100*x4 + 0*x5 + 62*x6 + 103*x7 = 73.9

Third equation: 679*x1 + 778.81*x2 + 745.42*x3 + 692.14*x4 + 683.94*x5 + 625.9*x6 + 867.03*x7 = 720.7

Note: The chosen values of x1,x2,x3,x4... should give the answers of all equations or with error +-5 ( i.e. 77 with +-5, 73.9 +-5 and etc).

I hope the idea is clear. If you help me how to do it with these 3 equations, I can proceed and add my other equations to the code.

Thanks again

[Star Strider]

My pleasure.

You have a linear system, so keep adding the ‘left-hand-side’ equation coefficients to the ‘LHS’ matrix, the right-hand-side values to the ‘RHS’ vector, then calculate the ‘X’ variables. This will solve for them in the least-squares sense.

The Code —

LHS = [42.4  83  100.1  100 0  62.6  118;  
     46.4  77.2  84.9  100  0  62  103;  
     679  778.81  745.42  692.14  683.94  625.9  867.03];

RHS = [77;  73.9;  720.7];

X = LHS\RHS;

You have to have at least seven rows in ‘LHS’ and ‘RHS’ to get a valid solution. You can have more.

[Ali Alnemer]

But is there anyway to write a code (with loops for example) to give me matrix of some possible solutions ?

This is because I don't want to see only one solution. Instead, I want to see matrix of possible solutions ( so that if I put one solution in one of the equations, it gives me an answer even with range +-5 to the shown answer. For example, if I got one possible answer from only equation 1 and try to put this solution in the second equation, it gives 73.9 +- 5) and same for following equations.

Hope it is clear.

[Star Strider]

There is only one optimal solution to a linear system of seven (or more) equations with seven unknowns, with real coefficients. You can use the Statistics and Machine Learning Toolbox regress function to get the confidence intervals of the ‘X’ values, since there will be some uncertainty in their estimation.

You can use a loop to experiment with other values for the ‘X’ vector. However, the result the least-squares solution will be the best estimate for it.