Calculation of mortgage interest from the amount of the mortgage, annuity monthly installments and number of years

I am having troubles to find a equation for interst (x) from the mortgage equation:

a=M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

x=? (Online equation solvers can't calculate it)

For Mortage 100,000 (M), interest x=10 and 1 year (r) monthly annuity a = 8,791.59

Answer 1

[Walter Roberson]

M = 100000;
a = 8791.59;
r = 1;
syms x
eqn = a == M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

eqn = 

sol = solve(eqn, x);
vsol = vpa(sol, 16)

vsol = 

vsol = vsol(imag(vsol)==0)

vsol = 

[Dan Richter]

Thank you very much but I am looking just for a general equation (variables without a real values) for x from the eqn above ;-).

[Walter Roberson]

MATLAB cannot express the general solution for unknown r values, as r controls the number of solutions.

There are some programming languages that have ways of expressing the general solution for unknown r, but they do not do so in any prescriptive way that can be used for calculations.

[Walter Roberson]

For that r value the general solution is

root(z^12 + (z^11*(14400*M - 1200*a))/M + (z^10*(95040000*M - 17280000*a))/M + (z^9*(380160000000*M - 114048000000*a))/M + (z^8*(1026432000000000*M - 456192000000000*a))/M + (z^7*(1970749440000000000*M - 1231718400000000000*a))/M + (z^6*(2759049216000000000000*M - 2364899328000000000000*a))/M + (z^5*(2837879193600000000000000*M - 3310859059200000000000000*a))/M + (z^4*(2128409395200000000000000000*M - 3405455032320000000000000000*a))/M + (z^3*(1135151677440000000000000000000*M - 2554091274240000000000000000000*a))/M + (z^2*(408654603878400000000000000000000*M - 1362182012928000000000000000000000*a))/M + (z*(89161004482560000000000000000000000*M - 490385524654080000000000000000000000*a))/M + (8916100448256000000000000000000000000*M - 106993205379072000000000000000000000000*a)/M, z, K)

where K is an integer 1 to 12.

This is a way of expressing roots of a polynomial of degree 12, but without calculating what those roots are. It has been proven that most polynomials of degree 5 or higher do not have "algebraic" solutions: the solutions cannot be expressed in terms of addiction, subtraction, division, multiplication, and square roots.

[Dan Richter]

OK, Thank you for detail explanation :-). Actually, I am interested to express "x" i.e. interest rate per year.

[Walter Roberson]

As you can see from the below, if you do not know r, there just isn't much you can do to get a useful expression of a solution.

You can also see that the value of r can substantially influence the number of solutions -- and that for some values of r, you can get fully explicit solutions.

%M = 100000;
%a = 8791.59;
%r = 1;
syms x
syms M a r positive
eqn = a == M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

eqn = 

X = solve(eqn, x);

Warning: Unable to find explicit solution. For options, see help.

r = 1;
eqn2 = a == M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

eqn2 = 

X2 = solve(eqn2, x)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

X2 = 

r = 2;
eqn3 = a == M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

eqn3 = 

X3 = solve(eqn3, x)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

X3 = 

r = sym(1)/3

r = 

eqn4 = a == M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

eqn4 = 

X4 = solve(eqn4, x, 'maxdegree', 4)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

X4 = 

M = 100000;
a = sym(879159)/sym(100);
%r = 1;
syms x
syms r positive
eqn5 = a == M*(x/1200*(1+x/1200)^(r*12))/((x/1200+1)^(r*12)-1)

eqn5 = 

X5 = solve(eqn5, x)

Warning: Unable to find explicit solution. For options, see help.

 
X5 =
 
Empty sym: 0-by-1