How Would You Solve For A Variable In An Equation?

You might want to wrap your math in swift via python...

See http://stackoverflow.com/questions/4449110/python-solve-equation-for-one-variable

You would need to write a parser to understand the equation :

either you let user write plain text as a * x + b = c

or provide a template to fill : [aField]. *. [x label.] + [b Field] = [c Field], where user would have to fill each field

at the end, you have to isolate a, b, c

Then the resolution is straightforward:

if a == 0, test if b == c : then any x is solution, otherwise, no solution

if a != 0, x = (c - b) / a

Once you have done this, you can extend the app with second order equation in the same way :

a * x² + b * x * c = 0

if a == 0, that's the previous case

if a != 0, then 2 cases :
if (b² - 4 a*c) >= 0, there are 2 values of x :
     (-b + √(b² - 4 a*c)) / 2 * a   and
     (-b - √(b² - 4 a*c)) / 2 * a

if (b² - 4 a*c) < 0, there are 2 values of x, but imaginary numbers :
     (-b + i √(-b² + 4 a*c)) / 2 * a   and
     (-b - i √(-b² + 4 a*c)) / 2 * a

Thanks. I'll definetly try this and let you know how it goes!

Have fun developing this.

Please note thjere was a typo in 2nd order equation :

should read:

a * x² + b * x + c = 0

instead of

a * x² + b * x * c = 0

Thanks! That's very helpful! Do you know how I might make a basic text parser that can extract the values of a, b, and c from one text field if written as something like this: a * x² + b * x + c = 0

Also, since I'm not quite as good at math as you are, I have to ask; does the above code solve for x in any instance with one instance of x?

For the parser, you should:

- search for = sign , to find the 2 members of equation

- then you can parse each member

- for the right, you get directly the value

- for the left member, first search for + sign to isolate each term

- then for each term, it will be easy to find the coefficients a, b, c.

For your second question

does the above code solve for x in any instance with one instance of x?

IN the first equqtion, you get x directly

FOr the second, you need to have x1 and x2 to hold the 2 solutions of the equqtion.