Solve an algebraic expression

Question

The Challenge

Your task is to create a program that can solve an algebraic equation.

Input

Input will consist of a String. The string will be an equality involving the variable x, and follows the following rules:

• the equation will always be linear
• The valid operations in the equation are + - * /
• Parenthesis will not be used
• Order of operations must be respected
• The coefficients of x will always have * in between the coefficient and x
• You can assume that there will be exactly one solution for x

Output

Output will be through the stdout (or an acceptable equivalent). The output should be the value of x that satisfies the equality. (You can round to the nearest thousandth if the decimal expansion is too long.)

---

Test cases

x=5
=> 5

2*x=5
=> 2.5

3*x/2=7
=> 4.667

2*x=x+5
=> 5

3*x+3=4*x-1
=> 4

3+x*2+x=x+x/2+x/2
=> -3

Scoring

Because this is a code-golf, the shortest solution (in bytes) wins. (NOTE: You may not use built-ins that trivialize the problem.)

Answer 1

Perl, 49 bytes

Ah, builtins got disallowed. I'll leave this solution up for reference though it is now non competing.

Includes +2 for -lp (-l can be dropped but is ugly)

Run with the equation on STDIN:

perl -lp algebra.pl <<< "x=5"

algebra.pl:

s'x'$.'g;$_=s/=(.*)/-($1)//(1-eval()*$.--/eval)

Answer 2

Javascript, 62 bytes

Without builtins, but with eval.

(s,f=eval("x=>"+s.replace(/=(.*)/,"-($1)")))=>f(0)/(f(0)-f(1))

Uses eval to transform foo(x) = bar(x) into a function f(x) = foo(x) - bar(x), then calculates x = f(0) / (f(0) - f(1)).

Answer 3

Python 3, 587 bytes

from re import sub as r,findall as f
a=input()
i=int;s=str;l=len;d=r"(\d+\.?\d*)";d=d,d
a=r(r"%s\*%s"%d,(lambda x:s(i(x.group(1))*i(x.group(2)))),a)
a=r(r"%s\/%s"%d,(lambda x:s(i(x.group(1))/i(x.group(2)))),a)
a=r(r"%s\-%s"%d,(lambda x:s(i(x.group(1))-i(x.group(2)))),a)
a=r(r"%s\+%s"%d,(lambda x:s(i(x.group(1))+i(x.group(2)))),a)
a=f(r"(\d+\.?\d*.)",a+"=")
if l(a)==0 or a[0][1]!="*":a[0:0]=["1"]
if l(a)==1 or a[1][1]!="=":a[1:1]=["0"]
if l(a)==2 or a[2][1]!="*":a[2:2]=["1"]
if l(a)==3 or a[3][1]!="=":a[3:3]=["0"]
a=[int(x[0])for x in a]
z=a[1]-a[3]
print(z/(z-a[0]-a[1]+a[2]+a[3]))

The only longer answers I've seen use 80% of the code to store a big data table. But unlike eval-based solutions, it only uses e twice! (In re and len.)

Log, Day 20: Used Solve. Solve got banned. Learned re. Used re. Slept.

Answer 4

Matlab, 61 bytes

x=[1 0];a=eval([strrep(input(''),'=','-(') ')']);a(2)/diff(a)

Uses the same algorithm as Ton Haspel to solve the equation. Replaces the = by -( and adds a ) at the end to make the string an equation that should equal zero, and solves (Newton's method is exact for linear functions).

Input is just a string, e.g.: '3+x*2+x=x+x/2+x/2'.

Matlab has a nice builtin for this too,

solve(input(''))