Python, 148 146 chars

import re I=raw_input a,b,p,c,d,q=eval(re.sub('x|y=','.,',I()+','+I())) D=a*d-b*c print'(%.2f,%.2f)'%((p*d-q*b)/D,(a*p-c*q)/D)if D else'parallel' 

sage - 162 chars

x,y=var('x,y') e=lambda:eval(raw_input().replace('x','*x').replace('y','*y=')) try:r=solve([e(),e()],[x,y])[0];print`r[0].rhs(),r[1].rhs()` except:print'parallel' 

Python+sympy - 161 chars

from sympy import* x,y=symbols('xy') e=lambda:eval(raw_input().replace('x','*x').replace('y=','*y-')) r=solve([e(),e()],[x,y]) print r and`r[x],r[y]`or'parallel' 

Python+sympy - 143 chars (different output format)

from sympy import* x,y=symbols('xy') e=lambda:eval(raw_input().replace('x','*x').replace('y=','*y-')) print solve([e(),e()],[x,y])or'parallel' 

output format for 143 char version is slightly different from the spec

In: 1x+1y=1 1x-1y=1 Out: {x: 1, y: 0} In: 0x+1y=0 1x+0y=0 Out: {x: 0, y: 0} 

J, 146 132 134 124

echo'parallel'"_`(1|.')(',[:(,',',])&(0j2&":)/[:,,.@{:%.}:)@.([:*@|[:-/ .*}:)|:".>{.`(<@}:@;@(1 2&{))`{:(`:0)"1;:;._2(1!:1)3 

Parsing this is awful.

Edit: Realized that my output is basically broken, although it works for the examples given and parallels...

Edit: Posting the shorter version I already had.

Edit: Fixed without too much pain...

Edit: Fixed a weird issue where the program was spread across multiple lines.

Somewhat ungolfed version:

words =: ;:;._2(1!:1)3 NB. Leverage J's parser itself to split each line into e.g. ax + by = c lines =: ".>{.`(<@}:@;@(1 2&{))`{:(`:0)"1 words NB. Use a 3-part gerund (`) to get boxed strings 'ax';'b';'c', unbox and parse calc_and_format =: 1|.')(',[:(,',',])&(0j2&":)/[:,,.@{:%.}: NB. %. (matrix multiply) first two rows (a d/b e) by third row turned (c/f) NB. And then format laboriously. 0j2&": formats numbers with 2 decimals. det_nonzero =: [:*@|[:-/ .*}: NB. 1 if determinant (-/ .*) of (a b/d e) is nonzero (*@|) echo'parallel'"_`[email protected]_nonzero|: lines NB. transpose parsed input to get (a d/b e/c f). NB. If det_nonzero, call calc_and_format, otherwise constant 'parallel' NB. Print 

OCaml + Batteries, 163 characters

As straightforward as it gets:

Scanf.scanf"%fx%fy=%f\n%fx%fy=%f\n"Float.(fun a b c d e f->Printf.(fun u->if u=0.then printf"parallel"else printf"(%.2f,%.2f)"((b*f-c*e)/u)((d*c-a*f)/u))(b*d-a*e)) 

Edit:

• Initial version, 169
• Use Batteries for delimited overloading of operators, 164
• Lambda binding of u, 163

C-149 bytes

Not much of golfing just the basics.

float a,b,c,d,m,n,t;main(){scanf("%fx%fy=%f%fx%fy=%f",&a,&b,&m,&c,&d,&n);t=b*c-a*d;t?printf("(%.2f,%.2f)",(b*n-d*m)/t,(c*m-a*n)/t):puts("parallel");} 

Here is the ideone link for testing.

Instead of printing 'parallel' an alternative could be to print the orthogonal distance between the lines when they are parallel.

Vyxal, 10 bytes

∆Q:[vt|`ċ₃ 

Try it Online!

8 if the format can be ⟨ ⟨ `x` | xval ⟩ | ⟨ `y` | yval ⟩ ⟩ . Returns fraction representation of all numbers if not integer.

Explained

∆Q:[vt|`ċ₃ ∆Q # Solve the two input equations simultaneously :[ # If there's a solution vt # get the last values of each two item tuple |`ċ₃ # else, push the string "parallel"