Respuesta :

a) 
|x + y = 5 
|2x - y = 7; 
b) 
|2x + y = 5 
|x - y = 2 
c) 
|3x + y = 6 
|4x - 3y = -5 
d) 
|1/(x - 1) = y - 3 
|x - y = -2 
e) 
|(9x + 4y)/3 - (5x - 11)/2 = 13 - y 
|13x - 7y = -8 
Answer: c and e has solution (1; 3)
altavistard
Best to enclose those fractions in parentheses:

(1/5)x + y = (2/5).

Otherwise, how would your reader know you didn't mean   1 / (5x) ?

Please note:  Your 1/0x involves division by zero and thus is not defined.  Make certain you've copied down the problem correctly.

If you present the following, instead, you can solve the system:

(1/5)x + y = (2/5).
       x + (1/3)y = (1/2)

Hint:  Determine what the LCD (lowest common denominator) is.  Multiply both equations by this LCD, to remove all fractions from the problem.


Otras preguntas