Two lines are given to be parallel. The equation of one of the lines is 4x + 3y = 14. Find the equation of the second line.

Answer : 

We know if we have two parallel lines a₁x + b₁y + c₁  =  0  And a₂x + b₂y + c₂  = 0  , Then the value of all cofficient are As : 

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

Here we have a line 4x + 3y = 14   , So  a₁  =  4  ,  b₁  =  3 and  c₁  = - 14  
let our line that is parallel to given line is a₂x + b₂y + c₂ = 0 , So 
We get 

$\frac{4}{a_2} = \frac{3}{b_2} \ne \frac{- 14}{c_2}$

So,

a₂ : b₂  =  4k : 3k                                 ( Where k is a ratio constant that could be any integer ) 

So,
We can form any equation that has the co efficients in the ratios 4k : 3k : n  (where n is any integer and n$\ne$ 14k) 

So,
Our parallel equation could be  

8x  + 6y  =  12                                                  ( Ans )