Question 563136
<pre>
No, they are consistent if they have a solution.
So we find what their solution must be by elimination:

2x + 4y = f
cx + dy = g

Multiply the first equation through by -d and the 
second equation through by 4

-2dx - 4dy = -df
 4cx + 4dy =  4g

(4c-2d)x = 4g-df

       x = {{{(4g-df)/(4c-2d)}}}

Start over:

2x + 4y = f
cx + dy = g

Multiply the first equation through by c and the second equation through
by -2

 2cx + 4cy =  cf
-2cx - 2cy = -2g

(4c-2d)y = cf-26

       y = {{{(cf-2g)/(4c-2d)}}}
 
Notice that for both x and y, the denominators
are the same, 4c-2d

In order for those expresions for x and y to be
defined solution, their denominators must not be
zero, because division by 0 is not defined. But
division by any other number is defined.  Therefore 
the only requirement is that their denominator must 
not be equal to 0, so

       4c-2d &#8800; 0

divide through by 2

        2c-d &#8800; 0

          2c &#8800; d

           d &#8800; 2c

As long as d is not equal to 2c, there will be a solution for
any values of f and g.   

The answer to: "What can you say about the coefficients c and d?" is

d is not equal to 2c.

[If you have studied Cramer's rule, you can just say the determinant of
coefficients D must not equal 0.  The above was done assuming you have not
studied or were not allowed to use Cramer's rule.]

Edwin</pre>