SOLUTION: Find the range of values of 'c' such that the two lines x - y = 2 and cx + y = 3 intersect in the first quadrant.

All points in the first quadrant have both
their x and y coordinates positive.

   x - y = 2
  cx + y = 3

Adding the two equations term by term:

  x + cx = 5
  x(1+c) = 5
       x = 5/(1+c)

This must be positive, so 1+c > 0 or c > -1

Substitute in the first:

   x - y = 2
      -y = 2 - x
       y = -2 + x
       y = -2 + 5/(1+c)
       y = -2(1+c)/(1+c) + 5/(1+c)
       y = (-2-2c)/(1+c) + 5/(1+c)
       y = (-2-2c+5)/(1+c)
       y = (3-2c)/(1+c)

This must be positive



This has critical numbers which are the
zeros of numerator and denominator.

Find the zero of the numerator:

3-2c=0 
 -2c=-3
   c=3/2 = 1.5

Find the zero of the denominator:

1+c=0
  c=-1

We place the critical numbers on a number line:

---------o---------o----------
-3  -2  -1   0   1   2   3   4   

Choose a test point -2 in the leftmost interval







This is false so the interval 
is not included in the range of values for c.

---------o---------o----------
-3  -2  -1   0   1   2   3   4   

Choose a test point 0 in the middle interval






This is true so the interval 
is included in the range of values for c.

---------o---------o----------
-3  -2  -1   0   1   2   3   4   

Choose a test point 2 in the rightmost interval







This is false so the interval 
is not included in the range of values for c.

Answer: 

Edwin