Question 1052325
Find the range of values of 'c' such that the two lines 
x - y = 2 and cx + y = 3 intersect in the first quadrant.
<pre>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

{{{(3-2c)/(1+c)>0}}}

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

{{{(3-2c)/(1+c)>0}}}
{{{(3-2(-2))/(1+(-2))>0}}}
{{{(3+4)/(1-2)>0}}}
{{{7/(-1)>0}}}
{{{-7>0}}}

This is false so the interval {{{(matrix(1,3,-infinity,",",-1))}}}
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

{{{(3-2c)/(1+c)>0}}}
{{{(3-2(0))/(1+(0))>0}}}
{{{(3)/(1)>0}}}
{{{3>0}}}

This is true so the interval {{{(matrix(1,3,-1,",",3/2))}}}
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

{{{(3-2c)/(1+c)>0}}}
{{{(3-2(2))/(1+(2))>0}}}
{{{(3-4)/(1+2)>0}}}
{{{(-1)/3>0}}}
{{{-1/3>0}}}

This is false so the interval {{{(matrix(1,3,3/2,",",infinity))}}}
is not included in the range of values for c.

Answer: {{{(matrix(1,3,-1,",",3/2))}}}

Edwin</pre>