Let c be a constant. For what values of c can the simultaneous equations #x-y=2; cx+y=3# have a solution (x, y) inside quadrant l?

2 Answers
Nov 24, 2016

In the first quadrant, both #x# values and #y# values are positive.

#{(-y = 2 - x), (y = 3 - cx):}#

#-(3 - cx) = 2 - x#

#-3 + cx = 2 - x#

#cx + x = 5#

#x(c + 1) = 5#

#x = 5/(c + 1)#

We need #x > 0# for there to be a solution in quadrant #1#.

#5/(c + 1) > 0#

There will be a vertical asymptote at #c = -1#. Pick test points to the left and to the right of this asymptote.

Let #c = -2# and #c= 2#.

#5/(3(-2) + 1) = 5/(-5)= -1#

#:. -1 >^O/ 0#

So, the solution is #c > -1#.

Hence, all values of #c# that are larger than #-1# will ensure that the intersection points are in the first quadrant.

Hopefully this helps!

Nov 25, 2016

Answer:

#-3/2 < c < 1#

Explanation:

The equation #x-y=2hArry=x-2# and hence this represents a line whose slope is #1# and intercept on #y#-axis is #-2#. Also intercept on #x#-axis can be obtained by putting #y=0# and is #2#. Equation of line appears as follows:
graph{x-2 [-10, 10, -5, 5]}

The other equation is #cx+y=3# or #y=-cx+3#, which represents a line with #y# intercept and slope #-c#. For this line to intersect above line in #Q1#,

(i) it should have a minimum slope that of line joining #(0,3)# and intercept of above line on #x#-axis i.e. at #(2,0)#, which is #(0-3)/(2-0)=-3/2#

and (ii) it should be passing through #(3,0)# but have slope not more than #1#, as it will then intersect the line #x-y=2# in #Q3#.

Hence, values of #c# for which simultaneous equations #x-y=2# and #cx+y=3# have a solution #(x,y)# inside #Q1# are given by

#-3/2 < c < 1#

graph{(x-y-2)(x-y+3)(3x+2y-6)=0 [-10, 10, -5, 5]}