Problem:
Let c be a constant. The simultaneous equations
xβycx+yβ=2=3β
have a solution (x,y) inside Quadrant I if and only if
Answer Choices:
A. 0<c<3/2
B. c=β1
C. c>β1
D. c<3/2
E. β1<c<3/2
Solution:
Algebraic solution. Solving simultaneously, we obtain
x=c+15β,y=c+13β2cβ
We wish to find all values of c for which x,y>0. First, x>0βc+1>0βc>β1. Next, given that c+1>0, then y>0β3β2c>0β3/2>c. Thus x,y>0ββ1<c<3/2.
OR
Geometric solution. We wish to find those values of c for which the lines xβy=2 and cx+y=3 intersect inside Quadrant I. The line xβy=2 has slope 1 and x-intercept 2, and is shown as L in the figure. The line cx+y=3 has y-intercept 3 and slope βc. Several possible choices for this line are shown dashed in the figure. For this line to intersect L in Quadrant I, it is necessary and sufficient to choose a slope between that of Lβ², which passes through (2,0), and that of Lβ²β², which is parallel to L. Thus
β23β<βc<1 or β1<c<23ββ