Question

The graphs of the equation x-y=2 and kx+y=3 where k is constant intersect at the point (x,y) in the first quadrant if and only if the value of k is?

Answer 1

Given:

The graphs of the equation x-y=2 and kx+y=3 where k is constant  

To find:

The graphs of the equations intersect at the point (x,y) in the first quadrant if and only if the value of k is?

Solution:

From given, we have the equations, x - y = 2 and kx + y = 3

They intersect at the point (x, y).  

This means that the solution of the equations is (x, y).

As they also intersect in the first quadrant, we have, both x and y are positive.

Hence k must be 0 or greater than 0 and lying between 1 and 3/2.

Therefore the value of k is k ≥ 0 and 1 ≤ k ≤ 3/2