Let’s start with an example:
Let R represent x2 + (y-3)2 – 4 = 0. Does R = Z have a real solution?
- Z is represented by y = 3
- Z is represented by mx + y = 3
Readers are encouraged to take some time to try to solve the above – challenge yourself to 30 seconds 🙂
Trick: For questions involving determination of real solutions, it is sometimes quicker to plot rough sketches of the equations on the xy axis
There are two critical ‘two variables’ equations we should get super familiar with
The first one ax + by = c represents a straight line on the xy axis.
The second: (x – a)2 + (y – b)2 = d2 represents a circle on the xy axis with center (a,b) and radius of d
R seems pretty similar to our circle equation
R can be rewritten as x2 + (y-3)2 = 22 (a circle with center at 0,3 and radius of 2 units)
Let’s plot this…
Now (1) tells us that Z is represented by y = 3.
y = 3 passes through the center our of circle and any line passing through the center of its circle will intersect its perimeter at some points – yielding real solutions as shown below.
Bingo! So (1) by itself is sufficient
Option (2) seems a bit more complex. We have an additional variables m and x to consider. Now, regardless of the value of m, the value of mx = 0 at x = 0.
So, at x = 0, mx + y = 3 simplifies to y = 3 so mx + y = 3 always passes through 0,3 which is the center of our circle.
Hence (2) by itself is also sufficient.
Hence, we go with option (D) 🙂