Question

A point (2x-3, -3x+5) is on a xy plane 1. If this point exists in the 2nd Quadrant, What is the value of x ? (Domain) 2. Is there any quadrant the point cannot exist regardless of the value of x? If so, which quadrant?​

Answer 1

The value of x for the point (2x-3, -3x+5) to be in the 2nd quadrant must be less than 3/2. There is no quadrant where the point cannot exist as x has no restrictions on being positive or negative.

Step-by-step explanation:

For a point (2x-3, -3x+5) to exist in the 2nd quadrant of an xy-plane, the x-coordinate of the point must be negative, and the y-coordinate must be positive. To find the value of x, we set the x-coordinate to be less than 0, so:

2x - 3 < 0

2x < 3

x < 3/2

Therefore, x must be less than 3/2 for the point to be in the 2nd quadrant.

Regarding whether there is a quadrant where the point cannot exist, we analyze the given point equation in terms of x. There is no restriction on x to prevent it from having both positive or negative values; hence, there is no quadrant where the point cannot exist, regardless of the value of x.