(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

Solution:

We’ll draw the equation’s graph and shade the side that has the inequality’s solutions.

You can choose any value, but you must always locate the two necessary values at x = 0 and y = 0, i.e., the x and y-intercepts.

2x + 3y ≤ 35

So when,

y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0