In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x > y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship cannot be established.
I. $$2x^{2}-11x+12=0$$
II. $$2y^{2}-19y+44=0$$
I.$$2x^{2} - 11x + 12 = 0$$
=> $$2x^2 - 8x - 3x + 12 = 0$$
=> $$2x (x - 4) - 3 (x - 4) = 0$$
=> $$(x - 4) (2x - 3) = 0$$
=> $$x = 4 , \frac{3}{2}$$
II.$$2y^{2} - 19y + 44 = 0$$
=> $$2y^2 - 8y - 11y + 44 = 0$$
=> $$2y (y - 4) - 11 (y - 4) = 0$$
=> $$(y - 4) (2y - 11) = 0$$
=> $$y = 4 , \frac{11}{2}$$
$$\therefore x \leq y$$
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