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 between x and y cannot be established.
I. $$x^{2}+12x+35=0$$
II. $$3y^{2} +19y+20=0$$
I.$$x^{2} + 12x + 35 = 0$$
=> $$x^2 + 7x + 5x + 35 = 0$$
=> $$x (x + 7) + 5 (x + 7) = 0$$
=> $$(x + 7) (x + 5) = 0$$
=> $$x = -5 , -7$$
II.$$3y^{2} + 19y + 20 = 0$$
=> $$3y^2 + 15y + 4y + 20 = 0$$
=> $$3y (y + 5) + 4 (y + 5) = 0$$
=> $$(3y + 4) (y + 5) = 0$$
=> $$y = -5 , \frac{-4}{3}$$
$$\therefore x \leq y$$
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