In each of the following questions, two equations I and II have been given. Solve these questions and answer
(1)if x < y
(2) if x ≤ y
(3) if x = y or the relation cannot be established
(4) if ≥ y
(5) if x > y
Statement I : $$30 x^{2} + 11x + 1 = 0$$
=> $$30x^2 + 6x + 5x + 1 = 0$$
=> $$6x (5x + 1) + 1 (5x + 1) = 0$$
=> $$(6x + 1) (5x + 1) = 0$$
=> $$x = \frac{-1}{6} , \frac{-1}{5}$$
Statement II : $$42 y^{2} + 13y + 1 = 0$$
=> $$42y^2 + 7y + 6y + 1 = 0$$
=> $$7y (6y + 1) + 1 (6y + 1) = 0$$
=> $$(7y + 1) (6y + 1) = 0$$
=> $$y = \frac{-1}{7} , \frac{-1}{6}$$
$$\therefore$$ $$x \leq y$$
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