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.$$3x^{2}-13x-10=0$$
II.$$3y^{2}+10y-8=0$$
I.$$3x^{2} - 13x - 10 = 0$$
=> $$3x^2 - 15x + 2x - 10 = 0$$
=> $$3x (x - 5) + 2 (x - 5) = 0$$
=> $$(x - 5) (3x + 2) = 0$$
=> $$x = 5 , \frac{-2}{3}$$
II.$$3y^{2} + 10y - 8 = 0$$
=> $$3y^2 + 12y - 2y - 8 = 0$$
=> $$3y (y + 4) - 2 (y + 4) = 0$$
=> $$(y + 4) (3y - 2) = 0$$
=> $$y = -4 , \frac{2}{3}$$
$$\therefore$$ No relation can be established.
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