In these questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.
$$3x^2-13x+12 = 0$$
$$(3x-4)(x-3) = 0$$
$$x = \frac{4}{3}, 3$$
$$3y^2-13y+14 = 0$$
$$(3y-7)(y-2) = 0$$
$$y = 2, \frac{7}{3}$$
relationship between x and y cannot be established
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