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Instructions
In the following questions two equations numbered I and
II are given. You have to solve both the equations and
a: if x > y
b: if x ≥ y
c: if x < y
d: if x ≤ y
e: if x = y or the relationship cannot be established.
Solution
$$x^2+x-12 = 0$$
$$(x-3)(x+4) = 0$$
$$x = -4, 3$$
$$y^2+2y-8 = 0$$
$$(y-2)(y+4) = 0$$
$$y = -4, 2$$
Comparing x and y,
-4 = -4
-4 < 2
3 > -4
3 > 2
Hence, relationship can not be established between $$x$$ and $$y$$
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