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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
$$4x^2-13x+9 = 0$$
$$(4x-9)(x-1) = 0$$
$$x = 1, \frac{9}{4}$$
$$3y^2-14y+16 = 0$$
$$(3y-8)(y-2) = 0$$
$$y = 2, \frac{8}{3}$$
Comparing x and y,
$$1 < 2$$
$$1 <\dfrac{8}{3}$$
$$\dfrac{9}{4}>2$$
$$\dfrac{9}{4}<\dfrac{8}{3}$$
Relationship between x and y cannot be established
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