In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer If
a:x > y
b: x ≥ y
c: x < y
d: x ≤ y
e: x = y or the relationship cannot be established
$$I. 9x^{2} - 27x + 20 = 0$$
$$II. 6y^{2} - 5y + 1 =0$$
I. 9$$x^{2}$$-27x+20=0
9$$x^{2}$$-15x-12x+20=0
3x(3x-5)-4(3x-5)=0
(3x-4)(3x-5)=0
x=($$\frac{4}{3})$$ or $$\frac{5}{3}$$
II .6$$y^{2}$$-5y+1=0
6$$y^{2}$$-3y-2y+1=0
3y(2y-1)-1(2y-1)=0
(3y-1)(2y-1)=0
y= ($$\frac{1}{3}$$) or $$\frac{1}{2}$$
Clearly x>y
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