In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

I. $\frac{x^{2} - 10 x + 16}{x^{2} - 12 x + 24} = \frac{2}{3}$
II. $y^{2} - y - 20 = 0$

if x > y

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if x < y

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if x ≥ y

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if x ≤ y

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if x = y or definite relationship cannot be established between ‘x’ and ‘y’

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Solution

The correct option is E if x = y or definite relationship cannot be established between ‘x’ and ‘y’
$x^{2} - 10 x + \frac{16}{x^{2}} - 12 x + 24 = \frac{2}{3} 3 x^{3} - 30 x + 48 = 2 x^{2} - 24 x + 8 x^{2} - 6 x = 0 x (x - 6) = 0 O n s o l v i n g t h e e q u a t i o n \to y = - 4, 5 H e n c e, d e f i n i t e r e l a t i o n s h i p c a n n o t b e e s t a b l i s h e d b e t w e e n x a n d y .$

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In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option: I. x2−10x+16x2−12x+24=23 II. y2−y−20=0

The correct option is E if x = y or definite relationship cannot be established between ‘x’ and ‘y’x2−10x+16x2−12x+24=233x3−30x+48 = 2x2−24x+8x2−6x=0x(x−6)=0On solving the equation→ y=−4,5Hence, definite relationship cannot be established between x and y.

In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

I. $\frac{x^{2} - 10 x + 16}{x^{2} - 12 x + 24} = \frac{2}{3}$
II. $y^{2} - y - 20 = 0$