Instructions

In each question, two equations numbered I and II are given.

You have to solve both the equations and mark an appropriate answer.
Give answer :

a: If x ≤ y
b: If x > y
c: If relationship between x and y cannot be established
d: x < y
e: x ≥ y

Question 126

I. $$9x^{2} - 21x + 10 =0$$
II. $$y^{2} - 8y+ 15 =0$$

Solution

I. $$9x^{2} - 21x + 10 = 0$$

=> $$9x^2 - 6x - 15x + 10 = 0$$

=> $$3x (3x - 2) - 5 (3x - 2) = 0$$

=> $$(3x - 5) (3x - 2) = 0$$

=> $$x = \frac{5}{3} , \frac{2}{3}$$

II. $$y^{2} - 8y + 15 = 0$$

=> $$y^2 - 3y - 5y + 15 = 0$$

=> $$y (y - 3) - 5 (y - 3) = 0$$

=> $$(y - 5) (y - 3) = 0$$

=> $$y = 3 , 5$$

$$\therefore x < y$$

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I. $$9x^{2} - 21x + 10 =0$$II. $$y^{2} - 8y+ 15 =0$$