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 127

I. $$2x^{2} -13x + 15 =0$$
II. $$2y^{2} - 11y + 12 =0$$

Solution

I. $$2x^{2} - 13x + 15 = 0$$

=> $$2x^2 - 10x - 3x + 15 = 0$$

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

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

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

II. $$2y^{2} - 11y + 12 = 0$$

=> $$2y^2 - 8y - 3y + 12 = 0$$

=> $$2y (y - 4) - 3 (y - 4) = 0$$

=> $$(2y - 3) (y - 4) = 0$$

=> $$y = 4 , \frac{3}{2}$$

$$\because 5 > \frac{3}{2}$$ and $$4 > \frac{3}{2}$$

$$\therefore$$ No relation can be established.

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I. $$2x^{2} -13x + 15 =0$$II. $$2y^{2} - 11y + 12 =0$$