Instructions

In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x >y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship between x and y cannot be established.

Question 138

I. $$3x^{2}-22x+40=0$$
II. $$5y^{2}-21y+16=0$$

Solution

I.$$3x^{2} - 22x + 40 = 0$$

=> $$3x^2 - 12x - 10x + 40 = 0$$

=> $$3x (x - 4) - 10 (x - 4) = 0$$

=> $$(x - 4) (3x - 10) = 0$$

=> $$x = 4 , \frac{10}{3}$$

II.$$5y^{2} - 21y + 16 = 0$$

=> $$5y^2 - 5y - 16y + 16 = 0$$

=> $$5y (y - 1) - 16 (y - 1) = 0$$

=> $$(y - 1) (5y - 16) = 0$$

=> $$y = 1 , \frac{16}{5}$$

$$\therefore x > y$$


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