• $$\Large x^{2}+x-20=0; y^{2}-y-30=0$$
 A) If x>y B) $$\Large If \ x \ge y$$ C) lfx D) $$\Large If \ x \le y$$

 D) $$\Large If \ x \le y$$

$$\Large x^{2}+x-20=0$$ [by factorisation method]

= $$\Large x^{2}+5x-4x-20=0$$

= $$\Large x \left(x+5\right)-4 \left(x+5\right)=0$$

= $$\Large \left(x+5\right) \left(x-4\right)=0$$

Therefore, x = -5 or 4

and $$\Large y^{2}-y-30=0$$

= $$\Large y^{2}-6y+5y-30=0$$

= $$\Large y \left(y-6\right)+5 \left(y-6\right)=0$$

= $$\Large \left(y-6\right) \left(y+5\right)=0$$

Therefore, y = 6 or -5

Hence, $$\Large y \ge x \ or \ x \le y$$

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