Question 29

Given that $$x^{2018}y^{2017}=\frac{1}{2}$$, and $$x^{2016}y^{2019}=8$$, then value of $$x^{2}+y^{3}$$ is

Solution

Given that $$x^{2018}y^{2017}=\frac{1}{2}$$  ... (1)

$$x^{2016}y^{2019}=8$$ ... (2)

Equation (2)/ Equation (1)

$$\dfrac{y^2}{x^2} = \dfrac{8}{1/2}$$

$$\dfrac{y}{x} = 4$$ or $$-4$$

Case 1: When $$\dfrac{y}{x} = 4$$

$$x^{2018}(4x)^{2017}=\dfrac{1}{2}$$

$$x^{2018+2017}(2)^{4034}=\dfrac{1}{2}$$

$$x^{4035}=\dfrac{1}{(2)^{4035}}$$

$$x=\dfrac{1}{2}$$

Since, $$\dfrac{y}{x} = 4$$, => y = 2

Therefore, $$x^{2}+y^{3}$$ = $$\dfrac{1}{4}+8$$ = $$\dfrac{33}{4}$$

Case 2: When $$\dfrac{y}{x} = -4$$

$$x^{2018}(-4x)^{2017}=\dfrac{1}{2}$$

$$x^{2018+2017}(2)^{4034}=\dfrac{-1}{2}$$

$$x^{4035}=\dfrac{1}{(-2)^{4035}}$$

$$x=\dfrac{-1}{2}$$

Since, $$\dfrac{y}{x} = -4$$, => y = 2

Therefore, $$x^{2}+y^{3}$$ = $$\dfrac{1}{4}+8$$ = $$\dfrac{33}{4}$$. Hence, option D is the correct answer.