CAT 2018 Slot 1 QA Question 9: Given that $x^{2018}y^{2017} = \dfrac{1}{2}$ and $x^{2016}y^{2019} = 8$, the value of $x^{2} + y^{3}$ is

Question

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

Accepted Answer

Divide the second equation by the first equation.
$\dfrac{x^{2016}y^{2019}}{x^{2018}y^{2017}} = \dfrac{8}{\dfrac{1}{2}}$
$\dfrac{y^{2}}{x^{2}} = 16$
$\dfrac{y}{x} = 4$ or $\dfrac{y}{x} = - 4$
If $y = 4x$, substitute in the first equation.
$x^{2018}(4x)^{2017} = \dfrac{1}{2}$
$x^{4035} 	imes 2^{4034

Given that x2018y2017=12x^{2018}y^{2017} = \dfrac{1}{2}x2018y2017=21​ and x2016y2019=8x^{2016}y^{2019} = 8x2016y2019=8, the value of x2+y3x^{2} + y^{3}x2+y3 is

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