An algebra problem by Wildan Bagus Wicaksono

Similar method with @Wildan Bagus W. Hafidz 's

$\begin{aligned} \frac {yx^{2017}}{x^{2019}+x^{2017}+x^{2015}} & = \frac 56 & \small \color{#3D99F6} \text{Divide up and down of LHS by } x^{2017}. \\ \frac y{x^2+{\color{#3D99F6}1}+\dfrac 1{x^2}} & = \frac 56 \\ \frac y{x^2+{\color{#3D99F6}2}+\dfrac 1{x^2}-{\color{#3D99F6}1}} & = \frac 56 \\ \frac y{\left({\color{#3D99F6}x+\dfrac 1x}\right)^2-1} & = \frac 56 & \small \color{#3D99F6} \text{Note that } x+\dfrac 1x = 7 \\ \frac y{{\color{#3D99F6}7}^2-1} & = \frac 56 \\ \frac y{48} & = \frac 56 \\ \implies y & = \boxed{40} \end{aligned}$

$\large x+\frac { 1 }{ x } =7\\ { \left( x+\frac { 1 }{ x } \right) }^{ 2 }={ 7 }^{ 2 }\\ { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2\cdot x\cdot \frac { 1 }{ x } =49\\ { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =47$

$\large \frac { y{ x }^{ 2017 } }{ { x }^{ 2019 }+{ x }^{ 2017 }+{ x }^{ 2015 } } =\frac { 5 }{ 6 } \\ \frac { y{ x }^{ 2017 } }{ { x }^{ 2019 }+{ x }^{ 2017 }+{ x }^{ 2015 } } \cdot \frac { \frac { 1 }{ { x }^{ 2017 } } }{ \frac { 1 }{ { x }^{ 2017 } } } =\frac { 5 }{ 6 } \\ \frac { y }{ { x }^{ 2 }+1+\frac { 1 }{ { x }^{ 2 } } } =\frac { 5 }{ 6 } \\ \frac { y }{ 47+1 } =\frac { 5 }{ 6 } \\ \frac { y }{ 48 } =\frac { 5 }{ 6 } \\ y=40$

6y x^2017 = 5(x^2019 + x^2017 + x^2015), 6y = 5(x^2 + 1 + 1/x^2) = 5[(x + 1/x)^2 - 1] = 5(7^2 - 1) = 5 48, so y = 5*(48/6) = 40.