What is [n]yf(y)(y)^2=50[n],f(y)(y)^1=20 ?

The solution to [n]yf(y)(y)^2=50[n],f(y)(y)^1=20 is

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[n]yf(y)(y)^2=50[n],f(y)(y)^1=20

[n] y f (y) (y)^{2} = 50 [n], f (y) (y)^{1} = 20

y = - \frac{5}{2}, y = \frac{5}{2}, f = - 410 f = 410

Solution steps

[[n] y f (y)^{2} = 50 [n] f (y)^{1} = 20]

Isolate

f

for

f (y)^{1} = 20 : f = \frac{20}{y}; y \neq = 0

Plug the solutions

f = \frac{20}{y}

into

[n] y f (y)^{2} = 50 [n]

For

[n] y f (y)^{2} = 50 [n]

, subsitute

f

with

\frac{20}{y} : y = - \frac{5}{2}, y = \frac{5}{2}; n \neq = 0

Plug the solutions

y = - \frac{5}{2}, y = \frac{5}{2}

into

f (y)^{1} = 20

For

f (y)^{1} = 20

, subsitute

y

with

- \frac{5}{2} : f = - 410

For

f (y)^{1} = 20

, subsitute

y

with

\frac{5}{2} : f = 410

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

[n] y f (y)^{2} = 50 [n], f (y)^{1} = 20

are

y = - \frac{5}{2}, y = \frac{5}{2}, f = - 410 f = 410

\frac{8}{x} + \frac{5}{y} = - 3, \frac{6}{x} + \frac{25}{y} = 2

y = (\frac{2}{3}) x + 4, - 2 x + 3 y = 6

2 x + y - 2 z = 1, x - 2 y + 3 z = 6, 3 x - 3 y - z = 5

x + \frac{y}{4} = - 1, x + y = 1

2 x + 3 y + 4 z = 3, 2 x + 6 y + 8 z = 5, 4 x + 9 y - 4 z = 4

What is [n]yf(y)(y)^2=50[n],f(y)(y)^1=20 ?
The solution to [n]yf(y)(y)^2=50[n],f(y)(y)^1=20 is