What is the answer to solvefor x,y^4xy^n=1 ?

The answer to solvefor x,y^4xy^n=1 is x= 1/(y^{4+n)};\quad x^4x^n\ne 0

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solvefor x,y^4xy^n=1

solve for

x, y^{4} x y^{n} = 1

x = \frac{1}{y ^{4 + n}}; x^{4} x^{n} \neq = 0

Solution steps

y^{4} x y^{n} = 1

Divide both sides by

y^{4} y^{n}; x^{4} x^{n} \neq = 0

x = \frac{1}{y ^{4 + n}}; x^{4} x^{n} \neq = 0

3 x = 29 - 8

y, 6 x - 3 y = 12

2 (x + 1) - 3 (2 x - 5) = 6 x - 3

\frac{2 - 3 x}{3} - \frac{x}{2} = \frac{2}{4} - \frac{x - 5}{6}

3 x + (- x + 1) - (x - 5) = 8 - (2 x - 6 + x)

What is the answer to solvefor x,y^4xy^n=1 ?
The answer to solvefor x,y^4xy^n=1 is x= 1/(y^{4+n)};\quad x^4x^n\ne 0