What is the answer to solvefor x,p^3x-p^2y-1=0 ?

The answer to solvefor x,p^3x-p^2y-1=0 is x=(p^2y+1)/(p^3);\quad p\ne 0

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solvefor x,p^3x-p^2y-1=0

solve for

x, p^{3} x - p^{2} y - 1 = 0

x = \frac{p ^{2} y + 1}{p ^{3}}; p \neq = 0

Solution steps

p^{3} x - p^{2} y - 1 = 0

Move

p^{2} y

to the right side

p^{3} x - 1 = p^{2} y

Move

1

to the right side

p^{3} x = p^{2} y + 1

Divide both sides by

p^{3}; p \neq = 0

x = \frac{p ^{2} y + 1}{p ^{3}}; p \neq = 0

x, 5 g - 3 x - 4 = 0

x, 4 x + 9 (\frac{y}{2}) = 69

\frac{p + 29}{8} = 10

x + 1.08 x + 1.16 x + 1.24 x = 4900

\frac{x}{a} = b + 3 x

What is the answer to solvefor x,p^3x-p^2y-1=0 ?
The answer to solvefor x,p^3x-p^2y-1=0 is x=(p^2y+1)/(p^3);\quad p\ne 0