Solving Algebra Problem: [3+ x5/3y]/x

[Imparcticle]

I'm not sure how to do this:

[3x^-1/3 + x^5/3y] / x^2/3

this is what i did:

[3 + x^5/3y ]/ x^1/3x^2/3

[3+ x^5/3y]/x

the answer to the problem is the same as what I have except instead of x^5/3, its x^2.

help.

 

[dextercioby]

Of course it is.Put it like that:

$$\frac{\frac{3}{x^{\frac{1}{3}}}+x^{\frac{5}{3}}y}{x^{\frac{2}{3}}}$$

Do you see where that x^{2} is coming from...?

Daniel.

 

[arildno]

Your 1.step, that is, extracting $$x^{-\frac{1}{3}}$$ from the parenthesis has been done incorrectly.
Try to figure out why.

 

[A_I_]

u can take x^2/3 in factor for the numerator, u will have:

[x^2/3(3x^-1 +xy)]/x^2/3

then u simplify and obtain

3/x + xy

which leads to (3 + yx^2)/x

 

[Imparcticle]

Of course it is.Put it like that:



Do you see where that x^{2} is coming from...?

no.

Your 1.step, that is, extracting from the parenthesis has been done incorrectly.
Try to figure out why.

I don't see it.

u can take x^2/3 in factor for the numerator, u will have:

[x^2/3(3x^-1 +xy)]/x^2/3

how are you factoring a x^2/3 when the smallest power is -1/3 ? if you factor that out, x^5/3 should be x^1/3.


help.

 

[Data]

$$\frac{\frac{3}{x^{\frac{1}{3}}} + x^{\frac{5}{3}}y}{x^{\frac{2}{3}}} = { \left( \frac{3}{x^{\frac{1}{3}}} + x^{\frac{5}{3}}y \right) \over x^{\frac{2}{3}}}\cdot \frac{x^{\frac{1}{3}}}{x^{\frac{1}{3}}} = \frac{3 + x^2y}{x}$$

I only posted this because I think you're getting more confused by the replies so far. Try to take a lesson from the method I used to simplify (that is, multiplying by 1 is a good idea sometimes!).