# 4/x - 7/x^2 over 7/x^3 - 2/x (simplify)I have no idea how this is simplify, havent taken a math class in 3 or 4 years. answers are provided but i do not know the proper steps.

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### 1 Answer

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4/x-7/x^2 over 7/x^3-2/x.

Simplification:

There is a difference between :

i)

4/x - 7/x^2 over 7/x^3 - 2/x

and

(ii)(4/x - 7/x^2) over( 7/x^3 - 2/x ).

i)

4/x - **7/x^2 over 7/x^3** - 2/x means

4/x-**(7/x^2)/(7/x^3)** - 2/x = 4/x -**(7/x^2)*(x^3/7)** -2/x.........(1)

Here 4/x and2/x are like terms and could be collected together and the middle term (7/x^2) over 7/x^3 or (7/x^2) divided by (7/x^3) or (7/x^2) / (7/x^3) or **(7/x^2)*(x^3/7)** are all same and is equal to 7x^3/7x^2 = x.

So the right side of (1) simplifies to:

4/x-2/x +x

=2/x+x = (2+x^2)/x

ii)

(4/x - 7/x^2) over( 7/x^3 - 2/x ) = (4/x - 7/x^2) / ( 7/x^3 - 2/x )

Numerator:

You should have common denominator for the denominators x and x^2. Taking the LCM of x and x^2, which is x^2, the equivalent fraction for4/x = 4x/x^2 and 7/x^2 remains as it is. So the numerator is:

(4/x - 7/x^2) = 4x/x^2 - 7/x^2 = (4x-7)/x^2............(2)

In a similar manner, denominator,7/x^3-2/x becomes:

7/x^3 - 2x^2/x^3 = (7-2x^2)/x^3..........................(3)

From (2) and (3),

(4/x - 7/x^2) over (7/x^3-2/x) = {(4x-7)/x^2}/{ (7-2x^2)/x^3}

=(4x-7)x^3/{(7-x^2)x^2}, as (a/b)/(c/d) = ad/(cd)

=(4x-7)x/(7-x^2).