MHB Simplify (Adding and Subtracting Rational Functions)

[eleventhxhour]

5c) Simplify.

$$\frac{2x}{3y} - \frac{x^2}{4y^3} + \frac{3}{5y^4} $$This is what I did, which is wrong according to the textbook. Could someone point out what I did wrong and how to correct it? Thanks.

$$\frac{(2x)(4y^3)-(x^2)(3y)}{(3y)(4y^3)} + \frac{3}{5y^4}$$

$$\frac{8xy^3-3x^2y}{(12y^3)} + \frac{3}{5y^4}$$

$$\frac{(8xy^3-3x^2y)(5y^4)+(3)(12y^4)}{(12y^3)(5y^4)}$$

$$\frac{(40xy^7-15x^2y^5)+36y^4}{60y^8}$$

 

[topsquark]

5c) Simplify.

$$\frac{2x}{3y} - \frac{x^2}{4y^3} + \frac{3}{5y^4} $$This is what I did, which is wrong according to the textbook. Could someone point out what I did wrong and how to correct it? Thanks.

$$\frac{(2x)(4y^3)-(x^2)(3y)}{(3y)(4y^3)} + \frac{3}{5y^4}$$

$$\frac{8xy^3-3x^2y}{(12y^3)} + \frac{3}{5y^4}$$

$$\frac{(8xy^3-3x^2y)(5y^4)+(3)(12y^4)}{(12y^3)(5y^4)}$$

$$\frac{(40xy^7-15x^2y^5)+36y^4}{60y^8}$$

Little things always mess you up.

Line 3, last term in the numerator. Check your power of y.

Last line, Check the power of y in the denominator.

Finally, there is some cancellation you can do.

-Dan

 

[MarkFL]

I would first observe that the lowest common denominator is $60y^4$ and to we may write the expression as:

$$\frac{2x}{3y}\cdot\frac{20y^3}{20y^3}-\frac{x^2}{4y^3}\cdot\frac{15y}{15y}+\frac{3}{5y^4}\cdot\frac{12}{12}$$

This is somewhat simpler than your method.

And so combining terms, what do we get?

As Dan stated, your expression is almost equivalent to this, you just need to divide each term in the numerator and denominator a common factor (after making the check Dan suggests).