# Simplify each expression: 2/1+y-2 /2x^3+x

1. Aug 28, 2011

### davie08

1. The problem statement, all variables and given/known data

2/1+y-2 /2x^3+x its all over top of 2x^3+x

2. Relevant equations

3. The attempt at a solution

should i multiply the top by 1+y, like the 2 and the -2.

2. Aug 28, 2011

### ArcanaNoir

can you use parentheses to make this clearer?

3. Aug 28, 2011

### davie08

heres a picture of it its d)

Last edited: Aug 28, 2011
4. Aug 28, 2011

### ArcanaNoir

you have $$\frac{\frac{2}{1+y}-2}{2x^3+x}$$

next you should do $$\frac{\frac{2}{1+y}-\frac{2(1+y)}{(1+y)}}{2x^3+x}$$

5. Aug 28, 2011

### davie08

so would this be the final answer. -2y/2x^3+x

6. Aug 28, 2011

### ArcanaNoir

I think you copied the problem wrong! And no that would not be the answer, you dropped a (1+y) somewhere.

7. Aug 28, 2011

### davie08

sorry im feeling a little stressed i got to 2y/1+y/2x^3+x I've forgotten how to do everything lol.

8. Aug 28, 2011

### ArcanaNoir

For problem d, $$\frac{\frac{2}{1+y}-2}{y}$$
Start the same way as before with $$\frac{\frac{2}{1+y}-\frac{2(1+y)}{(1+y)}}{y}$$

which becomes $$\frac{\frac{2y}{1+y}}{y}$$

then remember the denominator "y" is actually $$\frac{y}{1}$$
so you have $$\frac{\frac{2y}{1+y}}{\frac{y}{1}}$$

and to divide to fractions you multiply by the reciprocal like this: $$\frac{2y}{(1+y)}(\frac{1}{y})$$

then just cancel a y from the top and bottom and you're done

9. Aug 28, 2011

### davie08

god I wrote it down wrong thanks.

10. Aug 28, 2011

### davie08

so that would make it 2/y^2

11. Aug 28, 2011

### ArcanaNoir

eek! no.
its $$\frac{2y}{y(1+y)}$$

which gives $$\frac{2}{1+y}$$

12. Aug 28, 2011

### Ray Vickson

No: you have written $$\frac{-2y}{2x^3} + x$$. Did you mean to write
$$\frac{-2y}{2x^3 + x}?$$ If so, then USE BRACKETS, like this: -2y/(2x^3+x). Isn't that simple? It makes everything clear and removes all confusion.

RGV

13. Aug 28, 2011

### ArcanaNoir

that was pointless...