Question

I need help with adding this rational expression

Answer 1

\[\frac{ x }{ 2x^2-x } + \frac{ 1 }{ 2x }\]

Answer 2

@Data_LG2 @pooja195 Can you help me please?

Answer 3

What should the least common denominator be for this problem?

Answer 4

@Directrix @confluxepic @zepdrix @robtobey @wio Would anyone help me with this problem please?

Answer 5

We could first simplify the denominator of the first fraction.

If we factored \(\sf 2x^2-x\) we would get \(\sf 2x(x-1)\)

Answer 6

And then now we have these two fractions:

\(\huge\sf\frac{x}{2x(x-1)}+\frac{1}{2x}\)

And then to the second fraction just multiply (x-1) to both the numerator and denominator, and now we have a common denominator and we just add \(\Large\ddot\smile\)

Answer 7

If we factor that wouldn't it be
\[x(2x-1)\]

Answer 8

Yes, yes, my bad. OS was lagging and ugh >.<

Answer 9

Yeah so then how would we do the rest of the problem?

Answer 10

And then now we have these two fractions:

\(\huge\sf\frac{x}{x(2x-1)}+\frac{1}{2x}\)

So we can do this:

\(\LARGE\sf\left(\frac{x}{x(2x-1)}\times\frac{2}{2}\right)+\left(\frac{1}{2x}\times\frac{(2x-1)}{(2x-1)}\right)\)

Answer 11

So then the LCD is \[2x(2x-1)\] ?

Answer 12

Ok. So after adding and multiplying together the answer should be
\[\frac{ 4x-1 }{ 2x(2x-1) }\]

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @happyrosy
So then the LCD is \[2x(2x-1)\] ?
\(\color{blue}{\text{End of Quote}}\)

Yup!

Answer 14

\(\color{blue}{\text{Originally Posted by}}\) @happyrosy
Ok. So after adding and multiplying together the answer should be
\[\frac{ 4x-1 }{ 2x(2x-1) }\]
\(\color{blue}{\text{End of Quote}}\)

And correcto! :D

Answer 15

Thank you so much for the help! You were the only one helping me SMART ONE :)

Answer 16

No problem \(\ddot\smile\)