Question

Does anyone know how to add and divide rational expressions ?

Answer 1

\[8x+4=4(2x+1)\]
But that's not really important. Basically, in this type of question, just multiply both sides by the LHS's denominator first, then multiply by the RHS's denominator to achieve for both denominator=1.

Answer 2

In essence, nobody but calculators enjoys fractional division, so change it into the much easier polynomial multiplication by making the demoninator=1.

Answer 3

I have no idea what you are talking about ..

Answer 4

whats LHS ?

Answer 5

left hand side

Answer 6

\[\frac{x ^{2}-25 }{ 2x+1 }+\frac{ x ^{2} -10x+25}{ 8x+4 }\]
\[=\frac{x ^{2}-25 }{ 2x+1 }+\frac{ x ^{2} -10x+25}{ 4(2x+1) }\]
\[=\frac{4(x ^{2}-25) }{ 4(2x+1) }+\frac{ x ^{2} -10x+25}{ 4(2x+1)}\]
\[=\frac{4(x^2-25)+x^2-10x+25}{4(2x+1)}\] but my guess is that this was a division not and addition

Answer 7

\[\frac{x ^{2}-25 }{ 2x+1 }\div\frac{ x ^{2} -10x+25}{ 8x+4 }\]
\[=\frac{(x^2-25)\times (9x+4)} {(2x+1)(x^2-10x+25)}\]

Answer 8

yea I am suppose to divide this for one answer and then add it for my second

Answer 9

typo there
\[=\frac{(x^2-25)\times (8x+4)} {(2x+1)(x^2-10x+25)}\]

Answer 10

now it is a factor and cancel problem

Answer 11

yea I did all that for the divison part but it is suppose to be broken down

Answer 12

\[\frac{4(x+5)(x-5)(2x+1)}{(2x+1)(x-5)(x-5)}\] and cancel away

Answer 13

\[\frac{(x-5)(x+5) }{ 2x+1 }\times \frac{ 8x+4 }{ (x-5)(x+5) }\]

Answer 14

cancel common factors

Answer 15

ohk thank

Answer 16

yw

Answer 17

thank you *

Answer 18

Do you know how to find the LCD of the same problem ?