Question

8/121x + 9/11x^2
(those are fractions)
sorry my brain just isn't working right now

Answer 1

so... what do you think is the LCD anyhow?

Answer 2

\(\bf \cfrac{8}{121x}+\cfrac{9}{11x^2}\implies \cfrac{8}{11\cdot 11x}+\cfrac{9}{11x\cdot x}=\cfrac{}{{\color{brown}{ lcd?}}}\) any ideas?

Answer 3

(99x^2 + 8x)/121 You can simplify if you're supposed to

Answer 4

@M_e_g_a is your expression this?
\[\frac{ 8 }{ 121 }x+\frac{ 9 }{ 11 }x ^{2}\]

Answer 5

@Michele_Laino yes, the X's are on the denominators

Answer 6

it doesn't matter where the x's are...

Answer 7

@jdoe0001 The LCD would be 11x? or 11x^2...?

Answer 8

the lcd, would be, one that comprises both
11x alone wouldn't

Answer 9

@M_e_g_a lcd is this:
\[11^{2}*x ^{2}\]

Answer 10

@jdoe0001 1+11^2+x?

Answer 11

that one should be an eleven

Answer 12

No i wrote that completely wrong, I meant to put 11+11x+x

Answer 13

well... in this case one 11x doesn't by itself and \(11^2x\) doesn't by itself either
but \(11^2x^2\) would

keep in mind that, at the very least, one could use the product of all denominators as lcd

Answer 14

@M_e_g_a your result is:
\[\frac{ 8x+11 }{ 121x ^{2} }\]

Answer 15

but the 11x appears twice so you only have to use it once... that's what I thought

Answer 16

sorry I have made an error:
\[\frac{ 8x+99 }{ 121x ^{2} }\]

Answer 17

hmmm once you have the lcd... you'd add the fractions like you'd any other fraction
divide the denominator by the lcd
the quotient multiplies the numerator
and adds to the next fraction result

Answer 18

@Michele_Laino @jdoe0001 you guys were right I'm just a bit stupid xD thanks