Simplify.. need help asap...
8/ab + 6/b
and
(3t2/(t + 2)) x ((t + 2)/t2)
thanks!!

- anonymous

Simplify.. need help asap...
8/ab + 6/b
and
(3t2/(t + 2)) x ((t + 2)/t2)
thanks!!

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- jim_thompson5910

The first one is \[\Large \frac{8}{ab} + \frac{6}{b}\] right?

- anonymous

yes.

- jim_thompson5910

what is the LCD in this case?

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## More answers

- anonymous

i don't know.

- jim_thompson5910

we have ab as the first denominator
and b as the second denominator

- jim_thompson5910

do you see how the LCD would be ab?

- anonymous

no.

- anonymous

:( I'm really bad at math. I'm sorry..

- jim_thompson5910

maybe replace a and b with numbers

- jim_thompson5910

say
a = 2
b = 3

- anonymous

okay.

- jim_thompson5910

ab = 2*3 = 6

- anonymous

i understand that.

- jim_thompson5910

if we had 1/6 + 1/3, what would the LCD be?

- anonymous

6

- jim_thompson5910

yes because we can multiply the denominator 3 by 2 to get 6

- jim_thompson5910

we can multiply top and bottom of 1/3 by 2 to get 2/6
1/6 + 1/3 turns into 1/6 + 2/6

- jim_thompson5910

from there you add straight across and leave the denominator alone

- anonymous

mhm.

- jim_thompson5910

the same idea applies here
we multiply top and bottom of the second fraction by 'a'
\[\Large \frac{8}{ab} + \frac{6}{b}\]
\[\Large \frac{8}{ab} + \frac{6{\color{red}{a}}}{b{\color{red}{a}}}\]
\[\Large \frac{8}{ab} + \frac{6a}{ab}\]

- jim_thompson5910

then you add the numerators and place that over the denominator

- anonymous

okat.

- anonymous

now I'm lost.

- jim_thompson5910

where at?

- anonymous

how do we find the answer? or did we already find it and i was lost before then? i thought I've been following.

- jim_thompson5910

do you see why I multiplied top and bottom of the second fraction by 'a'?

- jim_thompson5910

I wanted to get every denominator equal to the LCD 'ab'

- anonymous

okay

- jim_thompson5910

there's one more step to go

- anonymous

alright.

- jim_thompson5910

What does
\[\Large \frac{8}{ab} + \frac{6a}{ab}\]
simplify to?

- anonymous

8/ab + 6/ab

- anonymous

i really am lost.

- anonymous

8/ab+6/b

- jim_thompson5910

just add up the numerators
\[\Large \frac{8}{ab} + \frac{6a}{ab}=\frac{8+6a}{ab}\]
this is possible because the denominators are both equal to ab

- anonymous

okay. now how would we do the second one?

- jim_thompson5910

The second one is \[\Large \left(\frac{3t^2}{t+2}\right)\times\left(\frac{t+2}{t^2}\right)\] right?

- anonymous

t+2 on the sec on one are in a parenthesis.

- anonymous

and in the first one. its like parenthesis insides of themselves.

- jim_thompson5910

ok so this?
\[\Large \left(\frac{3t^2}{(t+2)}\right)\times\left(\frac{(t+2)}{t^2}\right)\]

- anonymous

yes that is correct.

- jim_thompson5910

first notice this cancellation
\[\Large \left(\frac{3{\color{red}{\cancel{\color{black}{t^2}}}}}{t+2}\right)\times\left(\frac{t+2}{{\color{red}{\cancel{\color{black}{t^2}}}}}\right)\]
the t^2 terms will go away leaving just
\[\Large \left(\frac{3}{(t+2)}\right)\times\left(\frac{(t+2)}{1}\right)\]

- anonymous

ok
what is the next step

- jim_thompson5910

then the (t+2) terms also cancel
\[\Large \left(\frac{3}{{\color{red}{\cancel{\color{black}{(t+2)}}}}}\right)\times\left(\frac{{\color{red}{\cancel{\color{black}{(t+2)}}}}}{1}\right)\]

- jim_thompson5910

I'm sure you see how to finish

- anonymous

so its 3?

- jim_thompson5910

yep

- anonymous

thats the easiest thing I've ever seen

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