Question

OK I have a question. It is an algebraic expression

Answer 1

\[12(\frac{ 1 }{ 3^{a2} }- \frac{ 5 }{ 6{a} } + \frac{ 7 }{ 12 })\]

Answer 2

Its says to multiply

Answer 3

is that 3^(2a) or 3^(a^2) ?

Answer 4

I wrote it correctly

Answer 5

Its 12(1/3a2-5/6a+7/12)

Answer 6

sorry...just distribute the 12 to each term...reducing each fraction

Answer 7

You still have to combine the fractions

Answer 8

Inside the parenthesis is where I am having the issue

Answer 9

can you explain it

Answer 10

You can also combine the fractions first before multiplying the 12

Answer 11

Anytime you're trying to combine fractions, finding the LCD is necessary.

Answer 12

which is 12

Answer 13

but I get confused with the a2 on the bottom doesnt all the bottom numbers need it

Answer 14

the LCD is \(3^{a^2} \dot\ 6a \dot\ 12\)

Answer 15

now dont I have to get all them to 12a2

Answer 16

this thing will really just make sense *if* the question was written incorrectly..

Answer 17

I wrote it down like it said as the very top

Answer 18

are you sure it's not written as \[12 \left( \frac{1}{3a^2} - \frac{5}{6a} + \frac{7}{12} \right)\]

Answer 19

yes just like that

Answer 20

because i have never seen a notation such as \(3^{a2}\)

Answer 21

see? now it'll make sense lol

Answer 22

I hate how it writes it out

Answer 23

yes. darn autocorrect right?

Answer 24

ok so I know all the bottom has to be the same inside the parenthesis correct to get the lowest LCD which is 12

Answer 25

i think @dumbcow and @Hero can resume explaining now the right question. \[12 \left(\frac{1}{3a^2} - \frac{5}{6a} + \frac{7}{12} \right)\]

Answer 26

can you just help me @lgbasallote

Answer 27

well i have to go soon....

Answer 28

the others are no longer attached to this question

Answer 29

@Hero can you help now

Answer 30

@mandypruitt said that her original fraction was correct as written so I believed her.

Answer 31

I wish the book gave an example of how to do this one. @Hero I thought I wrote it right I am still learning how to write these correctly

Answer 32

It still doesn't change much since the lcd is \(3a^2 \dot\ 6a \dot\ 12\)

Answer 33

isnt it 12?

Answer 34

No, it's not 12

Answer 35

Can you work the problem out for me so I can see what i am doing wrong.

Answer 36

It's only 12 if you assume a = 2, but we can't assume that

Answer 37

I am so confused I always thought that if there were exponents in the denominator they all have to be the same so you would need to get them the same also

Answer 38

the only way to get the denominators the same is to find the lcd, which I've already provided you with.

Answer 39

wow when someone asks for your help you just leave @Hero

Answer 40

I'm helping other students as well.

Answer 41

u still need help @mandypruitt ?

Answer 42

Okay, I'll show you how to do it in vyew. Just this once....

Answer 43

I posted a link in vyew

Answer 44

yes @hartnn

Answer 45

so u got LCM of 3,6,12 as 12 right?
now whats the LCM of a and a^2?

Answer 46

a2 @hartnn

Answer 47

right ,its a^2
so the LCM of 3a^2,6a and 12 is actually 12a^2
ok?
u have to make THIS as common denominator...

Answer 48

did u get this much? @mandypruitt

Answer 49

@Hero thank u