Question

Find the lowest common denominator that would eliminate the fractions in this equation.

8/9x+2/3x=1/9x+45

Answer 1

How would you do this?

Answer 2

I dont know, thats why im asking

Answer 3

Can you just help me?

Answer 4

I thought you had this figured out

Answer 5

This is different

Answer 6

How?

Answer 7

The last one was just completly different and in a different unit

Answer 8

Im sorry if i disapointed you

Answer 9

You didn't disappoint anyone

Answer 10

Okay:/

Answer 11

Can someone just plzzz help me?

Answer 12

Wanna go back to vyew?

Answer 13

I would but its actually four in the morning here and i have no headphones and my parents are light sleepers:'(

Answer 14

What are you doing up at 4 in the morning?

Answer 15

Well i couldnt sleep and schools tom well today anyways, i just dont feel good and so i decided to do some school instead of drooling into a cup haha

Answer 16

5 & 12

Answer 17

n 3rd is 6

Answer 18

What do you mean 5&12?

Answer 19

we get 3 ans
5
12
6

Answer 20

@igbasalotte i need your help!

Answer 21

i can only have one

Answer 22

@adityanaik28 , there's only one value for x.

Answer 23

Yes, would it be 12? @Hero

Answer 24

Actually, it only wants the lowest common denominator that would eliminate the fractions.

Answer 25

So its none of the answers he said?

Answer 26

There's only one number that would work and it's not 12 and definitely not any of the answers the other guy said.

Answer 27

Oh okay

Answer 28

The LCD is 9x.

Answer 29

lol sorry @PaigeMya i did not notice the tag...next time please tag it to Lgbasallote (just lower case) :DD

if you havent had your answer yet look at your denominators

9x, 3x, and (9x+45)

we'd like to break down these denominators to a simpler form

the first denominator would be (3)(3)(x)
the second denominator would be (3)(x)
the third denominator would be 9(x+5) -> (3)(3)(x+5)

the LCD would be a value that contains ALL of the denominators

3 is surely in your LCD because all of them contains that..however there is still a 3x in your first denominator, an x in your second, and a 3(x+5) in the third and remember that the LCD must contain ALL of the values. so put another 3 in the LCD.

This now gives you a 9 in the LCD and leaves you with x in the first denominator, another x in the second denominator and (x+5) in the third. Obviously, we put x in the LCD now..

so... 9x is your LCD so far..9x contains the first denominator already, it also contains the second denominator. But it still leaves (x+5) in the third. so we add that to the LCD..

and so..with the given information..our LCD is 9x(x+5)

let us check...
1. does it contain the first denominator a.k.a 9x? \(\checkmark\)
2. does it contain the second denominator a.k.a 3x? 9x is 3x(3) so it does contain 3x \(\checkmark\)
3. Lastly, does it contain the third denominator a.k.a. 9(x+5)? \(\checkmark\)

therefore, our LCD as 9x(x+5) is correct...hope that helped ^_^