Question

Please help! I will give a medal
How do I find the LCD of two rational expressions?
Can you also give an example

Answer 1

@phi can you help?

Answer 2

do you have a specific problem?

Answer 3

No, I just help explaining it.

Answer 4

if you have two denominators, you would first factor them into their prime factors
then you would choose the prime factors that comprise both denominators

for example , if you have 1/2 and 1/4
and you want the least common denominator (which is 4) you would write
2
and 2*2
and 2*2 is the least common denominator

Answer 5

if you had
1/6 and 1/15
6= 2*3
15= 3*5

we need a 2 and 3 (to get the 6). we need 3 (already have one) and 5 to get the 15
so the LCD is 2*3*5= 30

Answer 6

Is that what you mean ?

Answer 7

if it's algebra, it's the same idea , but we have more complicated expressions

Answer 8

for example
x^2+x and x^2-1

factor x^2+x = x(x+1)
factor (x^2-1)= (x-1)(x+1)

from x^2+x we need x(x+1)
cross off the (x+1) from the 2nd denominator (because we already have it)
that leaves (x-1) that we need
the LCD is x(x+1)(x-1)

Answer 9

Yes

Answer 10

I am not sure how you describe this process clearly.

Answer 11

I'm not sure either. My teacher does not ever put examples, he just expects us to understand what he's saying.

Answer 12

are you doing algebra (with letters) or just numbers?

Answer 13

Is your problem to
write up an explanation on how to find the LCD

or are you just trying to understand how to do it?

Answer 14

I'm trying to figure out how to do it

Answer 15

and is it with algebra or just numbers?

Answer 16

For just numbers, say you had these two denominators (already factored)
2*3*5
3*5*5

from the first set we need 2*3*5 (that is all of them )
now cross off any of 2 , 3, or 5 from the second set
\[ \cancel{3} \cdot \cancel{5} \cdot 5\]
there is no 2, but we do have one 3 and one 5 we can cross off. we are left with one 5 that we need.
the LCD is 2*3*5 * 5

Answer 17

another example:
2*5*5*5*7
2*5*5*5*5*7*7

we take all the numbers from the first number: 2*5*5*5*7
now cross off any of those numbers from the second set: cross of one 2, three 5's and one 7:
\[ \cancel{2}\cdot \cancel{5\cdot 5 \cdot 5} \cdot 5\cdot \cancel{7} \cdot 7\]
the LCD is 2*5*5*5*7 * 5*7

Answer 18

If you are doing algebra, we would use different examples. Is that what you want to learn?