Help with an Algebra II question!
(Subtracting fractions with unlike denominator.)

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- DF001

Help with an Algebra II question!
(Subtracting fractions with unlike denominator.)

- jamiebookeater

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- DF001

\[\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]

- jim_thompson5910

what's the LCD in this case?

- DF001

I have trouble figuring out the LCD, I was never good at it. Please teach me :(

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## More answers

- DF001

I thought you multiple the fractions by the opposite denominators

- jim_thompson5910

the denominators are S and T
they have nothing in common except for 1, so we can multiply them to get the LCD
the LCD is simply `ST` or `TS`

- jim_thompson5910

to add or subtract fractions, the denominators must be the same. They aren't in this case, so we have to get them all equal to the LCD

- jim_thompson5910

the first denominator is S
we want it to be equal to ST or TS
it's missing a T, so multiply top and bottom of the first fraction by T
\[\Large \frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]
\[\Large \frac{\color{red}{T}}{\color{red}{T}}\times\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]
\[\Large \frac{ \color{red}{T}(S-1) }{ \color{red}{T}S } - \frac{ T+1 }{ T }\]

- DF001

If in other cases like one denominator is 3a+12 and the other is a+4, how can I find the LCD in this?

- jim_thompson5910

notice how the `T/T` is a fancy form of 1. Multiplying by 1 doesn't change the fraction

- DF001

As to why do we want it to equal ST or TS?

- jim_thompson5910

you would factor 3a+12 to get 3(a+4)
we really have the denominators 3(a+4) and a+4
so the LCD is 3(a+4). The unique factors are 3 and (a+4)

- DF001

Can I get a brief definition of what an lcd is

- jim_thompson5910

LCD = lowest common denominator
it's the LCM of the denominators

- DF001

Im not sure what it really means but, i thought it was the lowest number that can go into two numbers

- jim_thompson5910

for example
1/2 + 2/3
the LCD is 6 since the LCM of 2 and 3 is 6

- jim_thompson5910

6 is the lowest multiple both 2 and 3 have in common

- DF001

Did you add the denominators together in 1/2+2/3

- jim_thompson5910

no

- jim_thompson5910

list the multiples of 2
2, 4, 6, 8, 10, 12, ...
list the multiples of 3
3, 6, 9, 12, 15, 18, ...
the common multiples are
6, 12, 18, ...
the number 6 is the smallest multiple

- jim_thompson5910

so 6 is the LCM (lowest common multiple)

- DF001

Back to the image 5 minutes ago, must you look at the first fraction first to know what to times the numerator and denominator by?

- jim_thompson5910

you mean back with the one with S and T in it?

- DF001

I see the first fraction is times by T thus ST

- jim_thompson5910

yes

- DF001

But, if I do that to the first fraction, should I do that to the second fraction ?

- DF001

mulitple T by S

- jim_thompson5910

you'll do something similar, but with S instead

- DF001

thus, they have the same denominator ST

- jim_thompson5910

you'll multiply top and bottom of the second fraction by S

- jim_thompson5910

yes that's the ultimate goal: to get the denominators equal to the LCD (so they are all the same)

- DF001

Oh, The answer I got for the ST question is -1T+1S/ST

- DF001

If I am correct about the numerator

- jim_thompson5910

so we have this so far after getting each denominator equal to the LCD
\[\Large \frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]
\[\Large \frac{\color{red}{T}}{\color{red}{T}}\times\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]
\[\Large \frac{ \color{red}{T}(S-1) }{ \color{red}{T}S } - \frac{ T+1 }{ T }\]
\[\Large \frac{ T(S-1) }{ TS } - \frac{\color{red}{S}}{\color{red}{S}}\times\frac{ T+1 }{ T }\]
\[\Large \frac{ T(S-1) }{ TS } - \frac{ \color{red}{S}(T+1) }{ \color{red}{S}T }\]
\[\Large \frac{ T(S-1) }{ TS } - \frac{ S(T+1) }{ TS }\]
agreed?

- DF001

Yes but, I mines is T(S-1)-S(T+1)/TS

- DF001

im not sure if thats the same as the last step

- jim_thompson5910

yeah you'll have T(S-1)-S(T+1) all over TS as one of your steps

- DF001

I combined the denominator

- jim_thompson5910

\[\Large \frac{ T(S-1) }{ TS } - \frac{ S(T+1) }{ TS }\]
turns into
\[\Large \frac{ T(S-1) - S(T+1) }{ TS }\]

- DF001

If I multiply t and s into what's in the parenthesis do I get \[\frac{ -1t+1s }{ st }\]

- jim_thompson5910

good, which is equivalent to \[\Large \frac{S-T}{ST}\]

- jim_thompson5910

the TS terms up in the numerator will cancel (since 1TS - 1TS = 0TS = 0)

- DF001

Why is it S-T

- jim_thompson5910

oh wait, that S should be negative

- jim_thompson5910

\[\Large \frac{ T(S-1) - S(T+1) }{ TS }\]
\[\Large \frac{ TS-T - ST-S }{ TS }\]
\[\Large \frac{ -T -S }{ TS }\]

- jim_thompson5910

hopefully all that makes sense?

- DF001

Oh, I see now

- DF001

I got confused by the -S

- DF001

thank you man

- jim_thompson5910

sure thing

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