DF001
  • DF001
Help with an Algebra II question! (Subtracting fractions with unlike denominator.)
Mathematics
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DF001
  • DF001
Help with an Algebra II question! (Subtracting fractions with unlike denominator.)
Mathematics
jamiebookeater
  • jamiebookeater
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DF001
  • DF001
\[\frac{ S-1 }{ S } - \frac{ T+1 }{ T }\]
jim_thompson5910
  • jim_thompson5910
what's the LCD in this case?
DF001
  • DF001
I have trouble figuring out the LCD, I was never good at it. Please teach me :(

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DF001
  • DF001
I thought you multiple the fractions by the opposite denominators
jim_thompson5910
  • 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
  • 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
  • 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
  • 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
  • jim_thompson5910
notice how the `T/T` is a fancy form of 1. Multiplying by 1 doesn't change the fraction
DF001
  • DF001
As to why do we want it to equal ST or TS?
jim_thompson5910
  • 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
  • DF001
Can I get a brief definition of what an lcd is
jim_thompson5910
  • jim_thompson5910
LCD = lowest common denominator it's the LCM of the denominators
DF001
  • DF001
Im not sure what it really means but, i thought it was the lowest number that can go into two numbers
jim_thompson5910
  • jim_thompson5910
for example 1/2 + 2/3 the LCD is 6 since the LCM of 2 and 3 is 6
jim_thompson5910
  • jim_thompson5910
6 is the lowest multiple both 2 and 3 have in common
DF001
  • DF001
Did you add the denominators together in 1/2+2/3
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • 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
  • jim_thompson5910
so 6 is the LCM (lowest common multiple)
DF001
  • 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
  • jim_thompson5910
you mean back with the one with S and T in it?
DF001
  • DF001
I see the first fraction is times by T thus ST
jim_thompson5910
  • jim_thompson5910
yes
DF001
  • DF001
But, if I do that to the first fraction, should I do that to the second fraction ?
DF001
  • DF001
mulitple T by S
jim_thompson5910
  • jim_thompson5910
you'll do something similar, but with S instead
DF001
  • DF001
thus, they have the same denominator ST
jim_thompson5910
  • jim_thompson5910
you'll multiply top and bottom of the second fraction by S
jim_thompson5910
  • jim_thompson5910
yes that's the ultimate goal: to get the denominators equal to the LCD (so they are all the same)
DF001
  • DF001
Oh, The answer I got for the ST question is -1T+1S/ST
DF001
  • DF001
If I am correct about the numerator
jim_thompson5910
  • 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
  • DF001
Yes but, I mines is T(S-1)-S(T+1)/TS
DF001
  • DF001
im not sure if thats the same as the last step
jim_thompson5910
  • jim_thompson5910
yeah you'll have T(S-1)-S(T+1) all over TS as one of your steps
DF001
  • DF001
I combined the denominator
jim_thompson5910
  • 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
  • DF001
If I multiply t and s into what's in the parenthesis do I get \[\frac{ -1t+1s }{ st }\]
jim_thompson5910
  • jim_thompson5910
good, which is equivalent to \[\Large \frac{S-T}{ST}\]
jim_thompson5910
  • jim_thompson5910
the TS terms up in the numerator will cancel (since 1TS - 1TS = 0TS = 0)
DF001
  • DF001
Why is it S-T
jim_thompson5910
  • jim_thompson5910
oh wait, that S should be negative
jim_thompson5910
  • 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
  • jim_thompson5910
hopefully all that makes sense?
DF001
  • DF001
Oh, I see now
DF001
  • DF001
I got confused by the -S
DF001
  • DF001
thank you man
jim_thompson5910
  • jim_thompson5910
sure thing

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