anonymous
  • anonymous
q+ 4 q ------- + --------- = 2 q−1 q+1 show all work please
Mathematics
jamiebookeater
  • jamiebookeater
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johnweldon1993
  • johnweldon1993
First we need a common denominator to add these fractions...what would that be?
anonymous
  • anonymous
would it be (q-1)(q+10) ??
johnweldon1993
  • johnweldon1993
Can I assume that 10 was a typo for 1?

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anonymous
  • anonymous
oops yeah it was
johnweldon1993
  • johnweldon1993
No problem, just want to make sure you got it :) \[\large \frac{(q+4)(q+1)}{(q-1)(q+1)} + \frac{q(q-1)}{(q-1)(q+1)} = 2\] Do you see how I got that with the common denominator?
anonymous
  • anonymous
yup! so now would it be q^2+q+4q+4+q^2-q = 2q+2 +2q - 2 right ?
johnweldon1993
  • johnweldon1993
Not quite! \[\large \frac{q^2 + 5q + 4 + q^2 - q}{(q-1)(q+1)} = 2\] \[\large q^2 + 5q + 4 + q^2 - q = 2\color\red{(q-1)(q+1)}\] \[\large (q-1)(q+1) = q^2 - 1\] so \[\large 2q^2 + 4q + 4 = 2q^2 - 2\]

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