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anonymous
 one year ago
q+ 4 q
 +  = 2
q−1 q+1
show all work please
anonymous
 one year ago
q+ 4 q  +  = 2 q−1 q+1 show all work please

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johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0First we need a common denominator to add these fractions...what would that be?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it be (q1)(q+10) ??

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0Can I assume that 10 was a typo for 1?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0No problem, just want to make sure you got it :) \[\large \frac{(q+4)(q+1)}{(q1)(q+1)} + \frac{q(q1)}{(q1)(q+1)} = 2\] Do you see how I got that with the common denominator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yup! so now would it be q^2+q+4q+4+q^2q = 2q+2 +2q  2 right ?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.0Not quite! \[\large \frac{q^2 + 5q + 4 + q^2  q}{(q1)(q+1)} = 2\] \[\large q^2 + 5q + 4 + q^2  q = 2\color\red{(q1)(q+1)}\] \[\large (q1)(q+1) = q^2  1\] so \[\large 2q^2 + 4q + 4 = 2q^2  2\]
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