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

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1. johnweldon1993

First we need a common denominator to add these fractions...what would that be?

2. anonymous

would it be (q-1)(q+10) ??

3. johnweldon1993

Can I assume that 10 was a typo for 1?

4. anonymous

oops yeah it was

5. 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?

6. anonymous

yup! so now would it be q^2+q+4q+4+q^2-q = 2q+2 +2q - 2 right ?

7. 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 $\large 2q^2 + 4q + 4 = 2q^2 - 2$