Question

q+ 4 q
------- + --------- = 2
q−1 q+1

show all work please

Answer 1

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

Answer 2

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

Answer 3

Can I assume that 10 was a typo for 1?

Answer 4

oops yeah it was

Answer 5

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?

Answer 6

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

Answer 7

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
\[\large 2q^2 + 4q + 4 = 2q^2 - 2\]