Make 'n' the subject of the formula if w = \(\frac{v(2 + cn)}{1 - cn}\)

A.

\(\frac{1}{c}(\frac{w - 2v}{v + w})\)

B.

\(\frac{1}{c}(\frac{w - 2v}{v - w})\)

C.

\(\frac{1}{c}(\frac{w + 2v}{v - w})\)

D.

\(\frac{1}{c}(\frac{w + 2v}{v + w})\)

Correct answer is A

w = \(\frac{v(2 + cn)}{1 - cn}\)

2v + cnv = w(1 - cn)

2v + cnv = w - cnw

2v - w = -cnv - cnw

Multiply through by negative sign

-2v + w = cnv + cnw

-2v + w = n(cv + cw)

n = \(\frac{-2v + w}{cv + cw}\)

n = \(\frac{1}{c}\frac{-2v + w}{v + w}\)

Re-arrange...

n = \(\frac{1}{c}\frac{w - 2v}{v + w}\)