Question on: JAMB Mathematics - 2012
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
\(\frac{1}{c}(\frac{w + 2v}{v + w})\)
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Correct Option: 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 = \(\frac{1}{c}\frac{w - 2v}{v + w}\)
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 = \(\frac{1}{c}\frac{w - 2v}{v + w}\)
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