Make m the subject of the relation k = \(\frac{m - y}{m + 1}\)

  • A m = \(\frac{y + k^2}{k^2 + 1}\)
  • B m = \(\frac{y + k^2}{1 - k^2}\)
  • C m = \(\frac{y - k^2}{k^2 + 1}\)
  • D m = \(\frac{y - k^2}{1 - k^2}\)
waec 2020mathematics

The correct answer is B. m = \(\frac{y + k^2}{1 - k^2}\)

k = \(\frac{m - y}{m + 1}\)

k\(^2\) = \(\frac{m  - y}{m + 1}\)

k\(^2\)m + k\(^2\) = m - y 

k\(^2\) + y = m - k\(^2\)m

\(\frac{k^2 + y}{1 - k^2}\) = m\(\frac{(1 - k^2)}{1 - k^2}\)

m = \(\frac{y + k^2}{1 - k^2}\)

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