Make m the subject of the relation k = \(\frac{m - y}{m + 1}\)
MATHEMATICS
WAEC 2020
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 = \(\frac{y - k^2}{1 - k^2}\)
Correct Answer: B. m = \(\frac{y + k^2}{1 - k^2}\)
Explanation
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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