make u the subject in x =\(\frac{2u-3}{3u + 2}\)
a
u = \(\frac{2x + 3}{3x - 2}\)
b
u = \(\frac{2x - 3}{3x - 2}\)
c
u = \(\frac{2x + 3}{2 - 3x}\)
d
u = \(\frac{2x + 3}{3x + 2}\)
Explanation
Correct Option
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