JAMB - Mathematics (1979 - No. 22)
Make c the subject of the equation a(b + c) + \(\frac{5}{d}\) - 2 = 0
c = \(\frac{2d - 5 - b}{ad}\)
c = \(\frac{2d - 5 - abd}{ad}\)
c = \(\frac{2d - 5 - ab}{ad}\)
c = \(\frac{5 - 2d - b}{ad}\)
Giải thích
a(b + c) + \(\frac{5}{d}\) - 2 = 0
ab + ac + \(\frac{5}{d}\) - 2 = 0
abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d
acd = 2d - 5 - abd
c = \(\frac{2d - 5 - abd}{ad}\)
ab + ac + \(\frac{5}{d}\) - 2 = 0
abd + \(\frac{acd}{d}\) + 5 - 2 = ab + ac + 5 - 2d
acd = 2d - 5 - abd
c = \(\frac{2d - 5 - abd}{ad}\)