JAMB - Mathematics (1981 - No. 21)
Solve \(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
x = -1 or 3
x = 1 or 3
x = 1 or -5
x = -1 or 5
x = -1 or -3
Explanation
\(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
\(\frac{x + 3 - x - 1}{(x + 1)(x + 3)}\) = \(\frac{1}{4}\)
\(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\)
= x2 + 4x + 3 = 8
x2 + 4x - 5 = 0
= (x - 1)(x + 5) = 0
x = 1 or -5
\(\frac{x + 3 - x - 1}{(x + 1)(x + 3)}\) = \(\frac{1}{4}\)
\(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\)
= x2 + 4x + 3 = 8
x2 + 4x - 5 = 0
= (x - 1)(x + 5) = 0
x = 1 or -5
Comments (0)
