JAMB Mathematics Past Questions & Answers - Page 313

\(\frac{1}{2}\)x + \(\frac{1}{4}\) > \(\frac{1}{3}\)x + \(\frac{1}{2}\)

Multiply through by through by the LCM of 2, 3 and 4

12 x \(\frac{1}{2}\)x + 12 x \(\frac{1}{4}\) > 12 x \(\frac{1}{3}\)x + 12 x \(\frac{1}{2}\)

6x + 3 > 4x + 6

6x - 4x > 6 - 3

2x > 3

\(\frac{2x}{2}\) > \(\frac{3}{2}\)

x > \(\frac{3}{2}\)

x \(\alpha\) \(\frac{1}{y}\) .........(1)

x = k x \(\frac{1}{y}\) .........(2)

When x = 2\(\frac{1}{2}\)

= \(\frac{5}{2}\), y = 2

(2) becomes \(\frac{5}{2}\) = k x \(\frac{1}{2}\)

giving k = 5

from (2), x = \(\frac{5}{y}\)

so when y =4, x = \(\frac{5}{y}\) = 1\(\frac{1}{4}\)

x - y = 2 ...........(1)

x2 - y2 = 8 ........... (2)

x - 2 = y ............ (3)

Put y = x -2 in (2)

x2 - (x - 2)2 = 8

x2 - (x2 - 4x + 4) = 8

x2 - x2 + 4x - 4 = 8

4x = 8 + 4 = 12

x = \(\frac{12}{4}\)

= 3

from (3), y = 3 - 2 = 1

therefore, x = 3, y = 1

27x3 - 8y3 = (3x - 2y)3

But 9x2 + 6xy + 4y2 = (3x +2y)2

So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2

Hence the other factor is 3x - 2y

p = \(\frac{M}{5}\)(X + Q) + 1

P - 1 = \(\frac{M}{5}\)(X + Q)

\(\frac{5}{M}\)(p - 1) = X + Q

\(\frac{5}{M}\)(p - 1)- x = Q

Q = \(\frac{5(p -1) - Mx}{M}\)

= \(\frac{5p - 5 - Mx}{M}\)

= \(\frac{5p - Mx - 5}{M}\)