The Chief Examiner reported that this question was well attempted by majority of the
candidates. However, there were few cases where some candidates did not recognize 0 (zero)
as one of the solutions in part (b).

In part (a) candidates were expected to follow the rule of BODMAS while substituting for x and y in the operations. Taking the bracket (2 * 3), we have:2²/2+3 = 4/5

( 2 * 3) * 5 now becomes (4/5)²/4/5+5 = 16/25/29/5 = 16/25 x 5/29 x 16/145

In part (b), candidates were expected to substitute (x + 1) for x and (x+2) for y in the expression x²/x+y, equate it to 1/3 and solve the resulting equation to obtain the values of x as follows:

__(x ____+ ____1)__2 = 1/3

(x+1 )+(x+2)

3(x+1)² =2x+3 l.e 3x²+6x+3=2x+3 3x²+4x = 0

x(3x + 4) = 0 x = 0 or 3/4