The Chief Examiner reported that this question was attempted by majority of the candidates and they performed very well in it. According to the report, candidates’ performance indicted that candidates had a good understanding of surds and determinants.
In part (a), candidates were reportedly able to subtract the fractions to obtain:

After simplifying the fraction, they obtained which should be written as
0 + .
In part (b), some candidates were reported not to adhere to the rubrics of the question which stated that they should use determinants methods to solve the problem. They were reported to have found the inverse of the matrix and used it to solve the problem. They lost some marks.
Candidates were expected to express the equations in matrix form as
 = . Next, candidates would calculate the determinant,
 =  which was 3-1 = 3(5 – 6) + 1(1 – 4) – 1(3- 10) =  7 – 6  = 1.

To find x, candidates were expected to substitute   for the first column of the matrix and then find the determinant.

i.e. = - 2 = -2(5 – 6) + 1(5 – 0) – 1(15 – 0) = 2 + 5 – 15 = -8. Hence,  x =  =    =  −8.  

=  41. i.e. x = -8, y = 19 and z = -41.