Question

13(a) Find the equation of a straight line which passes through the point (2,-3) and is parallel to the line 2x+y=6
(b) The operation Δ is defined on the set T={2,3,5,7} by x Δ y = (x+y+xy) mod 8
(i) Construct modulo 8 table for the operation Δ on the set T.
(ii) Use the table to find: I. 2Δ(5Δ7) II. 2Δn=5Δ7

Accepted Answer

(a) Parallel lines have equal gradient. First find the gradient of the given line (2x+y=6). Expressing this in the form y=mx+c: y=-2x+6, so m=-2 (this is also the gradient of the required line).
Since the required line passes through (2,-3) with gradient -2:
y-y1=m(x-x1)
y-(-3)=-2(x-2)
y+3=-2x+4
y+2x=1, i.e. 2x+y=1

(b)(i) Modulo 8 table for x Δ y = (x+y+xy) mod 8 on T={2,3,5,7}:
Δ | 2  3  5  7
2 | 0  3  1  7
3 | 3  7  7  7
5 | 1  7  3  7
7 | 7  7  7  7
(computed as (x+y+xy) mod 8 for each pair)

(ii) I. 2Δ(5Δ7): From the table, 5Δ7=7. Then 2Δ7=7 (from table). So 2Δ(5Δ7)=7.
II. 2Δn=5Δ7=7. From the table, 2Δ7=7, so n=7.

WAEC Mathematics 2016 Theory — Question 13

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