Question 9:
Solve the simultaneous equations:
2 (x – y) = x + y – 1 = 2x2 – 11y2 [5 marks]
Solution:
2 (x – y) = x + y – 1
2x – 2y = x + y – 1
x = 3y – 1 —- (1)
x + y – 1 = 2x2 – 11y2
2x2– 11y2 – x – y+ 1 = 0 —- (2)
Substitute (1) into (2):
2 (3y – 1)2– 11y2 – (3y – 1) – y + 1 = 0
2 (9y2– 6y + 1) – 11y2 – 3y + 1 – y+ 1 = 0
18y2– 12y + 2 – 11y2 – 4y + 2 = 0
7y2– 16y + 4 = 0
(7y – 2)(y – 2) = 0
y=27ory=2
From (1):
when y=27,x=3(27)−1x=67−1=−17
When y = 2,
x = 3(2) – 1 = 5
Hence the solutions are, x=−17,y=27 or x=5,y=2