3.3.5 System of Equations, Long Questions (Question 9)

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 = \frac{2}{7} or y = 2

From (1):

\begin{array}{l} when y = \frac{2}{7}, \\ x = 3 (\frac{2}{7}) - 1 \\ x = \frac{6}{7} - 1 = - \frac{1}{7} \end{array}

When y = 2,

x = 3(2) – 1 = 5

\begin{array}{l} Hence the solutions are, \\ x = - \frac{1}{7}, y = \frac{2}{7} or x = 5, y = 2 \end{array}