Find the positive solution of the following simultaneous equations:

Answer:

14) x = 3 ; y = 1

2x – y = 5
y = 2x – 5

Substitute this expression of ‘y’ into the first equation.
4x² + (2x – 5)² + 10 (2x – 5) = 47
4x² + (4x² – 20x + 25) + (20x – 50) = 47
4x² + 4x² – 20x + 25 + 20x – 50 = 47
8x² – 25 = 47
8x² = 72
x² = 9
x = 3

Substitute this value of ‘x’ into the second equation.
y = 2 (3) – 5
y = 6 – 5 = 1