\(3x + y = 8 ... (i)\)
\(x^{2} + xy = 6 ... (ii)\)
From (i), \(y = 8 - 3x\)
From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\)
Equating the two values of y, we have
\(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\)
\(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\)
\(2x^{2} - 8x + 6 = 0\)
\(x^{2} - 4x + 3 = 0\)
\(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\)
\((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\)
\(y = 8 - 3x \)
When x = 1, \(y = 8 - 3(1) = 5\)
When x = 3, \(y = 8 - 3(3) = -1\)
\(\therefore \text{y = -1 or 5}\)