History of Hindu Mathematics: A Source Book v2.c17.s23.67

v2.c17.s23.67

The solution of the first equation has been given before to be $x &= 1/2a(y²)/(λ)(c - (b²)/(4a²)) - λ - by/(2a²), / u &= ½(y²)/(λ)(c - (b²)/(4a²)) + λ,$ where λ is an arbitrary rational number. Putting λ = y, we have $x &= y/2a(c - (b²)/(4a²) - 1) - by/(2a²) = α y, / u &= y/2(c - (b²)/(4a²) + 1);$ where $α = 1/2a(c - (b²)/(4a²) - 1) - b/(2a²).$