Discussion

Explanation:

2x² + 6x + 5y + 1 = 0
∴ 2x(x + 3) + 5y + 1 = 0   ...(1)

Also, 2x + y + 3 = 0
∴ 2x = – (y + 3)   ...(2)
∴ x = (-(y+3))/2   ...(3)

Substituting (2) & (3) in (1),
-(y + 3)$\left[\frac{-(y+3)}{2}+3\right]$ + 5y + 1 = 0

∴ $\frac{-(y+3)(-y-3+6)}{2}$ + 5y + 1 = 0

∴ –(y + 3) (–y + 3) + 10y + 2 = 0

∴ –(9 – y²) + 10y + 2 = 0

∴ y² + 10y – 7 = 0

Hence, option (c).

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