The values of $y$ which will satisfy the equations $2x^{2} + 6x + 5y + 1 = 0, 2x + y + 3 = 0$ may be found by solving.
- $y^{2} + 14y - 7 = 0$
- $y^{2} + 8y + 1 = 0$
- $y^{2} + 10y - 7 = 0$
- $y^{2} + y - 12 = 0$
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None of these
From the second equation, x = (-y - 3) / 2. Substitute this into the first equation: 2((-y-3)/2)^2 + 6((-y-3)/2) + 5y + 1 = 0. This simplifies to 2((y^2 + 6y + 9)/4) - 3y - 9 + 5y + 1 = 0, which is (y^2 + 6y + 9)/2 + 2y - 8 = 0. Multiplying by 2 gives y^2 + 6y + 9 + 4y - 16 = 0, resulting in y^2 + 10y - 7 = 0.
From the linear equation 2x + y + 3 = 0, express x in terms of y as x = (-y - 3) / 2. Substitute this expression for x into the quadratic equation 2x^2 + 6x + 5y + 1 = 0 to get 2[(-y - 3)/2]^2 + 6[(-y - 3)/2] + 5y + 1 = 0. Expanding the terms yields (y^2 + 6y + 9)/2 - 3y - 9 + 5y + 1 = 0. Multiplying the entire equation by 2 to clear the fraction gives y^2 + 6y + 9 - 6y - 18 + 10y + 2 = 0, which simplifies to the quadratic equation y^2 + 10y - 7 = 0.