The solutions of x and y are (1.105, -1.07) and (-1.547, 0.698)
Given equations;
2x + 3y + 1 = 0 -----(1)
y² + 6xy + 6 = 0 ---- (2)
The solutions of x and y are obtained as follows;
From equation (1) above, make x the subject of the formula;
[tex]2x +3y + 1 = 0\\\\2x = -3y-1\\\\x = \frac{-3y-1}{2} \\\\Substitute \ the \ value \ of \ x\ into \ equation \ (2)\\\\y^2 + 6xy + 6 = 0\\\\y^2 + 6( \frac{-3y-1}{2})y + 6 = 0\\\\y^2 -9y^2-3y+ 6 = 0\\\\-8y^2 - 3y + 6 = 0\\\\multiple \ through \ by \ - 1\\\\8y^2 +3y - 6 = 0\\\\solve \ the above \ quadratic \ equation \ using \ formula \ method\\\\a = 8, \ b = 3, \ c = - 6\\\\y = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\y = \frac{-3 \ \ +/- \ \ \sqrt{(3)^2 - 4(8\times -6)} }{2\times 8}\\\\[/tex]
[tex]y = \frac{-3 \ \ +/- \ \ \sqrt{201} }{16}\\\\y = \frac{-3 \ \ +/- \ 14.177}{16} \\\\y = \frac{-3 -14.177}{16} \ \ \ or \ \ \ \frac{-3 +14.177}{16} \\\\y = - 1.07 \ \ or \ \ \ 0.698[/tex]
Now solve for the values of x
[tex]x = \frac{-3y - 1}{2} \\\\when \ y = -1.07, \ \ \ x = \frac{-3(-1.07) -1}{2} = 1.105\\\\when \ y = 0.698, \ x = \frac{-3(0.698) -1}{2} = -1.547[/tex]
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