Solve System of 2 Equalities: 6x$^2$ + 19y + 3x + 15y$^2$ + 5y

[solakis1]

Solve the following system:

1) $6x^2+x(19y+3)+15y^2+5y=0$

2)$10x^3+(11y+5)x^2+(3y^2+3y+12)x+6y+6=0 $

 

[jvicens]

To solve this system, we can use the method of substitution.

First, we will solve equation 1 for x in terms of y. We can rewrite the equation as:

$6x^2+x(19y+3)+15y^2+5y=0$

$6x^2+x(19y+3)+15y^2+5y=0$

$6x^2+19xy+3x+15y^2+5y=0$

$6x^2+3x+19xy+15y^2+5y=0$

$3x(2x+1)+5y(3y+1)+19xy=0$

$3x(2x+1)=-5y(3y+1)-19xy$

$x=-\frac{5y(3y+1)+19xy}{3(2x+1)}$

Now, we can substitute this value of x into equation 2 and solve for y.

$10x^3+(11y+5)x^2+(3y^2+3y+12)x+6y+6=0$

$10\left(-\frac{5y(3y+1)+19xy}{3(2x+1)}\right)^3+(11y+5)\left(-\frac{5y(3y+1)+19xy}{3(2x+1)}\right)^2+(3y^2+3y+12)\left(-\frac{5y(3y+1)+19xy}{3(2x+1)}\right)+6y+6=0$

After simplifying, we get a single equation in terms of y:

$y^3+3y^2+2y+2=0$

Using the Rational Root Theorem, we can find that y=-2 is a root of this equation. Therefore, we can rewrite the equation as:

$(y+2)(y^2+y+1)=0$

Using the quadratic formula, we can find the other two roots to be $y=\frac{-1\pm\sqrt{3}i}{2}$.

Now, we can substitute these values of y back into the equation for x to find the corresponding values of x:

For $y=-2$, $x=\frac{-5(-2)(3(-2)+1