Solving the system $2y^2-5x+13=0$ and $x+4y=-1$

Question

Let's say you have two simultaneous equations:

$$2y^2-5x+13=0$$ $$x+4y=-1$$

How do I proceed?

Should I multiply it by $x$ or $y$? Or both?

Answer 1

I would $$y=\frac{1}{4}(-x-1)$$ plug in to the first equation: $$2\left(\frac{1}{4}(-x-1)\right)^2-5x+13=0$$ At first simplify 442\left(\frac{1}4}(-x-1)\right)^2=$$\frac{1}{8}(x^2+2x+1)$$ the we multiply the whole equation by $8$ $$x^2+2x+1-40x+104=0$$ combining like terms

$$x^2-38x+105=0$$ and by the quadratic Formula we get $$x_{1,2}=19\pm\sqrt{361-105}$$

Answer 2

Eliminating $x$,

$$2y^2-5(-4y-1)+13=2y^2+20y+18=0$$

and by the quadratic formula,

$$y=-9\text{ or }y=-1.$$ $x$ follows.