simultaneous equation help

Question

Can someone help me with to solve this system of equations ?

$$ \left\{ \begin{array}{c} y=x+1 \\ y^2+2x^2=2 \end{array} \right. $$

Answer 1

$$\begin{cases} y=x+1 \\ { y }^{ 2 }+2{ x }^{ 2 }=2 \end{cases}\Rightarrow \begin{cases} y=x+1 \\ { \left( x+1 \right) }^{ 2 }+2{ x }^{ 2 }=2 \end{cases}\Rightarrow 3{ x }^{ 2 }+2x-1=0\quad \Rightarrow x=\frac { -1\pm 2 }{ 3 } ,y=1+\frac { -1\pm 2 }{ 3 } \\ $$

Answer 2

We know from equation one that $y = x+1$. Plugging this value into equation two we get \begin{align*} y^2+2x^2 &= (x+1)^2+2x^2 \\ &=x^2+2x+1+2x^2\\ &=3x^2+2x+1 \\ \end{align*} We know that $3x^2+2x+1 = 2$ from equation two. Simplifying we get $$3x^2+2x-1 = 0$$$$(x+1)(3x-1) = 0 \tag{factoring}$$ $$x = -1, \frac{1}{3}$$Plugging in these values of $x$ into $y = x+1$ we get that $y=0, \dfrac43$ So the solutions are $$\boxed{\left(-1,0\right)\ \& \left(\frac13, \frac43\right)}$$