solve this simultaneous equation: $2x + y = 5$ and $4x^2 + y^2 = 17$ I already squared both sides of the first equation: $4x^2+4xy+y^2=25$
Then minused this from the 2nd equation to get $4xy=8$
Simplified to $xy=2$
But i don't know what to do from there on out
 A: Now, $2x+y=5$ and $2x\cdot y=4,$ which by Viete's theorem gives:
$$(2x,y)=(4,1)$$ or $$(2x,y)=(1,4)$$ and we got an answer:
$$\left\{(2,1),\left(\frac{1}{2},4\right)\right\}$$
A: So, you know $xy=2$. Notice this means $x=2/y$. (This is okay, because if $xy=2$, then $y \ne 0$.) In the first equation, then, this implies
$$2x+y = 5 \implies 2 \cdot \frac 2 y + y = \frac 4 y + y = 5$$
Multiply throughout by $y$:
$$4 + y^2 = 5y$$
You can solve this quadratic for $y$, and then you could substitute the known value to solve for $x$. (Be sure to note that you can introduce an extraneous solution in $y$ by doing this, so be sure to double-check your solution.)
A: You could instead, as suggested in sirous's question comment, from your first equation, determine
$$2x + y = 5 \implies y = 5 - 2x \tag{1}\label{eq1A}$$
Substitute this into your second equation to get
$$\begin{equation}\begin{aligned}
4x^2 + (5 - 2x)^2 & = 17 \\
4x^2 + 25 - 20x + 4x^2 & = 17 \\
8x^2 - 20x + 8 & = 0 \\
2x^2 - 5x + 2 & = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Now you can use the quadratic formula to solve for the values of $x$, and then determine $y$ from \eqref{eq1A}. Also, double-check to ensure you don't have any extraneous solutions. I'll leave it to you to do those final steps.
A: Hint: put $y=5-2x$  in the second equation and solve the quadratic equation you get.
A: Another neat solution is using the fact that
$$
\begin{aligned}
(2x+y)^{2}+(2x-y)^{2}&=2(4x^{2}+y^{2})\\
2x-y&=\pm 3
\end{aligned}
$$
From here it is straightforward
