solve this simultaneous equation: $2x + y = 5$ and $4x^2 + y^2 = 17$

Question

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

Answer 1

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.

Answer 2

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.)

Answer 3

Hint: put $y=5-2x$ in the second equation and solve the quadratic equation you get.

Answer 4

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: $$\left\{(2,1),\left(\frac{1}{2},4\right)\right\}$$

Answer 5

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