<s>[INST] Solving simultaneous equations involving a quadratic I have the question 
Solve the simultaneous equation  pair
$$x^2 + y^2 = 25\tag1$$
$$2x - y = 5\tag2$$

I have found the value of $y$ from the second equation which is $2x-5$ and substituted this into the first equations $y$ value.
I get $x^2 + (2x -5)^2 = 25$
When I expand the brackets I get the equation 
$$x^2+4x^2-20 =0\tag3$$
However, when I checked the solutions the equation should simplify to $x^2 - 4x = 0$ and I do not understand how this is achieved.
 [/INST] $$
\begin{cases}
\text{x}^2+\text{y}^2=25\\
2\text{x}-\text{y}=5
\end{cases}\space\Longleftrightarrow\space
\begin{cases}
\text{x}^2+\left(2\text{x}-5\right)^2=25\\
\text{y}=2\text{x}-5
\end{cases}
$$
Solving:
$$\text{x}^2+\left(2\text{x}-5\right)^2=25$$
Gives us $x=0$ or $x=4$
</s>