The system of equations $x^2 + y^2 - x - 2y = 0$ and $x + 2y = c$ I have
$(1.) \quad x^2 + y^2 - x - 2y = 0 \\
(2.) \quad x + 2y = c$
Solving for $y$ in $(2.)$ gives
$y = (c - x) / 2$
Is there a way to simplify equation $(1.)$?
Because at the end I arrive at 
$c^2 - 2x - c = 0$
and can't proceed. Need to get typical form of square equation $Ax^2 + Bx + C = 0$.
The solution for $c$ is $0$ or $5$.
EDIT 1:
For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?
(g) $ \quad x^2 + y^2 - x - 2y = 0, \quad x + 2y = c$,  Ans. c = 0 or 5.
The process is to solve for $y$, then to substitute that into second equation.
From that we get A, B and C. Delta being $B^2 - 4AC$ we can get parameter.
So
$y = (c - x) / 2$
$x^2 + y^2 - x -2y = 0$
$x^2 + ((c-x)/2)^2 - x - 2 * ((c-x)/2) = 0$
and i got lost...

If you substitute correctly, you will get the equation $5\,{x}^{2}-2\,c\,x+{c}^{2}-4\,c=0$.
If you solve this, you will get $x = \frac{1}{5}(c \pm 2\sqrt{c(5-c)})$. The other value is given by $y=\frac{1}{2}(c-x)$, but you don't need this to answer your question.
If the two sets of solutions have the same values, then we must have $c(5-c)=0$, hence $c=0$ or $c=5$.
Addendum: To get the above equation, replace $y$ in $x^2+y^2-x-2y=0$ by $y = \frac{1}{2}(c-x)$ (from the second equation). That is, the equation
\begin{eqnarray}
x^2+\frac{1}{4}(c-x)^2 -x -(c-x) &=& \frac{1}{4}(4x^2+(c-x)^2 -4c) \\
&=& \frac{1}{4}(5 x^2-2 c x +c^2-4c)
\end{eqnarray}
Then the solutions are (ignoring the $\frac{1}{4}$, of course):
\begin{eqnarray}
x &=& \frac{1}{10}\left(2c \pm \sqrt{4 c^2-20(c^2-4c)} \right) \\
&=& \frac{1}{10}\left(2c \pm \sqrt{16 c(5-c)} \right) \\
&=& \frac{1}{5}\left(c \pm 2\sqrt{ c(5-c)} \right) 
\end{eqnarray}
