For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$. Find the Maximum and Minimum Value of the expression $A=3x-2y$. For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$.Find the Maximum and Minimum Value of the expression $A=3x-2y$.

Using secondary school  "tools", you can replace
$$
x = \left( {A + 2y} \right)/3
$$
into the quadric equation, to get, after some simple simplifications
$$
25y^{\,2}  - \left( {8A + 60} \right)y + \left( {8A + 60} \right)^{\,2}  = 0
$$
whose solution is
$$
\eqalign{
  & y = {{2\left( {2A + 15} \right) \pm \sqrt {\left( {8A + 60} \right)^{\,2}  - \left( {10A + 30} \right)^{\,2} } } \over {25}} =   \cr 
  &  = {{2\left( {2A + 15} \right) \pm \sqrt {\left( {18A + 90} \right)\left( { - 2A + 30} \right)} } \over {25}} =   \cr 
  &  = {{2\left( {2A + 15} \right) \pm 3\sqrt {\left( {2A + 10} \right)\left( { - 2A + 30} \right)} } \over {25}} \cr} 
$$
For the solutions to be real, we must have
$$
0 \le \left( {2A + 10} \right)\left( { - 2A + 30} \right)\quad  \Rightarrow \quad  - 5 \le A \le 15
$$
corresponding to the $(x,y)$ values
$$
\left\{ \matrix{
  A =  - 5 \hfill \cr 
  y = 10/25 = 2/5 \hfill \cr 
  x = \left( { - 5 + 4/5} \right)/3 =  - 7/5 \hfill \cr}  \right.\quad \left\{ \matrix{
  A = 15 \hfill \cr 
  y = 90/25 = 18/5 \hfill \cr 
  x = \left( {15 + 36/5} \right)/3 = 37/5 \hfill \cr}  \right.
$$
