Strengthen Your Systems of Equations Skills

The system of equations $x^2+y^2-<!--\; -\; -->x-<!--\; -\; -->2y=0$ and $x+2y=c$
$$\begin{gathered} (1.)x^2+y^2-<!--\; -\; -->x-<!--\; -\; -->2y=0 \\ (2.)x+2y=c \end{gathered}$$
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 $A{x}^2+Bx+C=0$.
The solution for c is $0$ or $5$.
For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?
(g) $x^2+y^2-<!--\; -\; -->x-<!--\; -\; -->2y=0,x+2y=c$, 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
$y=(c-<!--\; -\; -->x)/2$
$x^2+y^2-<!--\; -\; -->x-<!--\; -\; -->2y=0$
$x^2+((c-<!--\; -\; -->x)/2{)}^2-<!--\; -\; -->x-<!--\; -\; -->2\ast <!--\; \ast \; -->((c-<!--\; -\; -->x)/2)=0$
and i got lost...

Consider the given system of polynomial equations, where all the coefficients are in $\mathbb{C}$:
$\{\begin{array}{l}{y}^n=P(x)\\ Q(x,y)=0\end{array}$
I would like to establish that either this system has solutions $(x,y)\in <!--\; \in \; -->\mathbb{C}$ for all $x\in <!--\; \in \; -->\mathbb{C}$, or either has solutions for countably (or even better, finitely) many $x\in <!--\; \in \; -->\mathbb{C}$. I am not completely sure this is true but haven't found any counterexamples yet.
So far, I have found that if we have solutions for infinitely many $x$, then we have solutions for infinitely many $y$.

Consider the system
$x^{\prime }=y^3-<!--\; -\; -->4x$
$y^{\prime }=y^3-<!--\; -\; -->y-<!--\; -\; -->3x$
What is the systemic way to find all the equilibrium points for this system?

A solution for a system of differential equations?
$x_1^{\prime }=1/5x_1+4/5x_2$
$x_2^{\prime }=4/5x_1+1/5x_2$
$x_1(0)=1$, $x_2(0)=3$

Let $A$ and $B$ be two $2\times <!--\; \times \; -->2$-matrices, and considering the equation
$AX=B$
Where $X$ is an unknown $2\times <!--\; \times \; -->2$-matrix. What I want to do is explain why the equation is equivalent to solving two $2\times <!--\; \times \; -->2$-systems of equations simultaneously and determine how this generalizes to $n\times <!--\; \times \; -->n$.
And, on the corollary, let $A,B,C$ be three $2\times <!--\; \times \; -->2$-matrices and consider the equation
$AX+XB=C$
What I want to do here explain why you can not solve this as two $2\times <!--\; \times \; -->2$-systems of equations simultaneously

Solve the following initial value problem:
$$\begin{gathered} u_t+u_x=v, \\ {v}_t+v_x=-<!--\; -\; -->u, \\ u(0,x)=u_0(x), \\ v(0,x)=v_0(x). \end{gathered}$$

Find the values of $k$ for which the simultaneous equations do not have a unique solution for $x,y$ and $z$.
Also show that when $k=-<!--\; -\; -->2$ the equations are inconsistent
$kx+2y+z=0$
$3x+0y-<!--\; -\; -->2z=4$
$3x-<!--\; -\; -->6ky-<!--\; -\; -->4z=14$
Using a determinant and setting to zero, then solving the quadratic gives
$\left|\begin{array}{ccc}k& 2& 1\\ 3& 0& -<!--\; -\; -->2\\ 3& -<!--\; -\; -->6k& -<!--\; -\; -->4\end{array}\right|=0$
$k=-<!--\; -\; -->2,k=\frac{1}{2}$
So far so good, but when subbing $-<!--\; -\; -->2$ for $k$
$-<!--\; -\; -->2x+2y+z=0...eqn1$
$3x+0y-<!--\; -\; -->2z=4...eqn2$
$3x+12y-<!--\; -\; -->4z=14...eqn3$
subbing eqn $2$ from eqn $3$ gives
$12y-<!--\; -\; -->2z=10$
$\Rightarrow <!--\; \Rightarrow \; -->y=\frac{5+z}{6}$
Subbing for $y$ into eqn $1$
$-<!--\; -\; -->2x+\frac{5+z}{3}-<!--\; -\; -->z=0$
$\Rightarrow <!--\; \Rightarrow \; -->x=\frac{5-<!--\; -\; -->2z}{6}$
Subbing for $y$ and $x$ into eqn $3$
$\frac{5-<!--\; -\; -->2z}{2}+10-<!--\; -\; -->2z=14$
$\Rightarrow <!--\; \Rightarrow \; -->z=\frac{-<!--\; -\; -->1}{2},x=1andy=\frac{3}{4}$
These values for $x,y$ and $z$ seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?

Systems of equations question
$\begin{array}{rl}{a}^a\cdot <!--\; \cdot \; -->b^b\cdot <!--\; \cdot \; -->c^c\cdot <!--\; \cdot \; -->d^d& =\frac{1}{2}\\ a+b+c+d& =1\end{array}$
How can we find solutions for this system of equations given that $a,b,c,d>0$?

Solve a set of non linear Equations on Galois Field
$M_1=\frac{y_1-<!--\; -\; -->y_0}{x_1-<!--\; -\; -->x_0}$
$M_2=\frac{y_2-<!--\; -\; -->y_0}{x_2-<!--\; -\; -->x_0}$
$M_1,M_2,x_1,y_1,x_2,y_2,$, are known and they are chosen from a $GF(2^m).$. I want to find $x_0,y_0$I ll restate my question. Someone chose three distinct $x_0,x_1,x_2$, as well as $y_0,y_1,y_2$, then computed $M_1,M_2$, and finally revealed $M_1,M_2,x_1,y_1,x_2,y_2$, but not $x_0,y_0.$ to us.All the variables are chosen from a Galois Field.
I want to recover the unknown $x_0,y_0.$. Is it possible to accomplish that?
If a set of nonlinear equations have been constructed with the aforementioned procedure e.g.
$M_1=\frac{k_1-<!--\; -\; -->(y_0+(\frac{y_1-<!--\; -\; -->y_0}{x_1-<!--\; -\; -->x_0})(l_1-<!--\; -\; -->x_0))}{(l_1-<!--\; -\; -->x_0)(l_1-<!--\; -\; -->x_1)}$
$M_2=\frac{k_2-<!--\; -\; -->(y_0+(\frac{y_1-<!--\; -\; -->y_0}{x_1-<!--\; -\; -->x_0})(l_2-<!--\; -\; -->x_0))}{(l_2-<!--\; -\; -->x_0)(l_2-<!--\; -\; -->x_1)}$
$M_3=\frac{k_3-<!--\; -\; -->(y_0+(\frac{y_1-<!--\; -\; -->y_0}{x_1-<!--\; -\; -->x_0})(l_3-<!--\; -\; -->x_0))}{(l_3-<!--\; -\; -->x_0)(l_3-<!--\; -\; -->x_1)}$
$M_4=\frac{k_4-<!--\; -\; -->(y_0+(\frac{y_1-<!--\; -\; -->y_0}{x_1-<!--\; -\; -->x_0})(l_4-<!--\; -\; -->x_0))}{(l_4-<!--\; -\; -->x_0)(l_4-<!--\; -\; -->x_1)}$
where $x_0,y_0{x}_1,y_1$ are the unknown GF elements. Can I recover the unknown elements?
My question was if the fact that the set of equations is defined on a Galois Field imposes any difficulties to find its solution.
If not I suppose that the set can be solved. Is this true?

Show that $x^{\ast <!--\; \ast \; -->}=(1,2)$ is a fixed point of the system
$x_1^{\prime }=2+3x_1-2x_2-x_1^2+2x_1{x}_2-x_2^2$
$x_2^{\prime }=3+4x_1-3x_2-x_1^2+2x_1{x}_2-x_2^2$
Determine $W^s(x)$ and $W^u(x)$, the global stable and unstable manifolds of the fixed point $x=(1,2)$ and give a parametric representation of those manifolds.

Find the solutions of this linear equation:
$\begin{array}{}\text{(}\star <!--\; \star \; -->\text{)}& X+(X^{\mathrm{\top <!--\; \top \; -->}}C)C=D\end{array}$
where $C=(c_1,\dots <!--\; \dots \; -->,c_n{)}^{\mathrm{\top <!--\; \top \; -->}}\in <!--\; \in \; -->{\mathbb{R}}^n$ and $D=(d_1,\dots <!--\; \dots \; -->,d_n{)}^{\mathrm{\top <!--\; \top \; -->}}\in <!--\; \in \; -->{\mathbb{R}}^n$ are given and non-zero.
I noticed that the equation $(\star <!--\; \star \; -->)$ is equivalent to:
$\begin{array}{}\text{(}\star <!--\; \star \; -->\star <!--\; \star \; -->\text{)}& \stackrel{~<!--\; ~\; -->}{C}X=D\end{array}$
where $\stackrel{~<!--\; ~\; -->}{C}\in <!--\; \in \; -->{\mathrm{M}\mathrm{a}\mathrm{t}}_n(\mathbb{R})$ such that: ${\displaystyle {\stackrel{~<!--\; ~\; -->}{C}}_{i,j}=\{\begin{array}{ll}{c}_j{c}_i& \text{if }i\ne <!--\; \ne \; -->j\\ 1+c_i^2& \text{if }i=j\end{array}}$. To do so, I just said that the $i$-th component of the LHS of $(\star <!--\; \star \; -->)$ is ${\displaystyle (1+c_i^2)x_i+\underset{j\ne <!--\; \ne \; -->i}{\sum <!--\; \sum \; -->}{x}_j{c}_j{c}_i}$
Is there another method to prove that $(\star <!--\; \star \; -->)$ is equivalent to $(\star <!--\; \star \; -->\star <!--\; \star \; -->)$?

Suppose I were to be given 3 equations involving variables a, b, and c. The goal of the problem is to find the number of solutions (they do not necessarily have to be real) for the system of equations. Would it be more logical to use Cramer's Rule to solve the linear system or to use basic algebra? A roadmap of the following example problem or the like would be helpful.
A basic system taken from the book would be:
$\{\begin{array}{l}3a+4y-<!--\; -\; -->z=4\\ 3a+2b-<!--\; -\; -->11c=-<!--\; -\; -->13\\ a+2b+3c=7\end{array}$

Set of ODE: how can I solve it?
I want to solve this system, but I have never solved a system of ODE, can you help me?
$\{\begin{array}{l}\frac{dA}{dt}=-<!--\; -\; -->aA\\ \frac{dB}{dt}=aA-<!--\; -\; -->bB\\ \frac{dC}{dt}=bB\end{array}$
I have solved the first equation:
$A(t)=A_0{e}^{-<!--\; -\; -->at}$

I understood that two systems of linear equations are equivalent if one can be obtained by the linear combination of the other system and vice versa. But can those two systems of equations be equivalent even if the solution $x_i=0?(1\le <!--\; \le \; -->i\le <!--\; \le \; -->n)$.Example set:
$x_1-<!--\; -\; -->x_2=0;$
$2x_1+x_2=0$
and
$3x_1+x_2=0;$
$x_1+x_2=0$
Are these systems of equations equivalent?

what are some typical systems of equations generating from practical problems?
Some examples or research articles would be good.

$\{\begin{array}{c}x-<!--\; -\; -->\arctan <!--\; \; -->y=y-<!--\; -\; -->\arctan <!--\; \; -->x\\ x^2-<!--\; -\; -->xy+4y^2=4\end{array}$

Given is the ODE system
$y^{\prime }=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 0& k& 0\\ 0& -<!--\; -\; -->k& k\end{array}\right)y$
with boundary conditions $y(T)=0$.
Why does for $y$ hold that
${\int <!--\; \int \; -->}_t^T\frac{d}{ds}(e^{K(T-<!--\; -\; -->s)}y(s))ds={\int <!--\; \int \; -->}_t^T{e}^{K(T-<!--\; -\; -->s)}\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)ds$
and using the boundary conditions
$y(t)=-<!--\; -\; -->e^{-<!--\; -\; -->K(T-<!--\; -\; -->t)}{\int <!--\; \int \; -->}_t^T{e}^{K(T-<!--\; -\; -->s)}\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)ds$

How to solve these two second-order coupled PDE?
$$\begin{gathered} 0=Af(x,y)+B\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}f(x,y)}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}+C\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}f(x,y)}{\mathrm{\partial <!--\; \partial \; -->}{y}^2}+D\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}g(x,y)}{\mathrm{\partial <!--\; \partial \; -->}x\mathrm{\partial <!--\; \partial \; -->}y} \\ 0=Ag(x,y)+B\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}g(x,y)}{\mathrm{\partial <!--\; \partial \; -->}{y}^2}+C\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}g(x,y)}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}+D\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}f(x,y)}{\mathrm{\partial <!--\; \partial \; -->}x\mathrm{\partial <!--\; \partial \; -->}y} \end{gathered}$$

Consider the following simultaneous equations in $x$ and $y$:
$x+y+axy=a$
$x-<!--\; -\; -->2y-<!--\; -\; -->x{y}^2=0$
where $a$ is a real constant. Show that these equations admit real solutions in $x$ and $y$.

Find the solutions for the linear equation system:
$\{\begin{array}{l}x+5y+11z=-<!--\; -\; -->5\\ 2x+3y+8z=4\\ -<!--\; -\; -->x+2y+3z=-<!--\; -\; -->9\end{array}$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 2& 3& 8& 4\\ -<!--\; -\; -->1& 2& 3& -<!--\; -\; -->9\end{array}\right]$
$2f_3+f_2$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 0& 6& 14& -<!--\; -\; -->14\\ -<!--\; -\; -->1& 2& 3& -<!--\; -\; -->9\end{array}\right]$
$f_1+f_3$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 0& 6& 14& -<!--\; -\; -->14\\ 0& 7& 14& -<!--\; -\; -->14\end{array}\right]$
$-<!--\; -\; -->f_2+f_3$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 0& 6& 14& -<!--\; -\; -->14\\ 0& 1& 0& 0\end{array}\right]$
$f_2{f}_3$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 0& 1& 0& 0\\ 0& 6& 14& -<!--\; -\; -->14\end{array}\right]$
$-<!--\; -\; -->6f_2+f_3$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 0& 1& 0& 0\\ 0& 0& 14& -<!--\; -\; -->14\end{array}\right]$
$\frac{1}{14}{f}_3$
$\left[\begin{array}{cccc}1& 5& 11& -<!--\; -\; -->5\\ 0& 1& 0& 0\\ 0& 0& 1& -<!--\; -\; -->1\end{array}\right]$
$-<!--\; -\; -->5f_2+f_1$
$\left[\begin{array}{cccc}1& 0& 11& -<!--\; -\; -->5\\ 0& 1& 0& 0\\ 0& 0& 1& -<!--\; -\; -->1\end{array}\right]$
$-<!--\; -\; -->11f_3+f_1$
$\left[\begin{array}{cccc}1& 0& 0& 6\\ 0& 1& 0& 0\\ 0& 0& 1& -<!--\; -\; -->1\end{array}\right]$
$6f_3+f_1$
$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& -<!--\; -\; -->1\end{array}\right]$
So the solution would be $(0,0,-<!--\; -\; -->1)$ which is NOT right. Just looking at the first equation we'd have
$0+5(0)+11(-<!--\; -\; -->1)=-<!--\; -\; -->11\ne <!--\; \ne \; -->-<!--\; -\; -->5$
What did I do wrong?