Strengthen Your Systems of Equations Skills

Solving a set of 3 Nonlinear Equations
In the following 3 equations:
$k_1{\cos }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})+k_2{\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})=c_1$
$2(k_2-<!--\; -\; -->k_1)\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})=c_2$
$k_1{\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})+k_2{\cos }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})=c_3$
$c_1$, $c_2$ and $c_3$ are given, and $k_1$, $k_2$ and $\mathrm{\theta <!--\; \theta \; -->}$ are the unknowns. What is the best way to solve for the unknowns?

I have an overdetemined system of linear equation and want to minimize overall error. Up to now, not a problem, I could use least squares. The problem is that I know that some equations in my system are more uncertain, while others are exact. Actually, I have a number of equations with different confidence levels ("low confidence","medium confidence", "high confidence" and so on). In a AX=B system, the solution should take this into account and keep unchanged the B coefficients of the "high confidence" equations, while the B coefficients of "low confidence" equations could be changed more drastically than the B coefficients of "mid confidence" equations.

How to solve the following pair of equation
1. $x^2+12x+y^2-<!--\; -\; -->4y=24$
2. $x^2-<!--\; -\; -->6x+y^2+8y=25$

find the solution to the problem $y^{\prime }=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)y,y(0)=\left(\begin{array}{c}4\\ 0\end{array}\right)$
I know i have to find the eigenvalues and eigenvectors of the matrix $A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$
there's only one eigenvalue which is 1 and only "one" eigenvector and we can choose $\left(\begin{array}{c}1\\ 0\end{array}\right)$
but now I dont know what to do.
what comes next?

Is this solution correct?
$\begin{array}{}\text{(A)}& \sqrt{x}+y=4\end{array}$
$$
Subtracting $A$ from $B$, we have
$x-<!--\; -\; -->y-<!--\; -\; -->\sqrt{x}+\sqrt{y}=2$
$(\sqrt{x}-<!--\; -\; -->\sqrt{y})(\sqrt{x}+\sqrt{y}-<!--\; -\; -->1)=2$
$(\sqrt{x}-<!--\; -\; -->\sqrt{y})(\sqrt{x}+\sqrt{y}-<!--\; -\; -->1)=(\sqrt{2})(\sqrt{2})$
Now we have,
$\begin{array}{}\text{(AA)}& \sqrt{x}-<!--\; -\; -->\sqrt{y}=\sqrt{2}\end{array}$
$\begin{array}{}\text{(BB)}& \sqrt{x}+\sqrt{y}-<!--\; -\; -->1=\sqrt{2}\end{array}$
This gives,
$\sqrt{x}=\sqrt{2}+\frac{1}{2}$
$\sqrt{y}=\frac{1}{2}$

Suppose we define a summation graph $G$ as follows:
Each vertex $v\in <!--\; \in \; -->G$ has a unique but unknown value ascribed to it. Each edge $e\in <!--\; \in \; -->G$ is labelled with the sum of the values of the two vertices it joins.
This construction corresponds to a system of equations where each equation is of the form $v_a+v_b=e_{ab}$ where $v_1$ and $v_2$ correspond to the unknown values of two vertices $ф$ and $b$, and $e_{ab}$ the value of the edge joining the two.
Now, if there is any odd cycle, it's known that there will always be a unique solution for just that cycle. But with an even cycle, there are either an infinite number of possible values, or an inherent contradiction in the values of the edges that make a solution impossible.
So my question is this: is it possible that there is some bipartite graph (which has no odd cycles, but can have a bunch of even cycles joined together) that has a configuration of values with a unique solution?

Is it wrong to write a linear system as below?
Suppose we have the following linear system
$\begin{array}{rl}2x_1+3x_2+4x_3& =1\\ 2x_2-<!--\; -\; -->3x_3& =6\\ 0& =0.\end{array}$
Is it wrong to write the above linear system by including all the zero coefficients as below
$\begin{array}{rl}2x_1+3x_2+4x_3& =1\\ 0x_1+2x_2-<!--\; -\; -->3x_3& =6\\ 0x_1+0x_2+0x_3& =0\end{array}$
instead?

On a system of PDE
Let $f_1(p_1,p_2)$ and $f_2(p_1,p_2)$ be two functions. The system of equations on ${\mathbb{R}}^2$ is just:
$p_1{\mathrm{\partial <!--\; \partial \; -->}}_{p_1}{f}_1+p_2{\mathrm{\partial <!--\; \partial \; -->}}_{p_1}{f}_2=0$
$p_1{\mathrm{\partial <!--\; \partial \; -->}}_{p_2}{f}_1+p_2{\mathrm{\partial <!--\; \partial \; -->}}_{p_2}{f}_2=0$
So nobody complain, I want the solutions to be at least of class $C^2$.

Solving a system of equations, why aren't the solutions preserved?
$6x^2+8xy+4y^2=3$
and
$2x^2+5xy+3y^2=2$
Multiply the second by $8$ to get: $16x^2+40xy+24y^2=16$
Multiply the first by $5$ to get: $30x^2+40xy+20y^2=15$
Subtract the two to get: $14x^2-4y^2=-1$
Later, the guy said to disregard his solution because the solutions to the first two equations do not satisfy the third equation. Why does this happen?

Technique to solve this equation of 2 unkowns
I was solving a problem of single phase eletrical circuits where I had to find the inductor $L$ and resistance $R$. I managed to get two equations containing the two unknowns.
$\frac{R}{R^2+(w\ast <!--\; \ast \; -->L{)}^2}=c_1$
and
$\frac{wL}{R^2+(w\ast <!--\; \ast \; -->L{)}^2}=c_2$
where $w,c_1\text{ and }{c}_2$ are known.How do I solve this?

Given that $a$, $b$ and $c$ are positive real numbers that satisfy
$b={\displaystyle \frac{64a}{a^2-<!--\; -\; -->64}}={\displaystyle \frac{81c}{2c^2-<!--\; -\; -->81}}=\sqrt{a^2+c^2}$, find $b$.

Solutions to
$x^3+y^3+z^3=x^2+y^2+z^2=x+y+z=0$
I need to prove that $xyz=1$. dealing trew this problem I get that $x+y=\frac{2}{3}$ and that results that $z=-<!--\; -\; -->\frac{2}{3}$. after all i got $xyz=-<!--\; -\; -->\frac{10}{27}$
When I was before dealing with this, I over and over get that $xyz=0$.

Determine the value of a if the system
$\begin{array}{rl}{x}_1+4x_2-3x_3+2x_4& =0\\ 2x_1+7x_2-4x_3+4x_4& =0\\ -x_1+a{x}_2+5x_3-2x_4& =0\\ 3x_1+10x_2-5x_3+(a²+4a+1)x_4& =0\end{array}$
has more then 1 solution

Solve for $X,Y,Z$ where :
$X^{²}=Y+a$
$Y^{²}=Z+a$
$Z^{²}=X+a$

$\frac{dy}{dt}=(x-<!--\; -\; -->y)y$
$\frac{dx}{dt}=-<!--\; -\; -->y$
How can I solve for $x(t),y(t)$? Is there a general method?

Solve the following system of equations
$\{\begin{array}{l}{x}_1^{{}^{\prime }}(t)=x_1(t)+3x_2(t)\\ x_2^{{}^{\prime }}(t)=3x_1(t)-<!--\; -\; -->2x_2(t)-<!--\; -\; -->x_3(t)\\ x_3^{{}^{\prime }}=-<!--\; -\; -->x_2(t)+x_3(t)\end{array}$
First, I create the column vectors $X$ and $X^{{}^{\prime }}$. Then the matrix
$A=\left[\begin{array}{ccc}1& 3& 0\\ 3& -<!--\; -\; -->2& -<!--\; -\; -->1\\ 0& -<!--\; -\; -->1& 1\end{array}\right]$
Now, I find the eigenvalues, $-<!--\; -\; -->4,3,1$ and their corresponding eigenvectors $(-<!--\; -\; -->3,5,1{)}^T(-<!--\; -\; -->3,-<!--\; -\; -->2,1{)}^T(1,0,3{)}^T$.
Then, what?

Finding $x^2+y^2+z^2$ given that $x+y+z=0$, $x^3+y^3+z^3=3$ and $x^4+y^4+z^4=15$

Find basis of solutions of this linear system
Supposed to find basis of the subspace of vector space ${\mathbb{R}}^3$ of solutions of this linear system of equations:
$y=\{\begin{array}{l}{x}_1+2x_2-<!--\; -\; -->x_3=0\\ 2x_1+7x_2-<!--\; -\; -->2x_3=0\\ -<!--\; -\; -->x_1+3x_2+x_3=0\end{array}$
I solve this system and I got: $x_1=x_3$ and $x_2=0$
$\stackrel{\to <!--\; \to \; -->}{x}=\left[\begin{array}{c}{x}_1\\ 0\\ x_1\end{array}\right]=x_1\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+0\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
Is the basis: $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$?

1) Do equations with $3$ variables require at least $3$ equation for us to solve without any dependent variable?2) if two equations of such kind equate to zero for eg: $a-<!--\; -\; -->b+2c=0$ and $3a+b+c=0$. Then using the matrice ,there is a method to solve them...explain that method.is there a proof or is it related to Gaussian method which is out of my scope?

Determining Whether a System of Linear Equations Is Inconsistent
1- Is getting a false equation during the normal process of solving a system of linear equations (e.g. substitution/elimination) is the only way we ensure this system has no solutions?
2- If no, what other ways available?