Strengthen Your Systems of Equations Skills

find any solution of this three equations
$2x+6y-<!--\; -\; -->z=10$
$3x+2y+2z=1$
$5x+4y+3z=4$

How to solve these simultaneous equations: $y=10x-<!--\; -\; -->3$ and $y=x^2-<!--\; -\; -->3x$?
What I got
$x=\frac{13\pm <!--\; \pm \; -->\sqrt{157}}{2}$
But I am not sure if this is right.
How would you go around solving this?

$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=x-<!--\; -\; -->y-<!--\; -\; -->x(x^2+y^2)$
$\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=y+x-<!--\; -\; -->y(x^2+y^2)$
1. Determine the equilibrium points
2. Show that this system has a periodic solution. Use the following substitution $x=r\cos <!--\; \; -->(\mathrm{\varphi <!--\; \varphi \; -->}),y=r\sin <!--\; \; -->(\mathrm{\varphi <!--\; \varphi \; -->})$.
3. Give the explicit solution for this system. Also determine the period of this solution.

For how many values of $\mathrm{\alpha <!--\; \alpha \; -->}$ does this system of equations have infinitely many solutions ?
$\left(\begin{array}{ccc}2& 1& -<!--\; -\; -->4\\ 4& 3& -<!--\; -\; -->12\\ 1& 2& -<!--\; -\; -->8\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}\mathrm{\alpha <!--\; \alpha \; -->}\\ 5\\ 7\end{array}\right)$

Determine the value of $b$ for which the system
$\begin{array}{rl}{x}_1+4x_2-3x_3+2x_4& =2\\ 2x_1+7x_2-4x_3+4x_4& =3\\ -x_1-5x_2+5x_3-2x_4& =b\\ 3x_1+10x_2-5x_3+6x_4& =4\end{array}$
is soluble, and determine the solution set.

What are the steps required to solve this system of equations?
The system that I need to solve is
$\begin{array}{rlr}& i_1+i_2+i_3& =0\\ & i_1+i_4+i_6& =0\\ & i_5+i_6& =i_2\\ & -<!--\; -\; -->v_{s1}+i_1{r}_1+i_3{r}_3-<!--\; -\; -->i_4{r}_4+v_{s4}& =0\\ & -<!--\; -\; -->i_2{r}_2+v_{s2}-<!--\; -\; -->i_5{r}_5-<!--\; -\; -->i_3{r}_3& =0\\ & -<!--\; -\; -->v_{s4}+i_4{r}_4+i_5{r}_5-<!--\; -\; -->i_6{r}_6-<!--\; -\; -->v_{s6}& =0\end{array}$
for the variables $i_k$.

Solve this system by rewriting in row-echelon form $x+y+z=6$, $2x-<!--\; -\; -->y+z=3$, $3x-<!--\; -\; -->z=0$
Here is what I did (which I didn't complete because I had no idea if I was doing the right thing):
Add (-2) times the first equation to the second equation to produce a new second equation
$\{\begin{array}{rl}x+y+z& =6\\ -<!--\; -\; -->3y-<!--\; -\; -->z& =9\\ 3x-<!--\; -\; -->z& =0\end{array}$
And next add (-3) times first equation to third equation to produce a new third equation:
$\{\begin{array}{rl}x+y+z& =6\\ -<!--\; -\; -->3y-<!--\; -\; -->z& =9\\ -<!--\; -\; -->3y-<!--\; -\; -->4z& =-<!--\; -\; -->18\end{array}$
And next add (-1) times second to third equation to produce a new third equation:
$\{\begin{array}{rl}x+y+z& =6\\ -<!--\; -\; -->3y-<!--\; -\; -->z& =9\\ -<!--\; -\; -->3z& =-<!--\; -\; -->27\end{array}$
Any help as to my journey in Linear Algebra and how to solve this is greatly appreciated.

system of 2 linear differential equations with variable coefficients
$$\begin{gathered} f^{″}(x)+a{f}^{\prime }(x)+(1+x)g^{\prime }(x)+bg(x)=0 \\ {g}^{″}(x)+a{g}^{\prime }(x)+(1-<!--\; -\; -->x)f^{\prime }(x)-<!--\; -\; -->bf(x)=0 \end{gathered}$$
where $a,b$ are some positive constants.

If $x$, $y$ and $z$ are positive integer and $3x=4y=7z$, then What is the least possible value of $x+y+z$?

This system of equations
$\begin{array}{rl}xy+yz+zx& =3\\ \\ x^4+y^4+z^4& =3\end{array}$
How to solve this system of equations?

Given two integer variables $x$ and $y$. We are given that each integer variable $x$ and $y$ can't be greater than a given integer $z$.
The problem: We are given the proportions $a$ and $b$ such that $a+b=1$, $a=\frac{x}{z}$, and $b=\frac{y}{z}$. Is it possible to reverse solve $x$ and $y$?

How to determine the rank of a linear system of equations involving parameters?
For instance, the following matrix:
$\left(\begin{array}{ccc}k& 1& 1\\ 1& k& 1\\ 1& 1& k\end{array}\right)$

wo identitical pendula each of length $l$ and with bobs of mass $m$ are free to oscillate in the same plane. The bobs are joined by a spring with spring constant $k$, by looking for solutions where $x$ and $y$ vary harmonically at the same angular frequency $\mathrm{\omega <!--\; \omega \; -->}$, form a simultaneous equation for the amplitudes of oscillation $x_0$ and $y_0$.
Considering the forces acting on each pendulum we can derive the following coupled-differential equations:
$\begin{array}{}\text{(1)}& m\stackrel{¨<!--\; ¨\; -->}{x}& =k(y-<!--\; -\; -->x)-<!--\; -\; -->\frac{mgx}{\mathrm{\ell <!--\; \ell \; -->}}\text{(2)}& m\stackrel{¨<!--\; ¨\; -->}{y}& =-<!--\; -\; -->k(y-<!--\; -\; -->x)-<!--\; -\; -->\frac{mgy}{\mathrm{\ell <!--\; \ell \; -->}}\end{array}$
Where $x$ and $y$ are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency $\mathrm{\omega <!--\; \omega \; -->}$ then we can write $\mathrm{\exists <!--\; \exists \; -->}\mathrm{\omega <!--\; \omega \; -->},{\mathrm{\varphi <!--\; \varphi \; -->}}_1,{\mathrm{\varphi <!--\; \varphi \; -->}}_2\in <!--\; \in \; -->\mathbb{R}$:
$\begin{array}{rl}x(t)& =x_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_1)\\ y(t)& =y_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_2)\end{array}$
Substituting these solutions back into $(1)$ and $(2)$ we get:
$\begin{array}{rl}-<!--\; -\; -->m{\mathrm{\omega <!--\; \omega \; -->}}^2{x}_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_1)& =k(x_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_1)-<!--\; -\; -->y_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_2))-<!--\; -\; -->\frac{mg}{\mathrm{\ell <!--\; \ell \; -->}}\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_1)\\ -<!--\; -\; -->m{\mathrm{\omega <!--\; \omega \; -->}}^2{y}_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_2)& =-<!--\; -\; -->k(x_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_1)-<!--\; -\; -->y_0\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_2))-<!--\; -\; -->\frac{mg}{\mathrm{\ell <!--\; \ell \; -->}}\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+{\mathrm{\varphi <!--\; \varphi \; -->}}_2)\end{array}$
However, without assuming that ${\mathrm{\varphi <!--\; \varphi \; -->}}_1={\mathrm{\varphi <!--\; \varphi \; -->}}_2$, in which case everything factors out nicely to leave a simultaneous equation in $x_0$ and $y_0$, I cannot see a way of making it linear in $x_0$ and $y_0$. So am I expected to use this assumption or is there a mathematical way of simplifying it?
If it is the former, then what would the physical justification for this assumption be?

Help me with this system of equations
$a+b=3-<!--\; -\; -->c$
$\frac{1}{a}+\frac{1}{b}=\frac{5}{12}-<!--\; -\; -->\frac{1}{c}$
$a^3+b^3=45-<!--\; -\; -->c^3$

Using Lyapunov function, prove that a critical poin (0,0) is asymptotically stable.
Let a linear system
$$\begin{gathered} x^{\prime }=-<!--\; -\; -->2x+x{y}^2 \\ {y}^{\prime }=-<!--\; -\; -->x^3-<!--\; -\; -->y \end{gathered}$$

System of equations with multiplication
I want to find all $x,y,z\in <!--\; \in \; -->\mathbb{R}$ such that
$(x+1)yz=12,(y+1)zx=4,(z+1)xy=4$.
I can multiply all three equations to get
$(x+1)(y+1)(z+1)x^2{y}^2{z}^2=192$
I can divide the first equation by the second to get
${\displaystyle \frac{(x+1)y}{(y+1)x}}=4$, which simplifies to $3xy+4x-<!--\; -\; -->y=0$
None of these seems to help. How can I solve the equations?

Three-variable system of simultaneous equations
$x+y+z=4$
$x^2+y^2+z^2=4$
$x^3+y^3+z^3=4$
Any ideas on how to solve for $(x,y,z)$ satisfying the three simultaneous equations, provided there can be both real and complex solutions?

Solve the system
$\{\begin{array}{cc}{x}_1^{\prime }(t)=3x_1(t)-<!--\; -\; -->2x_2(t)+e^{2t},x_1(0)=a& \\ x_2^{\prime }(t)=4x_1(t)-<!--\; -\; -->3x_2(t),x_2(0)=b& \end{array}$
by using the method of diagonalization. Substitution $x=Tz$

How to prove the existence of solution of a non linear system of equations
Writing the ortogonality condition for any element of $O(n)$, I've arrived to:
If we take $n=2$, we know that $\mathrm{\Lambda <!--\; \Lambda \; -->}{\mathrm{\Lambda <!--\; \Lambda \; -->}}^T=\mathbb{I}$, so:
$\left(\begin{array}{cc}x& y\\ z& t\end{array}\right)\left(\begin{array}{cc}x& z\\ y& t\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$
This yields:
$\{\begin{array}{l}{x}^2+y^2=1\\ xz+yt=0\\ z^2+t^2=1\end{array}$
Visually, these equations mean that we can find two orthonormal vectors.
We could generalise our reasoning to arbitrary dimension easily.
How could I prove rigorously, without plugging in any numbers nor functions, that this system of equations has solution (infinite, in fact)?
Is there any method to prove the existence of solution of non non linear equations (or systems)?

Solving systems of equations in 3 variables
$x+y+z=0$
$xy+yz+xz=c-<!--\; -\; -->c^2$
$xyz=3c^2$
where, $c$ is some fixed non-zero constant.