Strengthen Your Systems of Equations Skills

Explain method to solve two equations in two unknowns where one of the variables has a square term
Given:
$-<!--\; -\; -->\frac{96}{x^{2}y}+1+y=0$
$-<!--\; -\; -->\frac{96}{x{y}^2}+2+x=0$
Solve for $x$ and $y$
How should I find $x$ and $y$?

My problem:
$x_1+3x_2-<!--\; -\; -->x_3+x_4+2x_5=2,$
$2x_1-<!--\; -\; -->x_2+x_3-<!--\; -\; -->x_4+x_5=-<!--\; -\; -->2,$
$4x_1+2x_2+2x_3-<!--\; -\; -->x_4-<!--\; -\; -->x_5=0.$
I thought that I could multiply 1 by equation 2, and then add that to equation 1, which would cancel both the $x_3$ and the $x_4$ factors for that line. Is that permissible?

Solve $a=x^n,b=(x+1{)}^n$ for $x,n$

Consider the system of linear equations:
$\begin{array}{r}\{\begin{array}{l}x+ay=1\\ bx+5y=2,\end{array}\end{array}$
where $a$ and $b$ are parameters.
(a) Determine the conditions on $a$ and $b$ to get a unique solution.
(b) Determine the conditions on $a$ and $b$ to get infinitely many solutions.
(c) Determine the conditions on $a$ and $b$ such that the system has no solutions.

system of three equations
$-<!--\; -\; -->sx+sy=0$
$rx-<!--\; -\; -->y-<!--\; -\; -->xz=0$
$xy-<!--\; -\; -->bz=0$
where $s,r,$ and b are given parameters and $x,y,z$ are my three variables. Trouble is the "$xz$" and the "$xy$" terms since they are coupled between two different variables. Ignoring those two terms, my matrix equation would read:
$\begin{array}{r}\left[\begin{array}{ccc}-<!--\; -\; -->s& s& 0\\ r& -<!--\; -\; -->1& ?\\ 0& ?& -<!--\; -\; -->b\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\end{array}$
Where "?" represents the coupled terms

$x\equiv <!--\; \equiv \; -->11(\mathrm{mod}84)$
$x\equiv <!--\; \equiv \; -->23(\mathrm{mod}36)$
I have the bulk of the work done for this;
$x=11+84j$
$x=23+36k$
$\Rightarrow <!--\; \Rightarrow \; -->11+84j\equiv <!--\; \equiv \; -->23(\mathrm{mod}36)$
$\Rightarrow <!--\; \Rightarrow \; -->84j\equiv <!--\; \equiv \; -->12(\mathrm{mod}36)$
$\Rightarrow <!--\; \Rightarrow \; -->12j\equiv <!--\; \equiv \; -->12(\mathrm{mod}36)$
$\Rightarrow <!--\; \Rightarrow \; -->j\equiv <!--\; \equiv \; -->1(\mathrm{mod}36)$
$\Rightarrow <!--\; \Rightarrow \; -->j=1+36n$
Thus this system is true for any $x$ of the form
$x=11+84(1+36n)=95+3024n$

If $2^x=3^y=6^{-<!--\; -\; -->z}$ and $x,y,z\ne <!--\; \ne \; -->0$ then prove that:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
I have tried starting with taking logartithms, but that gives just some more equations.
Any specific way to solve these type of problems?

Simultaneous Equations
$\begin{array}{}\text{(1)}& 19s+12t=82\end{array}$
$\begin{array}{}\text{(2)}& 5s+4t=30\end{array}$
The point of the exercise is to find the values of t and s.
-So what I've done is label the first equation 1 and the second equation 2, as normal.
-Then I multiplied equation 2 by 3 to make the value of t the same on both equations and labelled this, new equation, equation 3:
$\begin{array}{}\text{(3)}& 15s+12t=90\end{array}$
-Then I took equation 3 from equation 2 to breaking it down to find the value of s.
(3) - (2)
$4s=8$
$s=2$
-So, at this point I have worked out that s=2.
-The problem, for me, is when I substitute the value of s back into equation 1. the value of t is a recurring number of 6.
-Have i done something wrong?

System of quadratic Diophantine equations
$\frac{4}{3}{x}^2+\frac{4}{3}x+1=y^2$
$\frac{8}{3}{x}^2+\frac{8}{3}x+1=z^2$

Condition number of system of non-linear equations. The system has only two unknowns but 6 equations (thus over-determined). Solving the system of equations are not a problem. However, I need an indication of how well-conditioned the system of equations is. I know the condition number is typically used to do this. Any advice on exactly how this procedure works will be appreciated.

Consider $a,b,c\in <!--\; \in \; -->\mathbb{R}$ such that $|a{x}^2+bx+c|\le <!--\; \le \; -->1\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->[0,1]$. Prove that $|a|\le <!--\; \le \; -->8,|b|\le <!--\; \le \; -->8$ and $|c|\le <!--\; \le \; -->1$.

Solving the system
$\{\begin{array}{l}18x{y}^2+x^3=12\\ 27x^{2}y+54y^3=38\end{array}$
and I wonder whether there is some slick method to find the only real root $(x,y)=(2,1/3)$ without relying on Cardano's formula, ideally giving some intuition. This closely reassembles some kind of elliptic curves, so I'm tagging it as such, please remove if wrong. Number theoretic approaches are welcome.

$(a_{11}x+a_{12}y)\mathrm{mod}8=b_1$
$(a_{21}x+a_{22}y)\mathrm{mod}8=b_2$
In the case of simple equations without modulo, the solution of $Ax=b$ is $x=A^{-<!--\; -\; -->1}b$, but how to handle this modulo?

Solve
$\{\begin{array}{c}5a^2-<!--\; -\; -->4ab-<!--\; -\; -->b^2+9=0\\ -<!--\; -\; -->21a^2-<!--\; -\; -->10ab+40a-<!--\; -\; -->b^2+8b-<!--\; -\; -->12=0.\end{array}$

How to solve the following pair of non-linear equations
$$\begin{gathered} x^3+y^3=152 \\ {x}^{2}y+x{y}^2=120 \end{gathered}$$

Finding solutions of a system of linear equations
$\begin{array}{rl}2x_1-<!--\; -\; -->4x_2+8x_4& =10\\ 3x_1+2x_2& =3\\ 3x_1+2x_3+6x_4& =11\\ 8x_1+6x_2+2x_3+14x_4& =24\end{array}$
Find the solution set $W$. Is $W$ a vector space?
I know I could express one variable from one of the equations and substitute into the remaining ones. For example, from the second equation I get $x_1=1-<!--\; -\; -->\frac{2}{3}{x}_2$.
After substituting this into the remaining equations I have
$\begin{array}{rl}-<!--\; -\; -->\frac{16}{3}{x}_2+8x_4& =8\\ -<!--\; -\; -->2x_2+2x_3+6x_4& =7\\ \frac{2}{3}{x}_2+2x_3+14x_4& =16\end{array}$
I could continue by expressing another variable and plug into other equations. But this approach seems to be rather cumbersome and if I have more than three equations I am prone to make mistakes. Is there a more elegant way to do this?

Solve non-linear ode system as a function of $t$.
$\{\begin{array}{l}\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=y\\ \stackrel{\dot{}<!--\; \dot{}\; -->}{y}=-<!--\; -\; -->x+x^2=x(x-<!--\; -\; -->1)\end{array}$
To solve it as a function $x(y)$ or $y(x)$ is trivial, but I need the solution as a function of time: $(x(t),y(t))$.
The system has a first integral: $H(x,y)=\frac{1}{2}(x^2+y^2)-<!--\; -\; -->\frac{1}{3}{x}^3$. In particular, if no general solution is possible, I'm searching for a solution for $H=\frac{1}{6}$.

When solving systems of equations for first-year physics, I am told that to be able to find the solution for a system of n unknowns (if one exists), I need at least n equations.
In this system, I am looking for $a$. The known quantities are ${\mathrm{\theta <!--\; \theta \; -->}}_0$ and ${\mathrm{\theta <!--\; \theta \; -->}}_1$.
$T_0\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_0=mg$
$T_0\sin <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_0=\frac{m{v}^2}{\mathrm{\ell <!--\; \ell \; -->}\sin <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_0}$
$T_1\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_1-<!--\; -\; -->mg=ma$
$T_1\sin <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_1=\frac{m{v}^2}{\mathrm{\ell <!--\; \ell \; -->}\sin <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_1}$
In this system of equations, I had 5 unknown quantities: $T_0$, $T_1$, $m$, $\mathrm{\ell <!--\; \ell \; -->}$, $a$. But I was able to successfully solve for $a$ with four equations. Where is the logical error in my understanding?

Solving system of multivariable 2nd-degree polynomials
$\begin{array}{c}{x}^2+3xy-<!--\; -\; -->9=0(1)\\ 2y^2-<!--\; -\; -->4xy+5=0(2)\end{array}$
where $(x,y)\in <!--\; \in \; -->{\mathbb{C}}^2$.
More generally, how would you solve any set of equations of the form:
$\begin{array}{c}{ax}^2+bxy+c=0\\ d{y}^2+exy+f=0\end{array}$
where $a,b,c,d,e,f\in <!--\; \in \; -->\mathbb{Q}$ and $(x,y)\in <!--\; \in \; -->{\mathbb{C}}^2$.

For $x^{\prime }=y$, and $y^{\prime }=-<!--\; -\; -->x-<!--\; -\; -->y$, Find all equilibrium points and decide whether they are stable, asymptotically stable, or unstable.
I found that the equilibrium points are $(0,0)$. Then I try to find the eigenvalues. The characteristic equation I found is $\mathrm{\lambda <!--\; \lambda \; -->}+{\mathrm{\lambda <!--\; \lambda \; -->}}^2+1=0$. I realize that they are complex eigenvalues...The real part of this complex eigenvalues are negative. Does this mean this is stable ? Is this also asymptotically stable? What are the differences between asymptotically stable and stable?