Solve the system of the following equations: $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$,$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$,and $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$.

The following system of equations $3x - 7y + 5z = 3$,$3x + y + 5z = 7$,and $2x + 3y + 5z = 5$ is:

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations
$2x - y + 2z = 2$
$x - 2y + \lambda z = -4$
$x + \lambda y + z = 4$
has no solution. Then the set $S$

Let $\alpha$ be a solution of $x^2+x+1=0$,and for some $a$ and $b$ in $\mathbb{R}$,$\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$. If $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$,then $m + n$ is equal to

The system of linear equations $x+y+z=6, x+2y+3z=10$ and $x+2y+az=b$ has no solutions when

For what values of $a$ will the system of equations $x+y+z=1$, $2x+3y+2z=2$, and $ax+ay+2az=4$ have a unique solution?