<s>[INST] Solve the system of equations: $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$ Given system of equations:
$$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$$
$$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$$
$$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$$
I'll use matrices to solve these. We can write the system as:
$$AX=B$$
$$A^{-1}AX=A^{-1}B$$
$$X=A^{-1}B$$ where $X$ is our variable matrix. For simplicity, let $\frac{1}{x}=\alpha, \frac{1}{y}=\beta,\frac{1}{z}=\gamma $
Thus we have to solve:
$$2\alpha+3\beta+10\gamma=4$$
$$4\alpha-6\beta+5\gamma=1$$
$$6\alpha+9\beta-20\gamma=2$$
$A$ is simply the matrix formed using the coefficients of the variables:
$$A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}$$
$X$ is the variable matrix
$$X=\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}$$
$B$ is the matrix formed by right hand side values of the respective equations:
$$B=\begin{bmatrix}4\\1\\2\end{bmatrix}$$
Thus, our system of equations becomes:
$$\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}$$
Now, we just need to find $A^{-1}.$ Finding cofactors for A
Cofactor $ A=\begin{bmatrix}75&110&72\\150&-100&0\\75&30&-24\end{bmatrix}$
$A^{-1}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}$
$\alpha=\frac{1}{2},\beta=\frac{1}{3},\gamma=\frac{1}{5}$
$\Rightarrow x=2,y=3,z=5$
My problem- This method seems unnecessarily lengthy. What better ways are there to solve such systems of equations?
 [/INST] Following the comments, subtracting thrice the first equation from the third gives $-50\gamma=-10$ or $\gamma=\frac15$. Subtracting twice the first equation from the second and substituting (and negating) gives
$12\beta=4$ or $\beta=\frac13$. Substituting the known values into the first equation finally gives $\alpha=\frac12$, so $(x,y,z)=(2,3,5)$.
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