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(C) Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$. The given system of equations becomes: $2p+3q+10r=4$ $4p-6q+5r=1$ $6p+9q-20r=2$ This system can

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$x=2, y=3, z=5$

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(C) Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$. The given system of equations becomes: $2p+3q+10r=4$ $4p-6q+5r=1$ $6p+9q-20r=2$ This system can be written as $AX=B$,where $A=\begin{bmatrix} 2 & 3 & 10 \ 4 & -6 & 5 \ 6 & 9 & -20 \end{bmatrix}, X=\begin{bmatrix} p \ q \ r \end{bmatrix}, B=\begin{bmatrix} 4 \ 1 \ 2 \end{bmatrix}$. Calculating the determinant $|A|$: $|A| = 2(120-45) - 3(-80-30) + 10(36+36) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200$. Since $|A| 
eq 0$,the inverse $A^{-1}$ exists. Calculating the adjoint of $A$: $A_{11} = 120-45 = 75, A_{12} = -(-80-30) = 110, A_{13} = 36+36 = 72$ $A_{21} = -(-60-90) = 150, A_{22} = -40-60 = -100, A_{23} = -(18-18) = 0$ $A_{31} = 15+60 = 75, A_{32} = -(10-40) = 30, A_{33} = -12-12 = -24$ $A^{-1} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24 \end{bmatrix}$. $X = A^{-1}B = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \ 1 \ 2 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 300+150+150 \ 440-100+60 \ 288+0-48 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \ 400 \ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \ 1/3 \ 1/5 \end{bmatrix}$. Thus,$p=1/2, q=1/3, r=1/5$. Therefore,$x=2, y=3, z=5$.

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 $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$.

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