Solve by Determinants$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=-2$ $\begin{aligned}
& \frac{2}{x}-\frac{3}{y}+\frac{2}{z}=13 \
& \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=9
\end{aligned}$

Question

Examlex · Accepted Answer

The Answer of Solve by Determinants$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=-2$ $\begin{aligned}
& \frac{2}{x}-\frac{3}{y}+\frac{2}{z}=13 \
& \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=9
\end{aligned}$

Solve by Determinants $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=-2$ $\begin{aligned}& \frac{2}{x}-\frac{3}{y}+\frac{2}{z}=13 \\& \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=9\end{aligned}$

Short Answer

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge

Solve by Determinants 1x+1y−1z=−2\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=-2x1​+y1​−z1​=−2 2x−3y+2z=131x+2y+3z=9\begin{aligned}& \frac{2}{x}-\frac{3}{y}+\frac{2}{z}=13 \\& \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=9\end{aligned}​x2​−y3​+z2​=13x1​+y2​+z3​=9​

Short Answer

Correct Answer:VerifiedUnlock this answer nowGet Access to more Verified Answers free of chargeAccess For Free

Solve by Determinants $\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=-2$ $\begin{aligned}& \frac{2}{x}-\frac{3}{y}+\frac{2}{z}=13 \\& \frac{1}{x}+\frac{2}{y}+\frac{3}{z}=9\end{aligned}$

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge