Question \#3: \( \left\{\begin{array}{l}x-2 y+3 z=-25 \\ x-2 y-3 z=5 \\ 5 x+2 y-3 z=17\end{array} \quad\right. \) Use cramers rule to

Solve the system of equations \( x-2y+3z=-25;x-2y-3z=5;5x+2y-3z=17 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-2y+3z=-25\\x-2y-3z=5\\5x+2y-3z=17\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-25+2y-3z\\x-2y-3z=5\\5x+2y-3z=17\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}-25+2y-3z-2y-3z=5\\5\left(-25+2y-3z\right)+2y-3z=17\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}-25-6z=5\\-125+12y-18z=17\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}z=-5\\-125+12y-18z=17\end{array}\right.\) - step5: Substitute the value of \(z:\) \(-125+12y-18\left(-5\right)=17\) - step6: Simplify: \(-35+12y=17\) - step7: Move the constant to the right side: \(12y=17+35\) - step8: Add the numbers: \(12y=52\) - step9: Divide both sides: \(\frac{12y}{12}=\frac{52}{12}\) - step10: Divide the numbers: \(y=\frac{13}{3}\) - step11: Substitute the values of \(y,z:\) \(x=-25+2\times \frac{13}{3}-3\left(-5\right)\) - step12: Simplify: \(x=-\frac{4}{3}\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-\frac{4}{3}\\y=\frac{13}{3}\\z=-5\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{4}{3}\\y=\frac{13}{3}\\z=-5\end{array}\right.\) - step15: Rewrite: \(\left(x,y,z\right) = \left(-\frac{4}{3},\frac{13}{3},-5\right)\) The solution to the system of equations using Cramer's rule is: \[ (x,y,z) = \left(-\frac{4}{3},\frac{13}{3},-5\right) \] or \[ (x,y,z) = (-1.3,4.3,-5) \]