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Solve the system of linear equations. \[ \left\{\begin{array}{r}x+y-2 z=3 \\ 3 x-2 y+4 z=1 \\ 2 x-3 y+6 z=8\end{array}\right. \]

Ask by Hodges Ramos. in the United States
Jan 22,2025

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The system of equations has no solution.

Solución

Solve the system of equations \( x+y-2z=3;3x-2y+4z=1;2x-3y+6z=8 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y-2z=3\\3x-2y+4z=1\\2x-3y+6z=8\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=3-y+2z\\3x-2y+4z=1\\2x-3y+6z=8\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}3\left(3-y+2z\right)-2y+4z=1\\2\left(3-y+2z\right)-3y+6z=8\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}9-5y+10z=1\\6-5y+10z=8\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=\frac{8+10z}{5}\\6-5y+10z=8\end{array}\right.\) - step5: Substitute the value of \(y:\) \(6-5\times \frac{8+10z}{5}+10z=8\) - step6: Simplify: \(-2=8\) - step7: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \\z \in \varnothing \end{array}\right.\) - step8: Rewrite: \((x, y, z) \in \varnothing\) The system of linear equations has no solution.

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To solve the system of equations, we will use the method of elimination or substitution. The given system is: 1. \( x + y - 2z = 3 \) (Equation 1) 2. \( 3x - 2y + 4z = 1 \) (Equation 2) 3. \( 2x - 3y + 6z = 8 \) (Equation 3) Let's first manipulate these equations for easier elimination. **Step 1: From Equation 1**, we can express \( y \) in terms of \( x \) and \( z \): \[ y = 3 - x + 2z \quad \text{(Equation 4)} \] **Step 2: Substitute Equation 4 into Equations 2 and 3**: Substituting \( y \) from Equation 4 into Equation 2: \[ 3x - 2(3 - x + 2z) + 4z = 1 \] Expanding it: \[ 3x - 6 + 2x - 4z + 4z = 1 \] Combine like terms: \[ 5x - 6 = 1 \implies 5x = 7 \implies x = \frac{7}{5} \] **Step 3: Now substitute \( x = \frac{7}{5} \) back into Equation 4 to find \( y \)**: \[ y = 3 - \frac{7}{5} + 2z = \frac{15}{5} - \frac{7}{5} + 2z = \frac{8}{5} + 2z \quad \text{(Equation 5)} \] **Step 4: Substitute both \( x \) and \( y \) into Equation 3**: Using Equation 5: \[ 2\left(\frac{7}{5}\right) - 3\left(\frac{8}{5} + 2z\right) + 6z = 8 \] Calculating: \[ \frac{14}{5} - \frac{24}{5} - 6z + 6z = 8 \] Simplifying: \[ -\frac{10}{5} = 8 \implies -2 = 8 \quad \text{(Contradiction)} \] We've reached a contradiction, indicating that the system of equations is inconsistent. This means there is no solution to the given system of equations. Thus, we conclude that the system does not have any solution.

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