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4 Risolvi il sistema di equazioni. \( \begin{array}{ll}\text { a) }\left\{\begin{array}{ll}x+3 y=9 & \text { b) }\left\{\begin{array}{l}x+2 y=9 \\ 2 x-3 y=0\end{array}\right. \\ x-2 y=1\end{array}\right. \\ {[x=3, y=2 ; x=5, y=}\end{array} \)

Ask by Lane Harper. in Italy
Feb 18,2025

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Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2y=9\\2x-3y=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=9-2y\\2x-3y=0\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(9-2y\right)-3y=0\) - step3: Simplify: \(18-7y=0\) - step4: Move the constant to the right side: \(-7y=0-18\) - step5: Remove 0: \(-7y=-18\) - step6: Change the signs: \(7y=18\) - step7: Divide both sides: \(\frac{7y}{7}=\frac{18}{7}\) - step8: Divide the numbers: \(y=\frac{18}{7}\) - step9: Substitute the value of \(y:\) \(x=9-2\times \frac{18}{7}\) - step10: Calculate: \(x=\frac{27}{7}\) - step11: Calculate: \(\left\{ \begin{array}{l}x=\frac{27}{7}\\y=\frac{18}{7}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=\frac{27}{7}\\y=\frac{18}{7}\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(\frac{27}{7},\frac{18}{7}\right)\) Solve the system of equations \( x+3y=9; x-2y=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+3y=9\\x-2y=1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=9-3y\\x-2y=1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(9-3y-2y=1\) - step3: Subtract the terms: \(9-5y=1\) - step4: Move the constant to the right side: \(-5y=1-9\) - step5: Subtract the numbers: \(-5y=-8\) - step6: Change the signs: \(5y=8\) - step7: Divide both sides: \(\frac{5y}{5}=\frac{8}{5}\) - step8: Divide the numbers: \(y=\frac{8}{5}\) - step9: Substitute the value of \(y:\) \(x=9-3\times \frac{8}{5}\) - step10: Calculate: \(x=\frac{21}{5}\) - step11: Calculate: \(\left\{ \begin{array}{l}x=\frac{21}{5}\\y=\frac{8}{5}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=\frac{21}{5}\\y=\frac{8}{5}\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(\frac{21}{5},\frac{8}{5}\right)\) Abbiamo due sistemi di equazioni da risolvere. ### Sistema a) Le equazioni sono: 1. \( x + 3y = 9 \) 2. \( x - 2y = 1 \) Risolvendo il sistema, otteniamo: \[ (x,y) = \left(\frac{21}{5}, \frac{8}{5}\right) \quad \text{oppure} \quad (x,y) = (4.2, 1.6) \] ### Sistema b) Le equazioni sono: 1. \( x + 2y = 9 \) 2. \( 2x - 3y = 0 \) Risolvendo il sistema, otteniamo: \[ (x,y) = \left(\frac{27}{7}, \frac{18}{7}\right) \quad \text{oppure} \quad (x,y) = (3.\dot{8}5714\dot{2}, 2.\dot{5}7142\dot{8}) \] ### Risultati finali - **Sistema a)**: \( (x,y) = (4.2, 1.6) \) - **Sistema b)**: \( (x,y) = (3.\dot{8}5714\dot{2}, 2.\dot{5}7142\dot{8}) \)

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Per risolvere il sistema \( a) \left\{\begin{array}{ll} x + 3y = 9 \\ x - 2y = 1 \end{array}\right. \), puoi sostituire \( x \) dalla seconda equazione nella prima. Risolvendo otterrai \( y = 2 \) e sostituendo nuovamente otterrai \( x = 3 \). Quindi, \( (x, y) = (3, 2) \). Per il sistema \( b) \left\{\begin{array}{l} x + 2y = 9 \\ 2x - 3y = 0 \end{array}\right. \), puoi risolvere la prima equazione per \( x = 9 - 2y \) e sostituirlo nella seconda. Dopo aver semplificato, ottieni \( y = 3 \) e sostituendo tornerai a \( x = 3 \). Quindi, \( (x, y) = (3, 3) \).

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