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Solución
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)**: \( (x,y) = (3.\dot{8}5714\dot{2}, 2.\dot{5}7142\dot{8}) \)
Respondido por UpStudy AI y revisado por un tutor profesional
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