\( \begin{array}{l}2 x+3 y=6 \\ 5 x+2 y=4\end{array} \) Which of the following equations could be the result of multiplication and addition to eliminate a variable in the system of equations? \( \begin{array}{l}19 x=24 \\ 11 y=22\end{array} \)

Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}19x=24\\11y=22\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x=\frac{24}{19}\\y=2\end{array}\right.\) - step2: Check the solution: \(\left\{ \begin{array}{l}x=\frac{24}{19}\\y=2\end{array}\right.\) - step3: Rewrite: \(\left(x,y\right) = \left(\frac{24}{19},2\right)\) Solve the system of equations \( 2 x + 3 y = 6; 5 x + 2 y = 4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+3y=6\\5x+2y=4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{6-3y}{2}\\5x+2y=4\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\times \frac{6-3y}{2}+2y=4\) - step3: Simplify: \(\frac{5\left(6-3y\right)}{2}+2y=4\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{5\left(6-3y\right)}{2}+2y\right)\times 2=4\times 2\) - step5: Simplify the equation: \(30-11y=8\) - step6: Move the constant to the right side: \(-11y=8-30\) - step7: Subtract the numbers: \(-11y=-22\) - step8: Change the signs: \(11y=22\) - step9: Divide both sides: \(\frac{11y}{11}=\frac{22}{11}\) - step10: Divide the numbers: \(y=2\) - step11: Substitute the value of \(y:\) \(x=\frac{6-3\times 2}{2}\) - step12: Calculate: \(x=0\) - step13: Calculate: \(\left\{ \begin{array}{l}x=0\\y=2\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=0\\y=2\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(0,2\right)\) To determine which of the given equations could result from multiplication and addition to eliminate a variable in the system of equations, we first need to solve the original system of equations: 1. **Original System of Equations:** \[ \begin{array}{l} 2x + 3y = 6 \\ 5x + 2y = 4 \end{array} \] Solving this system, we find: \[ (x, y) = (0, 2) \] 2. **Given Equations:** \[ \begin{array}{l} 19x = 24 \\ 11y = 22 \end{array} \] Solving these equations, we find: \[ (x, y) = \left(\frac{24}{19}, 2\right) \] 3. **Analysis of Results:** - The solution to the original system is \( (0, 2) \). - The solution to the given equations is \( \left(\frac{24}{19}, 2\right) \). Since both systems yield different values for \( x \) (0 vs. \(\frac{24}{19}\)), we can conclude that neither of the given equations \( 19x = 24 \) or \( 11y = 22 \) can be derived from the original system through multiplication and addition to eliminate a variable. Thus, the answer is that neither equation could be the result of the operations on the original system.