\( \left. \begin{array} { l l } { x + y = 4 } & { 3 x - 4 y = 1 } \\ { 2 x - y = 5 } & { 8 } \\ { a - 2 b = 9 } & { 3 x + 2 y = 10 } \\ { 2 a + 3 b = 4 } & { 4 x - y = 6 } \end{array} \right. \)

It looks like you have provided a system of equations. However, the formatting seems a bit unclear, as it appears to be a mix of different systems. Let's break it down and solve the systems of equations separately. ### System 1: 1. \( x + y = 4 \) 2. \( 3x - 4y = 1 \) ### System 2: 1. \( 2x - y = 5 \) 2. \( 8 \) (This seems to be a standalone number, not an equation) ### System 3: 1. \( a - 2b = 9 \) 2. \( 3x + 2y = 10 \) ### System 4: 1. \( 2a + 3b = 4 \) 2. \( 4x - y = 6 \) I will solve the first and second systems of equations, as the second system does not have a valid equation. The third and fourth systems will be solved together as they involve the variables \( a \) and \( b \). Let's proceed with solving these systems. ### Solving System 1: 1. \( x + y = 4 \) 2. \( 3x - 4y = 1 \) ### Solving System 3 and 4: 1. \( a - 2b = 9 \) 2. \( 2a + 3b = 4 \) I will solve these systems now. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=4\\3x-4y=1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=4-y\\3x-4y=1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(4-y\right)-4y=1\) - step3: Simplify: \(12-7y=1\) - step4: Move the constant to the right side: \(-7y=1-12\) - step5: Subtract the numbers: \(-7y=-11\) - step6: Change the signs: \(7y=11\) - step7: Divide both sides: \(\frac{7y}{7}=\frac{11}{7}\) - step8: Divide the numbers: \(y=\frac{11}{7}\) - step9: Substitute the value of \(y:\) \(x=4-\frac{11}{7}\) - step10: Calculate: \(x=\frac{17}{7}\) - step11: Calculate: \(\left\{ \begin{array}{l}x=\frac{17}{7}\\y=\frac{11}{7}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=\frac{17}{7}\\y=\frac{11}{7}\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(\frac{17}{7},\frac{11}{7}\right)\) Solve the system of equations \( a - 2b = 9; 2a + 3b = 4 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a-2b=9\\2a+3b=4\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=9+2b\\2a+3b=4\end{array}\right.\) - step2: Substitute the value of \(a:\) \(2\left(9+2b\right)+3b=4\) - step3: Simplify: \(18+7b=4\) - step4: Move the constant to the right side: \(7b=4-18\) - step5: Subtract the numbers: \(7b=-14\) - step6: Divide both sides: \(\frac{7b}{7}=\frac{-14}{7}\) - step7: Divide the numbers: \(b=-2\) - step8: Substitute the value of \(b:\) \(a=9+2\left(-2\right)\) - step9: Calculate: \(a=5\) - step10: Calculate: \(\left\{ \begin{array}{l}a=5\\b=-2\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}a=5\\b=-2\end{array}\right.\) - step12: Rewrite: \(\left(a,b\right) = \left(5,-2\right)\) Here are the solutions for the systems of equations: ### Solution for System 1: 1. \( x + y = 4 \) 2. \( 3x - 4y = 1 \) The solution is: \[ (x, y) = \left( \frac{17}{7}, \frac{11}{7} \right) \] ### Solution for System 3: 1. \( a - 2b = 9 \) 2. \( 2a + 3b = 4 \) The solution is: \[ (a, b) = (5, -2) \] ### Summary of Results: - For the first system, \( x \) and \( y \) are approximately \( 2.43 \) and \( 1.57 \) respectively. - For the second system, \( a \) is \( 5 \) and \( b \) is \( -2 \). If you have any further questions or need additional assistance, feel free to ask!