ASSIGNMENT \( \begin{array}{ll}\text { 1.) } x+y=3 E 1 & \text { 2.) } 2 x-y=4 E 1 \\ x-y=11 E 2 & x+3 y=5 E 2\end{array} \)

To solve the system of equations, we can start by working with the pairs of equations provided. For the first set: 1) \( x + y = 3 \) (Equation 1) 2) \( 2x - y = 4 \) (Equation 2) From Equation 1, we can express \( y \) in terms of \( x \): \( y = 3 - x \) Now substitute \( y \) in Equation 2: \( 2x - (3 - x) = 4 \) This simplifies to \( 2x - 3 + x = 4 \) So, \( 3x - 3 = 4 \) Now add 3 to both sides: \( 3x = 7 \) Divide by 3: \( x = \frac{7}{3} \) Using \( x \) to find \( y \): \( y = 3 - \frac{7}{3} = \frac{9}{3} - \frac{7}{3} = \frac{2}{3} \) Thus, the solution for the first set of equations is: \( x = \frac{7}{3}, y = \frac{2}{3} \). For the second set: 1) \( x - y = 11 \) (Equation 3) 2) \( x + 3y = 5 \) (Equation 4) From Equation 3, we can express \( x \) in terms of \( y \): \( x = y + 11 \) Now substitute \( x \) in Equation 4: \( (y + 11) + 3y = 5 \) This simplifies to \( 4y + 11 = 5 \) Now subtract 11 from both sides: \( 4y = 5 - 11 \) So, \( 4y = -6 \) Divide by 4: \( y = -\frac{6}{4} = -\frac{3}{2} \) Using \( y \) to find \( x \): \( x = -\frac{3}{2} + 11 = -\frac{3}{2} + \frac{22}{2} = \frac{19}{2} \) Thus, the solution for the second set of equations is: \( x = \frac{19}{2}, y = -\frac{3}{2} \). Final solutions are: 1) \( \left( \frac{7}{3}, \frac{2}{3} \right) \) 2) \( \left( \frac{19}{2}, -\frac{3}{2} \right) \)