Use Fausta's method to solve these simultaneous equations. \( \begin{array}{ll}\text { i } x+2 y=8 \text { and } & \text { Tip } \\ 5 x=2 y+4 \\ \text { ii } \\ 3 x+2 y=28 \text { and } \\ 3 y=3 x+12\end{array} \)

To tackle these simultaneous equations using Fausta's method, we first organize the equations: **For the first set:** 1. \( x + 2y = 8 \) 2. \( 5x = 2y + 4 \) From the first equation, express \( x \) in terms of \( y \): \[ x = 8 - 2y \] Substituting this into the second equation: \[ 5(8 - 2y) = 2y + 4 \] Expanding and simplifying gives: \[ 40 - 10y = 2y + 4 \] \[ 40 - 4 = 12y \] \[ 36 = 12y \] \[ y = 3 \] Now substitute \( y \) back into the equation for \( x \): \[ x = 8 - 2(3) = 2 \] Thus, the solution for the first set is \( x = 2 \) and \( y = 3 \). **For the second set:** 1. \( 3x + 2y = 28 \) 2. \( 3y = 3x + 12 \) From the first equation, express \( 2y \): \[ 2y = 28 - 3x \] Now substitute \( y \) from the second equation: From \( 3y = 3x + 12 \), we have: \[ y = x + 4 \] Substituting into \( 2y \): \[ 2(x + 4) = 28 - 3x \] \[ 2x + 8 = 28 - 3x \] Combining like terms gives: \[ 5x + 8 = 28 \] \[ 5x = 20 \] \[ x = 4 \] Now substitute back to find \( y \): \[ y = 4 + 4 = 8 \] Thus, the solution for the second set is \( x = 4 \) and \( y = 8 \). In summary: 1. For the first set: \( (x, y) = (2, 3) \) 2. For the second set: \( (x, y) = (4, 8) \)