Solve each pair of simultaneous equations. Use your favourite algebraic method. \( \begin{array}{lll}x+y=7 & \text { b } & y+x=19 \\ \text { a } \\ y=2 x-8 & & \text { c } \\ & & 2 x+1\end{array} \)

Certainly! Let's solve each pair of simultaneous equations labeled **a** and **c** using substitution, which is an effective algebraic method. --- ### **System a** \[ \begin{cases} x + y = 7 \quad \text{(1)} \\ y = 2x - 8 \quad \text{(2)} \end{cases} \] **Step 1:** Substitute equation (2) into equation (1): \[ x + (2x - 8) = 7 \] **Step 2:** Simplify and solve for \( x \): \[ 3x - 8 = 7 \\ 3x = 15 \\ x = 5 \] **Step 3:** Substitute \( x = 5 \) back into equation (2) to find \( y \): \[ y = 2(5) - 8 \\ y = 10 - 8 \\ y = 2 \] **Solution for System a:** \[ x = 5, \quad y = 2 \] --- ### **System c** \[ \begin{cases} x + y = 19 \quad \text{(1)} \\ y = 2x + 1 \quad \text{(2)} \end{cases} \] **Step 1:** Substitute equation (2) into equation (1): \[ x + (2x + 1) = 19 \] **Step 2:** Simplify and solve for \( x \): \[ 3x + 1 = 19 \\ 3x = 18 \\ x = 6 \] **Step 3:** Substitute \( x = 6 \) back into equation (2) to find \( y \): \[ y = 2(6) + 1 \\ y = 12 + 1 \\ y = 13 \] **Solution for System c:** \[ x = 6, \quad y = 13 \] --- **Summary of Solutions:** - **System a:** \( x = 5 \), \( y = 2 \) - **System c:** \( x = 6 \), \( y = 13 \) If there was a **System b** intended, please provide the equations, and I'd be happy to help solve it as well!