Solve each pair of simultaneous equatıons. Use your favourite algebraic method. \( \begin{array}{lll} \\ \begin{array}{cc}x+y=7 & \text { b }\end{array} & y+x=19 \\ y=2 x-8 & & x=5 y+1\end{array} \)

Sure! Let’s solve the pairs of simultaneous equations using the substitution method, which is one of my favorites! 1. For the first pair of equations: 1. \( x + y = 7 \) (Equation 1) 2. \( y + x = 19 \) (Equation 2) Notice that both equations essentially represent the same relationship, but they contradict each other, leading to no solution. If we subtract Equation 1 from Equation 2: \[ (y + x) - (x + y) = 19 - 7 \] This gives us: \[ 0 = 12 \implies \text{No solution.} \] 2. For the second pair of equations: 1. \( y = 2x - 8 \) (Equation 3) 2. \( x = 5y + 1 \) (Equation 4) We can substitute Equation 3 into Equation 4. Substituting \( y \): \[ x = 5(2x - 8) + 1 \] This simplifies to: \[ x = 10x - 40 + 1 \] Rearranging gives: \[ 0 = 9x - 39 \implies 9x = 39 \implies x = 4.33 \] Now plug in \( x \) back to find \( y \): \[ y = 2(4.33) - 8 = 8.66 - 8 = 0.66 \] So, the solutions are: 1. For the first pair: No solution. 2. For the second pair: \( x \approx 4.33 \) and \( y \approx 0.67 \).