Question B *4 с Time left 0:32:07 ХА А D 13 X2 5 F X2 E If the flow in, at intersection A, is 248 cars and the flow out, at Dis 248 cars, consider the following question: the total time it takes vehicles to travel along any stretch of road is proportional to the traffic along that stretch. For example, the total time it takes the vehicles along AB is kx minutes. Assuming that the constant is the same for all sections of road, the total time for all 248 vehicles to be in this network is kx; +2kx2 + kx3 + 2kx, + kas minutes What is the total time if k = 2?

\begin{array}{l}x 1+x 2=248 \\x 4+x 5=248 \\x 1-x 3-x 4=0 \\x 2+x 3-x 5=0\end{array}The augmented matrix is\mathrm{R} 3=\mathrm{R} 3-1 * \mathrm{R} 11-00\begin{array}{cc}0 & 248 \\1 & 248 \\0 & -248 \\-1 & 0\end{array}interchange R 2 and R 3\begin{array}{ll}1 & 1 \\ 0 & -1 \\ 0 & 0 \\ 0 & 1\end{array}\begin{array}{cc}0 & 0 \\ -1 & \\ 0 & \\ 1 & \end{array}\begin{array}{cc}0 & 0 \\ -1 & 0 \\ 1 & 1 \\ 0 & -1\end{array}\begin{array}{cc}0 & 248 \\ 0 & -248 \\ 1 & 248 \\ -1 & 0\end{array}\mathrm{R} 2=-1 * R 21000\begin{array}{ll}1 & 0 \\ 1 & 1 \\ 0 & 0 \\ 1 & 1\end{array}\begin{array}{ll}0 & 0 \\ 1 & 1 \\ 0 & 1 \\ 1 & 0\end{array}R 1=R 1-1^{*} R 2, R 4=R 4-1^{*} R 2R 1=R 1+1 * R 3, R 2=R 2-1 * R 3, R 4=R 4+1 * R 3\begin{array}{cccccc}1 & 0 & -1 & 0 & 1 & 248 \\0 & 1 & 1 & 0 & -1 & 0 \\0 & 0 & 0 & 1 & 1 & 248 \\0 & 0 & 0 & 0 & 0 & 0\end{array}It is reduced row echelon form of matrixHence1 \times 1-1 \times 3+1 \times 5=2481 \times 2+1 \times 3-1 \times 5=01 \times 4+1 \times 5=248Hencex 1=248+1 \times 3-1 \times 5\times 2=0-1 \times 3+1 \times 5x 4=248-1 \times 5We choose x 3=0 and x 5=0x 1=248x 2=0x 4=248hence\begin{aligned}\text { total time } & =2^{*} 248+2^{*} 2^{*} 248 \\& =1488 \text { minutes answer }\end{aligned}Please positively rate the solution. Thanks. ...