Answer
No solution.
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 2&-3&4&3\\4&-5&9&13\\ 2&0&7&0 \end{bmatrix}
\begin{array} \\(-R_1+R_3)/3\to R_1 \\-R_2+2R_3\to R_2\\R_3\leftrightarrow R_1 \\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 2&0&7&0\\0&1&1&-1\\ 0&5&5&-13 \end{bmatrix}
\begin{array} \\ \\ \\ R_3-5R_2\to R_3 \\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix: we get the third row $0=-8$ indicating that there is no solution to the system.
