#### Answer

$(98,2t-43, t)$

#### Work Step by Step

Step 1. Based on the system of equations, write an augmented matrix as
$\begin{bmatrix} -2 & -5 & 10 & | & 19 \\ 1 & 2 & -4 & | & 12 \end{bmatrix} \begin{array} ..\\R1+2R2\to R2 \end{array}$
Step 2. Perform row operations given to the right of the matrix:
$\begin{bmatrix} -2 & -5 & 10 & | & 19 \\ 0 & -1 & 2 & | & 43 \end{bmatrix} \begin{array} ..\\.. \end{array}$
Step 3. As the number of equations is less than the number of variables, the system is dependent. Let $z=t$; from the last equation, we have $-y+2t=43$. Thus $y=2t-43$
Step 4. From the first equation, we have $-2x-5(2t-43)+10t=19$, which gives $x=98$
Step 5. The solution set of the system can be written as $(98,2t-43, t)$