Answer
$(1,-1)$
Work Step by Step
To solve the system $\begin{cases}4x-y=5 \\ 3x+3y=0 \\ \end{cases},$ we perform elementary row operations on the corresponding augmented matrix to obtain an equivalent matrix with $1s$ along the main diagonal.
The corresponding augmented matrix is
$$\left[
\begin{array}{cc|c}
4 & -1 & 5 \\
3 & 3 & 0\\
\end{array}
\right].$$
We switch Row_1 and Row_2 to obtain the equivalent matrix
$$\left[
\begin{array}{cc|c}
3 & 3 & 0 \\
4 & -1 & 5\\
\end{array}
\right].$$
Next, we multiply Row_1 by $\dfrac{1}{3}$ to obtain the equivalent matrix
$$\left[
\begin{array}{cc|c}
1 & 1 & 0 \\
4 & -1 & 5\\
\end{array}
\right].$$
Thirdly, we replace Row_2 with Row_2-4*Row_1 to obtain the equaivalent matrix
$$\left[
\begin{array}{cc|c}
1 & 1 & 0 \\
0 & -5 & 5\\
\end{array}
\right].$$
Fourthly, we multiply Row_2 by $-\dfrac{1}{5}$to obtain the equivalent matrix
$$\left[
\begin{array}{cc|c}
1 & 1 & 0 \\
0 & 1 & -1\\
\end{array}
\right].$$
Now, we form the system of equations corresponding to this matrix
$$\begin{cases}x+y=0 \\ y=-1 \\ \end{cases}.$$
We see $y=-1$, and to find our value for $x$, we plug $y=2$ into the first equation and solve it for $x$.
Thus
$$x+(-1)=0 \\ x=1.$$
Thus we have exactly one solution, which is $(1,-1).$
