Answer
Consistent
Solution set: $\left\{\left(x,y\right)|x=\frac{1}{3}y+\frac{7}{3},y\text{ is any real number}\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
3x-y=7\\
9x-3y=21
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}3&-1&|&7\\9&-3&|&21\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_1=\dfrac{1}{3}r_1$
$\begin{bmatrix}1&-\frac{1}{3}&|&\frac{7}{3}\\9&-3&|&21\end{bmatrix}$
$R_1=-9r_1+r_2$
$\begin{bmatrix}1&-\frac{1}{3}&|&\frac{7}{3}\\0&0&|&0\end{bmatrix}$
The last row only contains zeroes, so the system is consistent, having infinitely many solutions.
Write the corresponding system of equations:
$\begin{cases}
x-\frac{1}{3}y=\frac{7}{3}\\
0=0
\end{cases}$
Express $x$ in terms of $y$:
$x-\frac{1}{3}y=\frac{7}{3}\Rightarrow x=\frac{1}{3}y+\frac{7}{3}$
The solution set is:
$\left\{\left(x,y\right)|x=\frac{1}{3}y+\frac{7}{3},y\text{ is any real number}\right\}$
