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Give the linear systems:

(i) $x_1+2x_2=2\3x_1+7x_2=8$

(ii) $x_1+2x_2=1\3x_1+7x_2=7$

Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of

$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$

Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.

Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.

$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$

Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$