Respuesta :

Using matrices, the system would be

[tex]\begin{bmatrix}{1} & {6} & {7} \\ {1} & {5} & {4}\end{bmatrix}[/tex]

We just have to make zero the second 1 of the first column. To do that, we subtract the first row with the second row.

[tex]\begin{bmatrix}{1} & 6 & {7} \\ {0} & {1} & {3} \\ {} & {} & {}\end{bmatrix}[/tex]

Since 1-1=0, 6-5=1, and 7-4=3.

Now we transform the matrices in equations again.

[tex]\begin{gathered} x+6y=7 \\ 0x+y=3 \end{gathered}[/tex]

As you can observe, we already got y=3. We use this value to find x

[tex]\begin{gathered} x+6(3)=7 \\ x+18=7 \\ x=7-18=-11 \end{gathered}[/tex]

Therefore, the solutions are y=3 and x=-11.

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