Answer:
The system
has infinitely many solutions
![x=(5)/(6)y +(7)/(6)\\y=arbitrary](https://img.qammunity.org/2020/formulas/mathematics/college/74walealsdh7s9oqrgzv1p4sl7xvbw5cvq.png)
Explanation:
We have the following system of equations:
![6x-5y=7\\12x-10y=14](https://img.qammunity.org/2020/formulas/mathematics/college/ezwc4b54rdauw0b2iomd98fe59lcak63xn.png)
The augmented matrix of the system is:
![\left[\begin{array}c6&-5&7\\12&-10&14\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/8n2ewm5jvbyu8xp2qfwuo7klvb67ki155s.png)
Transform the augmented matrix to the reduced row echelon form
- Row Operation 1: multiply the 1st row by 1/6
![\left[\begin{array}c1&-5/6&7/6\\12&-10&14\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/7c6bn2h2ybd011y2gijzlo76633qdzc8yf.png)
- Row Operation 2: add -12 times the 1st row to the 2nd row
![\left[\begin{array}cc1&-5/6&7/6\\0&0&0\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/college/doxlsoffm4eubdzbvfess89gw57qgnf8g7.png)
From the reduced row echelon form of the augmented matrix we have the corresponding system of linear equations:
![x-(5)/(6)y=(7)/(6)\\0=0](https://img.qammunity.org/2020/formulas/mathematics/college/sxt2sx8nap3m1n6u2x4hxunhxkxb1gzjzt.png)
The last row of the system (0 = 0) means that the system has infinitely many solutions.
![x=(5)/(6)y +(7)/(6)\\y=arbitrary](https://img.qammunity.org/2020/formulas/mathematics/college/74walealsdh7s9oqrgzv1p4sl7xvbw5cvq.png)