Answer:
x1 =1
x2 =2
x3 =4
Explanation:
Given is a systems of equations in 3 variables.
No of equations given = 3
![x1 + x2 + x3 = 7 ... I\\x1 - x2 + 2x3 = 7 ... II\\5x1 + x2 + x3 = 11 ... III](https://img.qammunity.org/2021/formulas/mathematics/college/ahv6lewsx7sroqeh5jvsohpxwwaay5wiss.png)
subtract equation 1 form equation 3
We get
![4x1=4\\x1=1](https://img.qammunity.org/2021/formulas/mathematics/college/naw1pudg5z1pm9a730k22q1m9hrvh0vqmq.png)
Substitute this value in 2 and 3
![x2-2x3 = -6 ... iv\\x2+x3 =6 ... v](https://img.qammunity.org/2021/formulas/mathematics/college/ds3ayio65mmkesevjofe0kmvbte3xa7nei.png)
subtract iv from v
3x3 = 12
x3=4
Substitute in v
x2 =2
solution is
x1 =1
x2 =2
x3 =4