x-3y =5 ..(1) and 2x+y =4....(2)

We eliminate x by (2)-2(1):

y- -6y = 4-2*5 = =-6

7y = -6

y = -6/7 .

Substituting y = -6/7 in (1) : x - 3y = 5.

x= 5+3y = 5+3(-6/7) = 17/7

x= 17/7 and y =-6/7.

(2x+y)-2(x-3y) = 4-2*5

**We'll try to solve the system using another method: substitution method.**

x - 3y = 5

x = 5 + 3y (1)

We'll substitute (1) in the second equation of the system:

2x + y = 4

2(5 + 3y) + y = 4

We'll remove the brackets:

10 + 6y + y = 4

We'll combine like terms:

10 + 7y = 4

We'll subtract 10:

7y = -6

We'll divide by 7:

**y = -6/7**

We'll substitute y in (1):

x = 5 - 18/7

x = (35-18)/7

**x = 17/7**

**The solution of the system is {(17/7;-6/7)}**

Given the two equations:

x - 3y = 5 ...(1)

2x + y = 4 ...(2)

(2) - 2* (1)

=> 2x + y - 2* (x - 3y) = 4 - 10

=> 2x + y - 2x + 6y = -6

=> 7y = -6

=> y = -6/7

susbstituting y = -6/7 in x - 3y = 5

=> x = 5 + 3*(-6 /7)

=> x = 5 - 18 /7

=> x = 17 / 7

**Therefore x is 17 / 7 and y is -6/7**