We have the two equations x-y = 5 and 2x +y = 4 to solve for x and y.

Now

x - y = 5 ...(1)

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

(1) + (2)

=> x - y + 2x + y = 5+4

=> 3x = 9

=> x= 9/3

=> x = 3

substitute x= 3 in (1)

x - y = 5

=> 3- y =5

=> -y = 5-3

=> -y = 2

=> y = -2

**Therefore x = 3 and y = -2**

To solve

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

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

Add the equations to eliminate y:

(x-y)+(2x+y) = 9

3x = 9

x = 9/3 = 3

From (2):

y =4-2x = 4-2*3 = 4-6 = -2.

x = 3 and y = -2.

We'll solve the system using substitution method:

x-y = 5

We'll add y both sides:

x = 5 + y (1)

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

2(5+y) +y = 4

We'll remove the brackets and we'll get:

10 + 2y + y = 4

We'll combine like terms form the left side:

3y + 10 = 4

We'll subtract 10:

3y = -6

We'll divide by 3:

**y = -2**

We'll plug in the value of y in (1):

x = 5 + (-2)

**x = 3**

**The solution of the system is {(3 , -2)}.**