You need to solve for x and y the system of simultaneous equations, hence, you may use elimination method, such that:

`{(x + y = 5),(2x - y = 1):} => x + y + 2x - y = 5 + 1`

Reducing duplicate members yields:

`3x = 6 => x = 2`

Replacing 2 for x in any of two equations yields:

`2 + y = 5 => y = 5 - 2 => y = 3`

**Hence, evaluating the system of simultaneous equations, using elimination method, yields **`x = 2, y = 3.`

We'll re-write the equations:

x + y = 5

2x - y = 1

We'll use the matrix to solve the system. We'll form the matrix of the system, using the coefficients of x and y:

[1 1]

A =

[2 -1]

We'll calculate the determinant of the system:

detA = -1 - 2 = -3

Since det A is not zero, the system is determinated and it will have only one solution.

x = det X/detA

|5 1|

det X =

|1 -1|

detX = -5 - 1 = -6

x = det X/detA

x = -6/-3

x = 2

We'll calculate y:

|1 5|

det Y =

|2 1|

det Y = 1 - 10

det Y = -9

y = detY/detA

y = -9/-3

y = 3

The solution of the system is: (2 , 3).

The set of equations x + y - 5 = 0 and 2x - y -1 = 0 has to be solved.

From x + y - 5 = 0 we can express x in terms of y as x = 5 - y

Substitute for x in 2x - y - 1 = 0

2(5 - y) - y - 1 = 0

10 - 2y - y - 1 = 0

-3y = -9

y = 3

x = 5 - y = 5 - 3 = 2

The solution of the given system of equations is x = 2 and y = 3

(1) x + y - 5 = 0

(2) 2x - y - 1 = 0

Add eq (1) and eq (2)

3x - 6 = 0

Now its a single variable problem and you can solve for x. Plug this value back into one of the original equations and solve for y.