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

2x-y = 1..........(2)

Let us use the elimination method to solve.

First let us add equation (1) + equation (2):

==> 3x = 6

Noe divide by 3:

==> x= 6/3

**==> x= 2**

Now, to find y, we will substitute in (1):

x + y= 5

==> y= 5-x

==> y= 5-2

**==> y= 3**

We'll use the matrix to solve the system. We'll form the matrix of the system:

1 1

A =

2 -1

We'll calculate the determinant of the system:

detA = -1 - 2 = -3

Since det A is not cancelling, 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). **

x+y = 5 and 2x-y = 1.

To find the solution for x and y.

Solution:

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

2x-y =1.....(2).

We solve the equation by the substitution method:

From the equation (2), 2x-y = 1, we get: y = 2x-1. We substitutes y = 2x-1 in (1):

x+2x-1 = 5.

3x -1 = 5.

Add 1 to both sides:

3x-1 + 1 = 5+1

3x= 6.

Divide by 3 both sides:

x = 6/3 = 2.

We substitute x= 2 in equation (1), x+y = 5:

2+y = 5.

y = 5-2 = 3.

Therefore x = 2 and y = 3.