2x + y = 3 ..............(1)

3y - 2x = 2

We will rewrite:

-2x + 3y = 2 ................(2)

Now , using the elimination method we will solve the system:

Let us add (1) and (2):

==> 4y = 5

Now divide by 4:

**==> y= 5/4**

now to find x, we will substitute in (1):

2x + y = 3

2x = 3- y

x = (3-y) /2

= ( 3- 5/4) / 2 = ( 12-5)/ 8 = 7/8

**=> x = 7/8**

**Then the solution to the system is the point : ( 7/8, 5/4)**

We'll re-write the equations:

2x + y = 3

-2x + 3y = 2

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

[2 1]

A =

[-2 3]

We'll calculate the determinant of the system:

detA = 6 + 2 = 8

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

x = det X/detA

|3 1|

det X =

|2 3|

detX =9 - 2 = 7

x = det X/detA

**x = 7/8**

**We'll calculate y:**

|2 3|

det Y =

|-2 2|

det Y = 4 + 6

det Y = 10

y = detY/detA

y = 10/8

**y = 5/4**

**The solution of the system is: (7/8 , 5/4). **

2x+y =3 .... ..(1)

3y-2x = 2......(2).

To solve the above system we can use elimination method. If we add both equations, in the left side the x terms 2x and -2x gets cancelled and we we get a sum of y and 3y equal to 4y . On the right sum of 3 and 2 = 5.

So 4y = 5. Or 4y = 5. This gives y = 5/4.

Putting this y = 5/4 in the first equation 2x+y = 3, we get 2x+5/4 = 3. So 2x = 3-5/4 = 7/4. Therefore x= 7/8.

Therefore solution to the system the given linear simultaneous equation is x = 7/8 and y =5/4.