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

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

First let us use the elimination method.

Let us multiply (1) by 2 and add to (2):

==> 7x = 14

==> x = 2

Now using the substitution method substitute x velue in (2)

x-2y = 4

2 - 2y = 4

2-4 = 2y

-2 = 2y

==> y= -1

The solution is:

**x= 2 and y= -1**

The given equations are:

3x+y=5.................(1) and

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

Multiplying (1) by 2 we get 6x+2y=10;

now adding (2) to this we get 6x+x+2y-2y=10+4, or 7x=14, x=14/7=2.

Substitute x=2 in (2). This gives 2-2y=4 or -2y=2 or y=2/(-2)= -1.

Therefore from the given equations we get x=2 and y=-1.

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

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

From the second equation x= 2y+4. Putting this in in (1)

3(2y+4)+y =5

6y+12+y =5

7y = 5-12 =-7,

y = -7/7 = -1.

From (1) : 3x+y =5,

3x+(-1) = 5

3x= 5+1= 6

3x= 6

x = 6/2=3.

So x = 3 and y = -1

We'll solve the system using the elimination method.

We'll note the equations of the system as:

3x+y = 5 (1)

x-2y = 4 (2)

We'll multiply the first equation by 2:

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

We'll remive the brackets:

6x + 2y = 10 (3)

We'll add (3) to (2):

6x + 2y+x-2y = 4+10

We'll eliminate like terms:

7x = 14

We'll divide by 7:

**x = 2**

We'll go into the second equation:

2-2y = 4

We'll subtract 2 both sides:

-2y = 2

We'll divide by -2:

**y = -1**

**The solution of the system is: {(2 , -1)}.**