how to solve a system of a linear equation by elimination?

To eliminate x you add the simultaneous equations ogether
3x - 2y = 9
-3x +6y = -21

4y = -12
y = -3

therefore x = 1

Lines intersect as (1, -3)

1. 3x - 2 y = 9

-2y = 9 - 3x

-y = (9 - 3x/2)

y = - (9-3x/2)

2.  -3x + 6 y = -21

6y = -21 + 3x

y = (-21 + 3x)/6

Notice that the coefficients for x (3 and -3) add up to 0. That means that if you add both equations (i.e. add both left hand sides and add both right hand sides), that x is eliminated, as follows:

3x - 2 y = 9
-3x + 6 y = -21
--------------------- +
3x - 2y - 3x + 6y = 9 - 21
4y = -12
y = -3

Now the only thing left to do is to find x from one of the equations:
3x - 2 y = 9
3x = 9 + 2y = 9 + 2*-3 = 3
x = 3/3 = 1

Or with the other equation:
-3x + 6 y = -21
-3x = -21 - 6y = -21 - 6 * -3 = -3
x = -3/-3 = 1

And trust me that as long as the process of solving for y was correct, both equations will always give the same value for x.

3x -2y =9
-3x +6y= -21
 solve them by adding then x gets eliminated from both equations.
The resultant equation is 4y= -12
 =>y= -3.

3x - 2y = 9 - (1)
-3x + 6y = -21 - (2)

(1) + (2),   (3x - 2y) + (-3x + 6y) = 9 + (-21)
                3x - 2y - 3x + 6y = -12
                -2y + 6y = -12
                4y = -12
                y = -12/4
                y = -3

Substitute y = -3 in (1),
3x - (2 x (-3)) = 9
3x - (-6) = 9
3x + 6 = 9
3x = 9 - 6
3x = 3
x = 3/3
x = 1

OR

Substitute y = -3 in (2),
-3x + (6 x (-3)) = -21
-3x +(-18) = -21
-3x -18 = -21
-3x = -21 + 18
-3x = -3
x = 1