You may write the first equation in terms of x moving all terms, except y, to the right, such that:

`y = 3 - 2x`

You should set the first transformed equation equal to the second, ` 3 - 2x = (1/2)x - 9/2`

You need to move the terms that contain x to the left side such that:

`-(1/2)x - 2x = -3 - 9/2`

Multiplying by -1 yields:

`(1/2)x+ 2x = 3+ 9/2 => 5x/2 = 15/2 => 5x = 15 => x = 3`

Substituting 3 for x in the first equation yields:

`y = 3 - 6 => y = -3`

You may solve the system of equations graphically such that:

Notice that the black line intersects the red line at (3,-3).

**Hence, evaluating the solution to the system of simultaneous equations yields (3,-3).**

Solve the system `2x+y=3;y=1/2 x-9/2` :

Since the second equation is solved in terms of y, we can substitute the expression `1/2 x-9/2` for y in the first equation:

`2x+1/2 x-9/2=3`

`5/2x=15/2`

`x=3`

If x=3 then `2(3)+y=3==>y=-3`

Checking in the second equation we get

`1/2(3)-9/2=-6/2=-3=y` as required.

--------------------------------------------------------------

**The solution is (3,-3)**

-------------------------------------------------------------