<PRE><B>Question 186588</B>
There are several methods to solve these equations. Matrix method, Cramers rule etc. Since I do not have the book mentioned, I shall solve it directly by manipulating the equations. Let us number these equations as
2x+y+z=9					(1)
-2x+3y+z=5					(2)
3x-y+2z=10					(3)

Eqn(1) – eqn(2) gives
4x -2y = 4   					let this be eqn(4)

Multiplying eqn(2) by 2 we get 

4x + 2y + 2z = 18    		Now subtract eqn(3) from this
3x – y   + 2z = 10
 x + 3y           = 8			let this be eqn (5)

Now multiply eqn (5)  by 4 and subtract eqn (4) from it

4x + 12 y = 32      That is 4 x Eqn(5)
4x -   2y   = 4 	rewrote eqn 4 to subtract
14 y  = 28		after subtraction
y = 2
			Plug-in the value of y in eqn (5)
That is x + 3(2) = 8,  ie.  x + 6 = 8, so x = 2
Now plug –in these values in eqn (1) 
2(2) + (2) + z = 9
4+2+z =9;    6+z =9 so z= 9-6 =3. Therefore x = 2, y= 2 and z = 3 are the solutions.
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