# solve system

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• Dec 13th 2006, 04:37 PM
mcooper2006
solve system
I need help understanding how to find solutions to equational systems such as 4x-y+3z=-2
3x+5y-z=15
-2x+y+4z=14
I have no idea how to find the solution, I have read over the examples in the book a thousand times and still have no clue as to how they come up with the solution, the formula they give is R1 + (-2)R2=R2
• Dec 13th 2006, 08:14 PM
CaptainBlack
Quote:

Originally Posted by mcooper2006
I need help understanding how to find solutions to equational systems such as 4x-y+3z=-2
3x+5y-z=15
-2x+y+4z=14
I have no idea how to find the solution, I have read over the examples in the book a thousand times and still have no clue as to how they come up with the solution, the formula they give is R1 + (-2)R2=R2

The method of solution of a system like this involves eliminating x from
equations 2 and 3 using equation 1.

So we proceed by replacing equation 2 by 3 times equation 1 minus 4 times
equation 2, which could be written as 3 R1 - 4 R2 = R2, that is the new
second row equation R2 is:

3(4x-y+3z)-4(3x+5y-z)=3(-2)-4(15)

or (check the arithmetic here please):

-23y + 13z = -66

Now we proceed in a similar manner with the first and third equations. We
add equation 1 to twice equation 3 to get a new equation 3, which we might
write as R1 + 2 R3 = R3:

(4x-y+3z) + 2(-2x+y+4z)=(-2) + 2(14),

or:

y + 11z = 26.

So the system of equations now is:

4x-y+3z=-2
-23y + 13z = -66
y + 11z = 26.

Now the last two equations are a pair of simultaneous equations in two
variables which you should be able to solve. Alternativly proceed as before
using the new equation 2 to eliminate y from equation 3.

Add equation 2 to 23 times equation 3 to get the new equation 3
(or R2+23 R3=R3):

(-23y + 13z) +23 (y + 11z) = -66 + 23 26,

or:

266 z = 532,

or z=2.

Now use this value of z in equation 2 to find y, and the values of z and y
in equation 1 to find x.

The full solution is shown in the attachment.

RonL