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Question 636581: Solve the system by using substitue or elimination method.
2w+x-3y=19
3w-x-3y-2z=14
w-2x-y-2z=-2
-2w+2x+3y+z=-7
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! (1) 2w + x - 3y + 0 = 19
(2) 3w - x - 3y - 2z = 14
(3) w - 2w - y - 2z = -2
(4)-2w + 2x + 3y + z = -7
To solve, the simplist method (used by most computer algorithms) is elimination.
Using (3) as the "anchor" polynomial we can eliminate w in each of the other three by performing the following operations;
(1) - 2*(3) yields (5) 5x - y + 4z = 23
(2) - 3*(3) yields (6) 5x + 4z = 20
(4) + 2*(3) yields (7)-2x + y - 3z = -11
Substituting (6) into (5) yields the value of the variable y;
y = -3
Now let y = -3 in (7) to result with
(7)' -2x - 3z = -8
Using (7)' and (6) we have
-2x - 3z = -8 and
5x + 4z = 20
The simultation solution of this pair is
x = 4
z = 0
Substitute the derived values of x, y, and z into (1) yields
2w + 4 - 3*(-3) = 19
2w = 6
w = 3
Are the values correct?
Substitute each into the four given equations and see if we get equality.
Is (6+4+9+0=19? yes
Is (9-4+9+0=14)? Yes
Is (3-8+3+0=-2)? Yes
Is (-6+8-9+0=-7)? Yes
The answer is w = 3, x = 4, y = -3 and z = 0
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