You didn't specify which of the many methods for solving systems of equations,
so I just picked one at random. Next time be sure to specify the method, for
there are more than one way to solve lots of problems in algebra. But
regardless, the answer will be (x,y,z,w) = (1,5,1,0).
(1) x − 2y + 3z + w = −6
(2) x − 3y + z − w = −13
(3) x − y = −4
(4) x + z + w = 2
Since w is already missing from (3), let's eliminate it from
(1) and (2) by adding them term by term:
(1) x − 2y + 3z + w = −6
(2) x − 3y + z − w = −13
---------------------------
(5) 2x - 5y + 4z = -19
Let's also eliminate w from (2) and (4) by adding them term by term:
(2) x − 3y + z − w = −13
(4) x + z + w = 2
---------------------------
(6) 2x - 3y + 2z = -11
Now we have a new system of only 3 equations in 3 unknowns:
(3) x − y = −4
(5) 2x - 5y + 4z = -19
(6) 2x - 3y + 2z = -11
Since z is also missing from (3), let's eliminate z from (5) and (6)
by first multiplying (6) through by -2
Multiplying (6) through by -2 gives
(7) -4x + 6y - 4z = 22
Now we add (5) and (7) term by term:
(5) 2x - 5y + 4z = -19
(7) -4x + 6y - 4z = 22
------------------------
(8) -2x + y = 3
Now we have a new system of only 2 equations in 2 unknowns:
(3) x − y = −4
(8) -2x + y = 3
We can eliminate y just by adding (3) and (8)
(3) x − y = −4
(8) -2x + y = 3
--------------------
(9) -x = -1
Now we have just one equation in one unknown, which we can solve
by dividing both sides by -1
(10) x = 1
We substitute 1 for x in (3)
(3) x − y = −4
1 − y = −4
-y = -5
y = 5
We substitute 1 for x and 5 for y in (6)
(6) 2x - 3y + 2z = -11
(6) 2(1) - 3(5) + 2z = -11
2 - 15 + 2z = -11
-13 + 2z = -11
2z = 2
z = 1
We substitute 1 for x and 1 for z in (4)
(4) x + z + w = 2
(1) + (1) + w = 2
2 + w = 2
w = 0
So (x,y,z,w) = (1,5,1,0)
Edwin