Solve this system of equations.
3y - z = -1
x + 5y - z = -4
-3x + 6y + 2z = 11
Since x has already been eliminated from the 1st
equation, to make things easier, we will eliminate
x also from the other two equations. To make the
x's cancel we multiply the 2nd equation through
by 3 and add it to the 3rd equation:
3[ x + 5y - z = -4]
1[-3x + 6y + 2z = 11]
3x + 15y - 3z = -12
-3x + 6y + 2z = 11
----------------------
21y - z = -1
Now we have this system of just two equations in
just two unknowns:
3y - z = -1
21y - z = -1
We can eliminate z by multiplying the 1st eq.
by -1 and adding to the 2nd:
-1[ 3y - z = -1]
1[21y - z = -1]
-3y + z = 1
21y - z = -1
---------------
18y = 0
y = 0
Substitute y = 0 in
3y - z = -1
3(0) - z = -1
-z = -1
z = 1
Finally substitute y = 0 and z = 1 into
one of the original equations which
contains x, say,
x + 5y - z = -4
x + 5(0) - 1 = -4
x - 1 = -4
x = -3
So the solution is (x, y, z) = (-3, 0 1)
Edwin
