Find the values of x,y, and z that satisfies the system: (2x-5y+2z=-18, -2x+3y-3z= 14, x-y+5z= -6)
1) 2x - 5y + 2z = -18
2) -2x + 3y - 3z = 14
3) x - y + 5z = -6
Start by eliminating the x. We need two equations with
only y and z.
Add equation 2 to equation 1.
2x - 5y + 2z = -18
+(-2x + 3y - 3z = 14)
---------------------------
-2y - z = -4
4) -2y - z = -4
Multiply equation 3 by 2.
2 * (x - y + 5z) = -6 * 2
5) 2x - 2y + 10z = -12
Add equation 5 to equation 2.
-2x + 3y - 3z = 14
+(2x - 2y + 10z = -12)
---------------------------
y + 7z = 2
6) y + 7z = 2
Multiply equation 6 by 2.
2 * (y + 7z) = 2 * 2
7) 2y + 14z = 4
Add equation 7 to equation 4.
-2y - z = -4
+(2y + 14z = 4)
---------------------
13z = 0
8) 13z = 0
z = 0 <<<<<<<<<<<<<<<<<<<<
Substitute that into equation 4 and solve for y.
-2y - z = -4
-2y - 0 = -4
-2y = -4
y = 2 <<<<<<<<<<<<<<<<<<<<
Substitute both y and z into equation 3 to solve for x.
x - y + 5z = -6
x - 2 + 5(0) = -6
x - 2 + 0 = -6
x - 2 = -6
x = -4 <<<<<<<<<<<<<<<<<<<<
Verify by substituting all three values into equation 1.
2x - 5y + 2z = -18
2(-4) - 5(2) + 2(0) = -18
-8 - 10 + 0 = -18
-18 = -18
Answer: x = -4, y = 2 and z = 0