given the system of equations
x + 2y = -z +2
z - 8 = -x
y - z = -5
find value of y
+37084 From third equation y = z-5 sunstitute into first equation to get
x + 2(z-5) = -z +2
x + 2z -10 = -z+2
x = 12 - 3z Sub into Second eq
z-8 = -(12-3z)
z-8 = -12 + 3z
z=2 Sub into Third equation
y-2= -5 to get y= -3 and x=6 z = 2
F
+39 Write down the equations as follows.
x + 2y + z = 2
x + z = 8
y - z = -5
Then you can create the matrix for this system
1 2 1 2
1 0 1 8
0 1 -1 -5
Subtract row 2 from row 1, then interchange these two rows.
1 0 1 8
0 2 0 -6
0 1 -1 -5
Divide row 2 by two, then subtract row 2 from row 3.
1 0 1 8
0 1 0 -3
0 0 -1 -2
Add row 3 to row 1 then multiply row 3 by -1.
1 0 0 6
0 1 0 -3
0 0 1 2
This matrix reads your solution.
x = 6
y = -3
z = 2
+37084 From third equation y = z-5 sunstitute into first equation to get
x + 2(z-5) = -z +2
x + 2z -10 = -z+2
x = 12 - 3z Sub into Second eq
z-8 = -(12-3z)
z-8 = -12 + 3z
z=2 Sub into Third equation
y-2= -5 to get y= -3 and x=6 z = 2
F
(substitution) (go down before going to the next column) x+2((6-2z)/2)=z+2
x+2y=-z+2 x+2((6-6)/2)=3+2
z-8=-x x+0=5
y-z=-5 x=5
x=-z+8 (x=5) y=(6-2(3))/2
z=-x+8 (z=3) y=6-6/2
y=(6-2z)/2 (y=0) y=0
(-z+8)+2y=z+2 Check by substituting back into the forumulas
-2z+6+2y=0 z-8=-x
-2z+2y=-6 3-8=-(5)
2y=6-2z -5=-5
-z+8+2((6-2z)/2)=z+2
-z+8+(12-4z)/4=z+2
-2z+11=z+2
-3z+11=2
-3z=-9
z=3
