SOLUTION: Solve the system by back substitution. 2x+4y+z+3w=1 -2y-12z-4w=-16 z+w=2 -5w=-15 The solution set is

Question 1104946: Solve the system by back substitution.
2x+4y+z+3w=1
-2y-12z-4w=-16
z+w=2
-5w=-15
The solution set is
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Start from the bottom equation and work your way up
Solve the fourth equation for w
-5w=-15
-5w/(-5)=-15/(-5)
w = 3

Then use that value of w to plug into the third equation. Notice we're going backwards or up the chain of equations. After w is replaced, solve for z
z+w = 2
z+3 = 2
z+3-3 = 2-3
z = -1

Now use z = -1 and w = 3 to plug into the second equation. Solve for y
-2y-12z-4w = -16
-2y-12(-1)-4(3) = -16
-2y+12-12 = -16
-2y = -16
-2y/(-2) = -16/(-2)
y = 8

Use y = 8, z = -1, w = 3 to plug into the first equation. Solve for x
2x+4y+z+3w=1
2x+4(8)+(-1)+3(3)=1
2x+32-1+9=1
2x+40=1
2x+40-40=1-40
2x=-39
2x/2=-39/2
x = -39/2

The value of x as a fraction is x = -39/2
The value of x in decimal form is x = -19.5

In summary we got the following values
x = -39/2
y = 8
z = -1
w = 3

So the solution is (x,y,z,w) = (-39/2, 8, -1, 3)

Note: you can optionally express -39/2 as -19.5
If you do so, then the solution would be (x,y,z,w) = (-19.5, 8, -1, 3)

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