November 14, 2006

Systems of Linear Equations in Three Variables

Today in class, (for all you naughty kids that skipped, haha JK) we worked on equations with three variables. We did work on equations with two variables too. Well, here are the questions that we worked on:

3x+5y= 17
4x-y= -8
23x= -23
x= -1

4(-1)-y= -8
(-4)-y= -8
y= 4

L3= L1+5L2
L4 solve L3

L5 solve L2 w/ x= -1
L6 solve L6
Therefore lines intersect at (-1, 4)

4x+3y= 1
3x+2y= 2

-x= -4

3(4)+2y= 2
12+ 2y= 2
2y= -10
y= -5

L3= 2L1-3L2
L4 Solve L3
L5 solve L2 w/ x=4
L6 solve L5
Therefore, lines intersect at (4, -5)

-x+y+2z= -3
2x-y+ z= -7
x+2y-z= 7

3x+y= 0
14x= -11
x= -11/14

3(-11/14)+y= 0
y= 33/14

2(-11/14)-(33/14)+z= 98/14
-22/14-33/14+z= 98/14
z= 153/14

L3= L1+5L2
L4 solve L3
L5 solve L2 w/ x= -11/14
L6 solve L5
L7 found x
L8 solve L4 for y w/ x= -11/14
L9 solve L8
L10 solve L2 for z with x=-11/14 and y= 33/14
L11 solve L10
L12 found z
Therefore lines interesect at (-11/14, 33/14, 153/14)

2x+y-3z= 9
x+y-2z= 7
zx-y+z= -7

3x-z= 0
4x-2z= 2
2x= -2
x= -1

3(-1)- z= 0
-3= z

x(-1)+y-2(3)= 7
-1+y+6= 7
y= -2

L4= L2+L3
L5= L1+L3
L6= 2L4-L5
L7 solve L6
L8 solve L4 / x= -1
L9 solve L2 w/ x= -1 and z= -3
Therefore, the solution is (-1, 2, -3)

3x-2y+4z= 11
2x+3y-z= 5
x+4y-3z= 3

(1, 2, 3)
Are these solutions to that equation?

To solve this equation, all you have to do is sub in the numbers 1 for x, 2 for y, and 3 for z! Then, ta-dah there you have it.

3(1)-2(2)+4(3)= 11
2(1)+3(2)-3= 5
1+4(2)-3(3)= 3

As you can see, both L1 and L2 work out. However, the third line does not. Zero does not equal three, well last time I checked.

And, Ta-dah. That's what happened in math class today. Our homework for tonight is the numbers that were omitted from yesterday and number 26. Once again, our homework for tonight is numbers 1 and 2 from 25 and all of 26! Before I leave, I hold in my hand, the name of who will be the next scribe! The next scribe is.................................BENOFSCHOOL! Well, see you tomorrow everyone!


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