Today's Class

Hey everyone, it's Raveen
        
   I hope everyone did a good job on yesterdays mental math quiz!  
Well in today's class, we continued the lesson from yesterday and talked about solving systems of equations in three variables. In our last class we solved these equations using the substitution method and in today's class we solved the same type of equations but instead we used the elimination method. These equations are very similar to solving equations in two variables but just require a few extra steps. These equations are not very complicated, but take a long time to solve because of the various steps involved in the process. 


  I will show you how to do both of these methods (SUBSTITUTION AND ELIMINATION) step by step. I will use the examples Mr.P did with us in class in our yellow booklets because they will be a bit easier to follow and will be easy to refer to if you run into any problems.


Okay, so first I will do an example of solving for three variables using the substitution method.


if we are given the following equations:


x + y - z = 2  [#1]
x - 2y + z = -1 [#2]
3x + y - 2z = 4 [#3]


We should first number off the equations as #1,2 and 3 (shown above) to make it less confusing and easier to follow.


STEP 1
Next, we must solve any one of the equations above for one variable.
(usually go for the easier one because it will save you a lot of time, believe me)   
  So let's use equation #1 because it is the easiest and solve for x.

x + y - z = 2   transforms into.....  x = 2 - y + z 


STEP 2  
Now we substitute the equation we just figured out ( x = 2 - y + z ) into equations # 2 and 3 for x. We are going to call these new equations #4 and 5.

first let's sub it into equation #2 which is  x - 2y + z = -1: 


(2 - y + z) -2y + z = -1
 2 - y + z - 2y +z = -1
we combine like terms....
-3y + 2z = -1 -2 
-3y + 2z = -3
So, the new equation -3y + 2z = -3 will be called equation #4.


Now let's do the same thing for equation #3 which is 3x + y - 2z = 4:


3(2 - y + z) + y - 2z = 4
multiply what is inside the brackets by three because it is the coefficient...
6 - 3y + 3z + y - 2z = 4 
again, combine the like terms.....
-2y + z = 4 - 6
-2y + z = -2 
This new equation -2y + z = -2 will be called equation # 5.


STEP 3
Now that we have two new equations,(#4 and 5), we can solve this system and get two of the three variables.
our two new equations are:
-3y + 2z = -3 [#4]
-2y + z = -2 [#5]


Now again, we solve one of the equations to solve for one of the variables.
let's solve for z in equation #5 because it is the easiest.


So, -2y + z = -2 transforms into... z = 2y - 2.


Now let's sub this value of (z = 2y - 2) for z in equation #4 to get y.
-3y + 2z = -3
-3y + 2(2y -2) = -3
multiply what is in the brackets by 2....
-3y + 4y - 4 = -3
combine like terms.....
y = -3 + 4
y = 1 
Now we plug this value of y into equation #5 to get z
-2y + z = -2
-2(1) + z = -2
-2 + z = -2
take the -2 to the other side to isolate z...
z = -2 + 2
z = 0
  
STEP 4
Now lets take these two values and sub them into any of the original equations to get x.
 lets take equation #1 and plug in the y and z value and solve for x.
y = 1 , z = 0
x + y - z = 2
x + (1) -(0) = 2 
x + 1 - 0 = 2
x = 2 - 1 + 0
x = 1


STEP 5
 now we write these as an ordered triple and check:
(1,1,0)

CHECK:
 x + y - z = 2 [#1]     
1 + 1 - 0 = 2
2 = 2


x - 2y + z = -1 [#2] 
1 - 2(1) + 0 = -1
1 - 2 + 0 = -1
-1 = -1


3x + y - 2z = 4 [#3]
3(1) + 1 - 2(0) = 4
3 + 1 - 0 = 4
4 = 4

We know that the answer is correct because it all checks, so finally we are done!!!


Now I will solve a similar equation using the elimination method:
we have the equations:
x + 4y + 3z = 5 [#1]
x + 3y + 2z = 4 [#2]
x + y - z = -1 [#3]
 (again labeled as #1, 2 and 3 to reduce confusion) 


STEP 1
Group the equations into pairs (#1 and #2) and (#2 and #3). Eliminate the variables which are the same from each of these pairs of equations. This will create two new equations which we will again label as equations #4 an 5.  
 lets start off with equations #1 and #2
in this case , we can subtract and get rid of the x values
    x + 4y + 3z = 5 [#1]
 -  x + 3y + 2z = 4 [#2]
          y  + z = 1 [#4]


We do the same thing for equations #2 and #3 
again, we can subtract and eliminate the x values
   x + 3y + 2z = 4 [#2]
 - x + y - z = -1 [#3]
       2y + 3z = 5 [#5]


STEP 2
 Now we can take equations #4 and 5 and solve the system by elimination.
 y  + z = 1 [#4]
2y + 3z = 5 [#5]
we have to make one of the coefficients equal in order to eliminate it.
lets  multiply y by 2 in equation #4 to make it equal.
(2)y  + (2)z = (2)1 [#4]
so, now we can eliminate the y value by subtracting....
      2y + 2z = 2[#4]
  -   2y + 3z = 5 [#5]
             -z = -3 
             -1   -1
              z = 3
We can now find y by substituting z in to equation #4
y  + z = 1 [#4]
y + 3 = 1
y = 1 - 3
y = -2


STEP 3
 Since we now have two of our three values we can plug them into any original equation to  find x.
 I will use equation #3 to find x.
x + y - z = -1 [#3]
x + (-2) - 3 = -1
x - 2 - 3 = -1
x = -1 + 2 + 3
x = 4


STEP 4
write as an ordered triple and check.
(4,-2,3)


CHECK:
x + 4y + 3z = 5 [#1]
4 + 4(-2) + 3(3) = 5
4 - 8 + 9 = 5
5 = 5


x + 3y + 2z = 4 [#2]
4 + 3(-2) + 2(3) = 4
4 - 6 + 6 = 4
4 = 4


x + y - z = -1 [#3]
4 + (-2) - 3 = -1
4 -2 -3 = -1
4 - 5 = -1
 -1 = -1 
 The equations all checked, so we are done!!!!!

Now that these examples are done, I would just like to remind everyone that the homework for today was the booklet we got on solving systems of equations in three variables using both the elimination and substitution method. Also the wiki solutions manual is due tomorrow for everyone who is working on it.


I hope this scribe helped you learn more in detail and I am sorry if I made any mistakes while explaining it.

On a brighter note, we only have 4 more days of school left until winter break!!!!   


Anyways, thanks for reading, see you guys tomorrow.