1. SYSTEM OF EQUATIONS SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES
WEEK 7
SYSTEM OF EQUATIONS
SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES

2. OBJECTIVES At the end of this session , you will be able to:
Verify the solution of a system of linear equations in three variables.
Solve systems of linear equations in three variables.
Solve problems using systems in three variables.

3. INDEX System of linear equations in three variables.
Solving system of linear equations in three variables by eliminating variables.
Application.
Summary.

4. 1. SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES
Equations such as 2x + 3y – z = 6; x + 3y – 8z = 12 are called linear equations in three variables.
In general, any equation of the form
Ax + By+ Cz = D
where A, B, C and D are real numbers such that A, B and C are not zero, is a linear equation in three variables.
If we graph a linear equation in three variables we would end up with a figure of a plane in a three dimensional coordinate system.  An example of what a plane would look like is a floor or a desk top. 
In the previous section we have studied systems of linear equations in two variables, in this section we will be looking at systems that have three linear equations and three unknowns.
Example: x – 2y + 3z = 22
2x – 3y – z = 5
3x + y – 5z = -3 Y Z x
System of linear equations in three variables

5. 1. SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES(Cont…)
A solution of a system in three variables is an ordered triple (x, y, z) that makes all three equations true.  In other words, it is what they all three have in common.  So if an ordered triple is a solution to one equation, but not another, then it is not a solution to the system.
Let us check if the ordered triple (-1, -4, 5) is a solution of the system
x – 2y + 3z = 22
2x – 3y – z = 5
3x + y – 5z = -32
We replace x with –1, y with -4 and z with 5 in each of the three given equations.
(-1) – 2(-4) + 3(5) = (-1) – 3(-4) – (5) = (-1) + (-4) – 5(5) = -32
= – 5 = – 25 = -32
22 =22 (True) = 5 (True) = -32 (True)
The ordered triple (-1, -4, 5) satisfies all the three equations. It makes all three equations true. Thus, the ordered triple (-1, -4, 5) is a solution of the system.

6. 1. SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES(Cont…)
Let us recall some important points from the previous section:
A consistent system is a system that has at least one solution.
An inconsistent system is a system that has no solution.
The equations of a system are dependent if all the solutions of one equation are also solutions of the other two equations.  In other words, they end up being the same line.
The equations of a system are independent if they do not share all solutions.  They can have just one point in common, not all of them.
A system of linear equations in three variables represents three planes in three dimensional coordinate system.
If these planes intersect at one point, then this point of intersection gives the solution of the system and the system is consistent and has exactly one solution.
The three planes may not have a common point of intersection and represent an inconsistent system with no solution.
The planes may coincide or intersect along a line and have many points in common and represent systems with infinitely many solutions.
   

7. 1. SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES(Cont…)
There are three possible outcomes that you may encounter when working with these systems:
Exactly One Solution
If the system in three variables has one solution, it is an ordered triple (x, y, z) that is a solution to all three equations. 
In other words, when you substitute the values of the ordered triple, it makes all three equations true.
The system is consistent and equations are independent.
No Solution
If the three planes are parallel to each other, they will never intersect.  This means they do not have any points in common.  In this situation, we would have no solution.
The system is inconsistent and equations are independent.
Infinite Solutions
If the three planes end up lying on top of each other, then there is an infinite number of solutions.  In this situation, they would end up being the same plane, so any solution that would work in one equation is going to work in the other.
The system is consistent and the equations are dependent. 

8. 2. SOLVING SYSTEMS OF EQUATIONS BY ELIMINATING VARIABLES
The method for solving a system of linear equations in three variables is similar to that used on systems of linear equations in two variables. We use addition to eliminate any variable, reducing the system to two equations in two variables.
Then we can use either addition or substitution to eliminate another variable and reduce the system to one equation in single variable.
We then solve this equation as linear equation in one variable to get the value of the remaining variable and back-substitute its value in the original equations to find the values of other variables.
Let us understand the procedure of solving a system of linear equations in three variables with the help of an example.
x + 4y – z = (1)
3x + 2y + z = (2)
2x – 3y + 2z = (3)
Remark: The numbers in (  ) are equation numbers.  They will be used throughout the problems for reference purposes.
NOTE: 1. There is more than one way that to solve this type of system.
2. Graphing is not used for solving system of linear equations in three and more variables as it involves more dimensions and plotting the graph becomes complicated.
3. We use addition, substitution, and combination of both the methods to solve a system of linear equations in three variables.

9. 2. SOLVING SYSTEMS OF EQUATIONS BY ELIMINATING VARIABLES(Cont…)
STEP 1: Simplify and put all three equations in the form Ax + By  + Cz = D if needed .
This step would involve things like removing the brackets () by applying distributive property and removing fractions if any in the given system of equations.
As all the three equations of given system are already in this form, we can skip this step.
STEP 2: Choose to eliminate any one of the variables from any pair of equations.
We will proceed exactly as we did for eliminating a variable by addition method with two linear equations and two variables.
NOTE: At this point, we are working with only two of the equations.  In the next step we will incorporate the third equation into the mix. 
In this step we need to make sure that one of the variables gets eliminated, leaving one equation and two unknowns.
It doesn't matter which variable we choose to eliminate,  we want to keep it as simple as possible. If a variable already has opposite coefficients then we just add the two equations together. If they don't, then we need to multiply one or both equations by a number that will create opposite coefficients in one of the variables.

10. 2. SOLVING SYSTEMS OF EQUATIONS BY ELIMINATING VARIABLES(Cont…)
For the given system of equations we will eliminate variable z from the first two equations. We just got to add the two equations as the variable already has opposite coefficients.
x + 4y – z = (1)
3x + 2y + z = (2)
  x + 6y = (4)
STEP 3: Eliminate the same variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.
Basically, we are going to do another elimination step, eliminating the same variable z we did in step 2, just with a different pair of equations. For example, if we used equations 1 and 2 in step 2, then we can use either 1 and 3 or 2 and 3 in this step. As long as we are using a different combination of equations its fine. This will get that third equation into the mix. We need to do this to give us two equations to go with our two unknowns that are left after the first elimination. We can use any method for eliminating the second variable - substitution or addition.
In this case we will eliminate variable z from equations 2 and 3 by multiplying equation 2 by -2 and adding the resulting equation to equation 3
-6x - 4y - 2z = -16
2x – 3y + 2z = -16
-4x – 7y = (5)
z has opposite coefficients, so it gets eliminated.
Adding the resulting equation to equation 3

11. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
STEP 4: Solve the remaining system found in step 2 and 3
After steps 2 and 3, we have a system of two equations and two unknowns given by equations 4 and 5.
4x + 6y = (4)
-4x – 7y = (5)
When we solve this system that has two equations and two variables, we will have the values for two of our variables.
Now we will proceed exactly as we did in the last section, by solving this system of equations by eliminating a variable by either substitution or addition.
Remember that if both variables get eliminated and we have a false statement, that means our answer is no solution. 
If both variables get eliminated and we get a true statement, that means our answer is infinite solutions.
In the above system, we observe that coefficients of x have opposite values, so we will eliminate x in the above system by adding equations 4 and 5.
4x + 6y = 28
-4x – 7y = -32
-y = -4
y = 4
Variable x gets eliminated and we get the value for variable y.

12. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
Now we will substitute the value of y in equation 4 and get the value for x.
Note: We can substitute the value of y in either of the two equations 4 or 5 to get the value for x.
4x + 6y = (Equation 4)
4x + 6(4) = (Replace y by 4)
4x + 24 = (Simplify)
4x = (Subtract 24 from both sides)
x = (Divide both sides by 4)
STEP 5: Solve for the third variable.
As we have found values for the two variables in step 4, that means the three equations have one solution. Now we substitute the values found in step 4 into any of the equations in the problem that have the missing variable in it and solve for the third variable.
Note: We can choose any of the original equations to find the value of the third variable.
In this case we substitute the values of x and y in equation 2.
3x + 2y + z = (Given equation 2)
3(1) + 2(4) + z = (Substitute x = 1 and y = 4)
z = (Simplify)
z = (Subtract 11 from both sides)
Thus, the solution set for the given system of equation is {(1, 4, -3)}.

13. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
STEP 6: Check
Now we substitute the proposed solution in all the three equations to check if the ordered triple is a solution to the given system of equations.
x + 4y – z = x + 2y + z = x – 3y + 2z = -16
(1) + 4(4) – (-3) = (1) + 2(4) + (-3) = (1) – 3(4) + 2(-3) = -16
= = – = -16
20 = 20 (True) = 8 (True) – 16 = (True)
Thus, ordered triple (1, 4, -3) is a solution to the given system of equations.

14. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
Now let us summarize the steps we have followed to solve a system of linear equations in three variables.
STEP 1: Simplify and put all three equations in the form Ax + By  + Cz = D if needed .
STEP 2: Choose to eliminate any one of the variables from any pair of equations.
STEP 3: Eliminate the same variable chosen in step 2 from any other pair of equations, creating a
system of two equations and two unknowns.
STEP 4: Solve the remaining system found in step 2 and 3.
STEP 5: Solve for the third variable.
STEP 6: Check the proposed solution.

15. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
ADDITIONAL EXAMPLE 1:
We have to solve the system 3(2x + y) + 5z = (1)
2(x – 3y + 4z) = (2)
4(1 + x) = -3(z – 3y) (3)
STEP 1: Simplify and put all three equations in the form Ax + By  + Cz = D if needed .
We need to simplify the given system of equations by removing the brackets using distributive property.
6x + 3y + 5z = (4)
2x – 6y + 8z = (5)
4 + 4x = -3z + 9y
4x – 9y + 3z = (6)
Equations 4, 5, and 6 give the simplified system of equations and is equivalent to the original system of equations. In step 5 we can substitute the values in this system to obtain the value of the third equation.
Next, we proceed in the same manner as explained in the previous example to solve this system of equations.
Solution set for the this system of equations is {(1/2, 1/3, -1)}

16. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
ADDITIONAL EXAMPLE 2:
Solve the system of equations x + y = (1)
x + y - z = (2)
3x + 2y + z = (3)
STEP 1: Simplify and put all three equations in the form Ax + By  + Cz = D if needed .
Since the above system is already in simplified form, we can skip this step.
STEP 2: Choose to eliminate any one of the variables from any pair of equations.
In this case, this step is already done for us as equation 1 contains only two variables x and y.
STEP 3: Eliminate the same variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.
We just need to eliminate z from equations 2 and 3. This will give us the second equation in two variables.
We will add the two equations as the variable already has opposite coefficients.
x + y – z = 4
3x + 2y + z = 0
4x + 3y = (4)

17. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
STEP 4: Solve the remaining system found in step 2 and 3.
After steps 2 and 3, we have a system of two equations and two unknowns given by equations 1 and 4.
2x + y = (1)
4x + 3y = (4)
Now we need to solve this system in two variables by eliminating another variable by either substitution or addition.
We will eliminate y in equation 4 by substituting the value of y from equation 1.
(Note: In this question we are using a combination of addition and substitution method to solve the given system of equations)
From equation 1, we get y = 2 – 2x
Substituting this value of y in equation 4 we get,
4x + 3(2 – 2x) = 4
4x + 6 – 6x = (Apply distributive property)
-2x = (Simplify)
x = (Divide both sides by -2)
Now we substitute this value of x in any of the equations 1 or 4 to get the value for y.
2(1) + y = (Substituting x =1 in equation 1)
y = (Simplify)

18. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
STEP 5: Solve for the third variable.
We have found the values of two variables in step 4, now we need to find the value of the third variable z by substituting the values of x and y in any of the original equations.
We substitute 1 for x and 0 for y in equation 2 and solve for z.
x + y - z = (Equation 2)
z = (Substitute the values of x and y)
-z = (Subtract 1 from both sides)
z= -3
The solution set for this system of equations is {(1, 0, -3)}
STEP 6: Check
We check the proposed solution by substituting the values of x, y and z in all the three equations to check if the ordered triple is a solution to the given system of equations.
2x + y = x + y - z = x + 2y + z = 0
2(1) + 0 = – (-3) = (1) + 2(0) + (-3) = 0
2 + 0 = = = 0
2 = 2 (True) = (True) = 0 (True)
Thus, ordered triple (1, 0, -3) is a solution to the given system of equations.

19. 2. SOLVING SYSTEMS OF EQUATIONS IN BY ELIMINATING VARIABLES(Cont…)
Now we will solve a word problem based on the system of linear equations in three variables.
ADDITIONAL EXAMPLE 3:
The sum of three numbers is 16. The sum of twice the first number, 3 times the second number, and 4 times the third number is 46. The difference between 5 times the first number and the second number is 31. Find the three numbers.
SOLUTION: In such word problems we let First number = x
Second number = y
Third number = z
Condition 1: Sum of three numbers is 16.
x + y + z = (1)
Condition 2: Sum of twice the first number(2x), three times the second number(3y) and four times the third number(4z) is 46.
2x + 3y + 4z = (2)
Condition 3: Difference of five times the first number(5x) and second number is 31.
5x – y = (3)
Now equations 1, 2 and 3 give the required system of equations in three variables and we need to find the values of x, y and z. This system is similar to the one solved in additional example 2. You can now solve this system on the same lines.
Solution set for this system is {(7, 4, 5)}.

20. 3. APPLICATION
Till now we have learnt how to solve a given system of equations in three variables. Now we learn how to solve problems using systems in three variables.
Find the quadratic function y = ax2 + bx + c whose graph passes through the points (1, 4), (2, 1) and (3, 4).
SOLUTION: We need to find values of a, b and c in the given equation y = ax2 + bx + c. We can do so by solving a system of equations in three variables a, b and c.
It is given that the graph of the quadratic function passes through three points (1, 4), (2, 1) and (3, 4), so we obtain the system of equations by substituting these values of x and y in the quadratic equation.
Substituting 1 for x and 4 for y in the quadratic function y = ax2 + bx + c, we get
4 = a(1)2 + b(1) + c
4 = a + b + c (1)
Now we substitute 2 for x and 1 for y in the quadratic function y = ax2 + bx + c, we get
1 = a(2)2 + b(2) + c
1 = 4a + 2b + c (2)
Next substituting 3 for x and 4 for y in the quadratic function y = ax2 + bx + c, we get
4 = a(3)2 + b(3) + c
4 = 9a + 3b + c (3)
Equations 1, 2 and 3 give us the required system of equations in three variables.

21. 3. APPLICATION(Cont…)
Now we will solve the system of linear equations in three variables as explained before.
STEP 1: Simplify and put all three equations in the form Ax + By  + Cz = D if needed .
Given equations are in simplified form, so we can skip this step.
STEP 2: Choose to eliminate any one of the variables from any pair of equations.
We choose to eliminate c from equations 1 and 2 by multiplying equation 2 by –1and adding the resulting equation to equation 1.
4a + 2b + c = (Original equation 2)
-4a - 2b – c = (Resulting equation)
a + b + c = (Equation 1)
-3a – b = (4)
STEP 3: Eliminate the same variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.
We will eliminate c from equation 2 and 3 by multiplying equation 3 by –1 and adding the resulting equation to equation 2.
9a + 3b + c = (original equation 3)
-9a –3b – c = (Resulting equation)
4a + 2b + c = (Equation 2)
-5a –b = (5)

22. 3. APPLICATION(Cont…)
STEP 4: Solve the remaining system found in step 2 and 3
After steps 2 and 3, we have a system of two equations and two unknowns given by equations 4 and 5.
-3a – b = (4)
-5a – b = (5)
Now we have to solve equations 4 and 5 for a and b. Multiply equation 5 by –1 and adding the resulting equation to equation 4.
5a + b = (Resulting equation)
-3a – b = (Equation 4)
2a = (Divide both sides by 2)
a = 3
Substitute the value of a in equation 4 and solve for b.
-3(3) – b = 3
-b = (Add 9 to both sides)
b = (Simplify)

23. 3. APPLICATION(Cont…) STEP 5: Solve for the third variable.
We substitute the values of a and b in equation 1 and solve for c.
a + b + c = (Original equation 1)
3 + (-12) + c = 4
c = – 3
c = 13
STEP 6: Check
You can check if the solution set {(3, -12, 13)} is a solution to the given system by substituting the values of a, b and c in each of the original equations.
Substituting the values a = 3, b = -12 and c = 13 in the quadratic function , we get y = ax2 + bx + c we get the required equation.
The required quadratic equation is y = 3x2 - 12x + 13.

24. 4. SUMMARY Let us recall what we have learnt so far:
In general, any equation of the form Ax + By+ Cz = D where A, B, C and D are real numbers such that A, B and C are not zero, is a linear equation in three variables.
A system of linear equations is two or more linear equations that are being solved simultaneously.
A solution of a system in three variables is an ordered triple (x, y, z) that makes all three equations true. 
A system of linear equations in three variables represents three planes in the three dimensional coordinate system.
The method for solving a system of linear equations in three variables is similar to that used on systems of linear equations in two variables. We use addition to eliminate any variable, reducing the system to two equations in two variables.