1. 7.3 Solving Linear Systems by Linear Combinations (Elimination) Method
Objective: Use linear combinations to solve a system of linear equations.

2. Other Methods Substitution Method
There are several other methods of solving systems of linear equations. Each is best used in different situations. These methods are:
Substitution Method
Elimination (Linear Combination) Method
Matrix Algebra

3. What is a System? A system of linear equations is: A set of parabolas
A set of two or more lines
A stereo component

4. Linear Combinations (Elimination) Method
Linear Combinations (Elimination) method is used when it appears easy to eliminate one variable from the system through transformation. Remember that linear transformations do not change the solutions of a system.

5. Linear Combinations (Elimination) Method
Step 1 Make sure the variables are lined up properly by their names
-2x + y = 4
-6x + y = 0
Step 2 Make the coefficients of one of variables opposites (Multiply one or both equations by appropriate numbers so that the coefficients in one variable are opposites). Notice that the y coefficients are 1, therefore we can multiply either equation by -1 and add the system, thus eliminating the y variable.

6. Linear Combination (Elimination) Method
Step 2 Let’s transform the second equation:
-1(-6x + y = 0)
6x – y = 0
Step 3 Add the new two equations up:
-2x + y = 4
+ 6x – y = 0
4x = 4

7. Linear Combination (Elimination) Method
Step 4 Solving the resulted one variable equation for x
4x = 4
yields x = 1.
Step 5 Back substitute to one of the equation in the system to find the value for the other variable. Once we have the x value, we can plug it into either of our original equations and solve for y:
-2x + y = 4
-6x + y = 0

8. Linear Combination (Elimination) Method
Step 5 Plugging x = 1 into the second equation yields:
-2 (1) + y = 4
-2 + y = 4, y = -6
Step 6 Check the answer.
So our solution is (1,6).

9. Linear Combination (Elimination) Method
Now you try
Use the elimination method to solve the system:
x – 5y = –2
3x + 2y = 11
The first step is:
Add the equations together
Transform the first equation by multiplying it by –3
I’m not sure. I need to review the elimination method.

10. Linear Combination (Elimination) Method
Adding the two equations together yields:
x – 5y = -2
+ 3x + 2y = 11
4x – 3y = 9
We have not eliminated any variables. We must transform one of the equations, and then add them together.
BACK

11. Linear Combination (Elimination) Method
Yes, notice that if we multiply the first equation by -3, we obtain additive inverses for the x coefficient. Our system is now:
-3x + 15y = 6
3x + 2y = 11
The next step is:
Add the two equations
Solve the second equation for y
I’m not sure. I need to review.

12. Linear Combination (Elimination) Method
Solving the second equation for y yields:
y = 0.5(11 – 3x) = ½ (11 – 3x)
This does not help us solve the system. Notice that since we have additive inverses, we can eliminate the x variable by adding the equations.

13. Linear Combination (Elimination) Method
Yes, by adding the two equations we get:
– 3x + 15y = 6
+ 3x + 2y = 11
17y = 17
Solving for y, we get y = 1. The next step is:
Plug y = 1 into either equation and solve for x
Plug x = 1 into either equation and solve for y
I’m not sure. I need to review.

14. Linear Combination (Elimination) Method
We found that y=1, not x=1. So we must plug y=1 into either equation to solve for x not for y.
BACK

15. Linear Combination (Elimination) Method
Yes, now we can plug y = 1 into either of the original equations, or the transformed equation. Let’s choose the first original equation:
x – 5y = –2
x – 5(1) = –2
x = 3
So our solution is (3,1), which must be the only solution to a system of two lines.
Next, we must check the solved order pair is the solution to the system. (omitted here)

16. You Try Solve the system using elimination method: 2x + 5y = 7
The solution is:
(12, -4)
(-4, 12)
(4, -21)
No Solution

17. Sorry!

18. YES!
The solution is (4, -21). You can verify this by plugging it into the system:
2(4) + 5(-21) = 7
3(4) + (-21) = -9

19. You Try Use any method to solve the system: -2x + 3y = 10
The solution is:
(2, 3)
(2, -3)
(3, -2)
No Solution

20. YES!
The answer is “no solution”. We multiply -1 on the first equation and add two equations up:
2x – 3y = –10
+ -2x + 3y = –10
0 = –20
This is a contradiction and the system is inconsistent. Therefore, the system has no solution.

21. Linear Combination (Elimination) Method
Example
Use any method to solve the system:
3g – 24 = –4h
–2 + 2h = g
The first step is(are):
Add the equations together
Use the subsitution method, replace the first variable g by the expression in the second.
Line up the variables.
I’m not sure. I need to review the both methods.

22. Any Method (Substitution)
Use any method to solve the system:
3g – 24 = –4h
–2 + 2h = g
Plug the second variable g into the g in the first equation:
3(-2 + 2h) – 24 = -4h
-6 + 6h – 24 = -4h
6h – 30 = -4h
10h – 30 = 0
10h = 30
h = 3
Go to the second equation to find out g:
g = (3) = = 4
The solution is (4, 3).

23. Any Method (Elimination)
Use the any method to solve the system:
3g – 24 = –4h
–2 + 2h = g
Line up the variable:
3g + 4h = 24
g – 2h = -2
Multiply 2 on both sides of the second equation and add onto the first equation:
2g – 4h = -4
+ 3g + 4h = 24
5g = 20 g = 4
Plug into second equation to find h:
4 – 2h = -2
-2h = -6 h = 3
The solution is (4, 3)

24. You Try A Challenge One
Your company currently uses widgets and gadgets to produce your best selling product, the Ultimate. Looking over your books you see that in May you bought 200 widgets and 400 gadgets for $500, and in June you bought 250 widgets and 250 gadgets for the same cost, $500. How much does one widget cost? One gadget? If a new supplier offered to sell you widgets for 75% cost of what you currently pay, but gadgets would cost 10% more than what you currently pay, should you switch to this new supplier or stay with your current supplier?

25. You Try A Challenge One
Let x and y be your current cost for each widget and gadget, respectively. You can set up the equation system like:
200x + 400y = 500 (1)
250x + 250y = 500 (2)
We divided 100 in the equation (1) and 250 in the equation (2). Then the system is simplified to:
2x + 4y = 5 (3)
x + y = 2 (4)

26. You Try A Challenge One x + y = 2 (4)
In the equation (4), we multiply -2:
-2x – 2y = -4 (5)
2y = 1
yields y = ½ = Plug into equation (4) to solve for x:
x + ½ = 2
yields x = 3/2 = 1.5
So the solution is (3/2, ½ ), or (1.5, 0.5).

27. You Try A Challenge One Now the challenge part is the new price deal.
xnew = 1.5  0.75(1.5) = 1.125
ynew = 0.5  1.1(0.5) = 0.55
We must recalculate the cost under the new supplier is a better
deal than the current cost.
200xnew + 400ynew = 200(1.125) + 400(0.55)
= = 445
250xnew + 250ynew = 250(1.125) + 250(0.55)
= =
Based on the May and June’s data, we have no reason not to
switch to the new supplier.

28. Summary
The key concepts of the linear combinations (elimination) method are to
(a) variables are lined up properly
(b) find out if there are any opposite coefficients for the same variable. If yes, directly add two equations up and eliminate one variable. If not,
(c) multiply (or divide) some number(s) to equation(s) so that two opposite coefficients for the same variable show up. Then follow (b).
Some problems can be applied both substitution and linear combination (elimination) methods. However, it is better to use linear combination (elimination) method than to use the substitution method to some other problems. Choosing an appropriate method is important.

29. Assignment
P 414 #’s