1. 4
Chapter
Chapter 2
Solving Systems of Linear Equations

2. Systems of Linear Equations and Problem Solving
Section
4.5
Systems of Linear Equations and Problem Solving

3. Using a System of Equations for Problem Solving
Objective 1
Using a System of Equations for Problem Solving

4. Problem Solving Steps Problem-Solving Steps
1. UNDERSTAND the problem. During this step, become comfortable with the problem. Some ways of doing this are to
Read and reread the problem.
Choose two variables to represent the two unknowns.
Construct a drawing.
Propose a solution and check Pay careful attention to check your proposed solution. This will help when writing equations to model the problem.
Continued

5. Problem Solving Steps
2. TRANSLATE the problem into two equations. 3. SOLVE the system of equations. 4. INTERPRET the results: Check the proposed solution in the stated problem and state your conclusion.

6. Example
One number is 4 more than twice the second number. Their total is 25. Find the numbers.
1. UNDERSTAND
Read and reread the problem. Suppose that the second number is 5. Then the first number, which is 4 more than twice the second number, would have to be (4 + 2•5).
Is their total 25? No: = 19. Our proposed solution is incorrect, but we now have a better understanding of the problem.
Continued

7. Example (cont)
Since we are looking for two numbers, we let x = first number
y = second number
2. TRANSLATE
One number is 4 more than twice the second number.
x = 4 + 2y
Their total is 25.
x + y = 25
Continued

8. Example (cont) 3. SOLVE We are solving the system x = 4 + 2y
Using the substitution method, we substitute the solution for x from the first equation into the second equation.
x + y = 25
(4 + 2y) + y = Replace x with 4 + 2y.
4 + 3y = Simplify.
3y = Subtract 4 from both sides.
y = Divide both sides by 3.
Continued

9. Example (cont) Now we substitute 7 for y into the first equation.
x = 4 + 2y = 4 + 2(7) = = 18
4. INTERPRET
Check: Substitute x = 18 and y = 7 into both of the equations.
First equation: x = 4 + 2y
18 = 4 + 2(7) True
Second equation:
x + y = 25
= True
State: The two numbers are 18 and 7.

10. Example
Hilton University Drama club sold 311 tickets for a play. Student tickets cost 50 cents each; non-student tickets cost $ If the total receipts were $385.50, find how many tickets of each type were sold.
1. UNDERSTAND
Read and reread the problem. Suppose the number of students tickets was 200. Since the total number of tickets sold was 311, the number of non-student tickets would have to be 111 (311 – 200).
Continued

11. Example (cont) 1. UNDERSTAND (continued)
The total receipts are $ Admission for the 200 students will be 200($0.50), or $100. Admission for the 111 non-students will be 111($1.50) = $ This gives total receipts of $100 + $ = $ Our proposed solution is incorrect, but we now have a better understanding of the problem.
Since we are looking for two numbers, we let
s = the number of student tickets
n = the number of non-student tickets
Continued

12. Example (cont) 2. TRANSLATE s + n = 311 total receipts were $385.50
Hilton University Drama club sold 311 tickets for a play.
s + n = 311
total receipts were $385.50
Admission for students
Total receipts =
385.50
1.50n
Admission for non-students +
0.50s
Continued

13. Example (cont) s + n = 311 s + n = 311 –s – 3n = –771
3. SOLVE
We are solving the system
s + n = 311
0.50s n =
Since the equations are written in standard form (and we might like to get rid of the decimals anyway), we’ll solve by the addition method. Multiply the second equation by –2.
s + n = 311
–s – 3n = –771
s + n = 311
–2(0.50s n) = –2(385.50)
–2n = –460
n = 230
Continued

14. Example (cont)
Now we substitute 230 for n into the first equation to solve for s.
s + n = 311
s = 311
s = 81
4. INTERPRET
Check: Substitute s = 81 and n = 230 into both of the equations.
s + n = First Equation
= True
0.50s n = Second Equation
0.50(81) (230) =
= True

15. Example (cont)
State: There were 81 student tickets and 230 non student tickets sold.

16. Example
Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current.
Continued

17. Example (cont) 1. UNDERSTAND
Read and reread the problem. We are going to propose a solution, but first we need to understand the formulas we will be using. Although the basic formula is d = r • t (or r • t = d), we have the effect of the water current in this problem. The rate when traveling downstream would actually be r + w and the rate upstream would be r – w, where r is the speed of the rower in still water, and w is the speed of the water current.
Continued

18. Example (cont) 1. UNDERSTAND (continued)
Suppose Terry can row 9 km/hr in still water, and the water current is 2 km/hr. Since he rows for 1 hour in each direction, downstream would be (r + w)t = d or (9 + 2)1 = 11 km
Upstream would be (r – w)t = d or (9 – 2)1 = 7 km
Our proposed solution is incorrect (hey, we were pretty close for a guess out of the blue), but we now have a better understanding of the problem.
Since we are looking for two rates, we let
r = the rate of the rower in still water
w = the rate of the water current
Continued

19. Example (cont) 2. TRANSLATE distance downstream rate downstream
time downstream
(r + w) • 1 =
10.6
rate upstream
time upstream
distance upstream
(r – w) • 1 =
6.8
Continued

20. Example (cont) 3. SOLVE We are solving the system r + w = 10.6
Since the equations are written in standard form, we’ll solve by the addition method. Simply add the two equations together.
r + w = 10.6
r – w = 6.8
2r = 17.4
r = 8.7
Continued

21. Example (cont) Now we substitute 8.7 for r into the first equation.
r + w = 10.6
8.7 + w = 10.6
w = 1.9
4. INTERPRET
Check: Substitute r = 8.7 and w = 1.9 into both equations.
(r + w)1 = First equation
( )1 = True
(r – w)1 = 1.9 Second equation
(8.7 – 1.9)1 = True
State: Terry’s rate in still water is 8.7 km/hr and the rate of the water current is 1.9 km/hr.

22. Example
A Candy Barrel shop manager mixes M&M’s worth $2.00 per pound with trail mix worth $1.50 per pound. How many pounds of each should she use to get 50 pounds of a party mix worth $1.80 per pound?
1. UNDERSTAND
Read and reread the problem. We are going to propose a solution, but first we need to understand the formulas we will be using. To find out the cost of any quantity of items we use the formula
price per unit •
number of units =
price of all units
Continued

23. Example (cont) 1. UNDERSTAND (continued)
Suppose the manage decides to mix 20 pounds of M&M’s. Since the total mixture will be 50 pounds, we need 50 – 20 = 30 pounds of the trail mix. Substituting each portion of the mix into the formula,
M&M’s $2.00 per lb • 20 lbs = $40.00
trail mix $1.50 per lb • 30 lbs = $45.00
Mixture $1.80 per lb • 50 lbs = $90.00
Continued

24. Example (cont) 1. UNDERSTAND (continued)
Since $ $45.00 ≠ $90.00, our proposed solution is incorrect (hey, we were pretty close again), but we now have a better understanding of the problem.
Since we are looking for two quantities, we let
x = the amount of M&M’s
y = the amount of trail mix
Continued

25. Example (cont) 2. TRANSLATE Fifty pounds of party mix x + y = 50
price per unit •
number of units =
price of all units
Using
Price of M&M’s
Price of trail mix
Price of mixture = 2x
1.5y +
1.8(50) = 90
Continued

26. Example (cont) 3. SOLVE We are solving the system x + y = 50
Since the equations are written in standard form, we’ll solve by the addition method. Multiply the first equation by 3 and the second equation by –2 (which will also get rid of the decimal).
3(x + y) = 3(50)
–2(2x y) = –2(90)
3x + 3y = 150
–4x – 3y = –180
–x = –30
x = 30

27. Example (cont) Now we substitute 30 for x into the first equation.
x + y = 50
30 + y = 50
y = 20
4. INTERPRET
Check: Substitute x = 30 and y = 20 into both of the equations.
x + y = First equation
= True
2x y = Second equation
2(30) (20) = 90
= True

28. Example (cont)
State: The store manager needs to mix 30 pounds of M&M’s and 20 pounds of trail mix to get the mixture at $1.80 a pound.

29. Solving Problems with Cost and Revenue Functions.
Objective 2
Solving Problems with Cost and Revenue Functions.

30. Example
A company that manufactures boxes recently purchased $2000 worth of new equipment to make gift boxes to sell to its customers. The cost of producing a package of gift boxes is $1.50 and it is sold for $4.00. Find the number of packages that must be sold for the company to break even.
Continued

31. Example
1. UNDERSTAND
The company will include a one-time cost of $2000 for equipment and then $1.50 per packaged product. The revenue will be $4.00 per package sold.
x = number of packages of gift boxes
C(x) = total cost for producing x packages of gift boxes
R(x) = total revenue for selling x packages of gift boxes
Continued

32. Example (cont) 2. TRANSLATE The revenue equation: R(x) = 4 · x
The cost equation:
C(x) = · x
Continued

33. Example (cont)
3. SOLVE
Since the break-even point is when R(x) = C(x), we solve the equation 4x = 1.5x
4x = 1.5x
2.5x = 2000
x = 800
Continued

34. Example (cont) 4. INTERPRET
Check: To see whether the break-even point occurs when 800 packages are produced and sold, we check to see if revenue equals cost when x = 800.
R(x) = 4x = 4(800) = 3200
C(x) = 1.5x = 1.5(800) = 3200
State: The company must sell 800 packages of gift boxes to break even.

35. Solving Problems Modeled by Systems of Three Equations.
Objective 3
Solving Problems Modeled by Systems of Three Equations.

36. Example
The measure of the largest angle of a triangle is 90° more than the measure of the smallest angle, and the measure of the remaining angle is 30° more than the measure of the smallest angle. Find the measure of each angle.
Continued

37. Example (cont) 1. UNDERSTAND
Read and reread the problem. We are going to propose a solution.
Suppose the measure of the smallest angle is 30º.
Then the largest angle is 30º + 90º = 120º (90º more than the smallest), and the remaining angle is 30º + 30º = 60º (30º more than the smallest).
In a triangle, the sum of the measures of the 3 angles is 180º.
Since 30º + 120º + 60º  180º, our proposed solution is incorrect, but we now have a better understanding of the problem.
Continued

38. Example (cont) 2. TRANSLATE If we let
s = the measure of the smallest angle, then
m = the measure of the middle angle, and
l = the measure of the largest angle, then
we are solving the system
s + m + l = 180
l = 90 + s
m = 30 + s
Continued

39. Example (cont)
3. SOLVE
Substitute the last two equations into the first equation.
s s s = 180
3s = 180
3s = 60
s = 20
Substitute this value for s into the 2nd and 3rd equations from the original set.
l = 90 + s = = 110
m = 30 + s = = 50
Continued

40. Example (cont) 4. INTERPRET
Check: Substitute s = 20, m = 50, and l = 110 into all three equations.
= true
110 = true
50 = true
State: The 3 angles of the triangle are 20º, 50º and 110º.