The document provides examples and instructions for setting up systems of equations to solve word problems involving two variables. It defines key words that represent different mathematical operations and encourages organizing multiple entities into tables. An example problem is given where the costs of hamburgers and fries must be determined based on information about two customers' orders and total costs. Systems of equations are set up with variables representing the price of a hamburger and order of fries to solve for the unknown prices.

1. Linear Word-Problems II
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2. Following are key-words translated into mathematic operations.
Linear Word-Problems II

3. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
Linear Word-Problems II

4. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
Linear Word-Problems II

5. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
Linear Word-Problems II

6. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Linear Word-Problems II

7. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Linear Word-Problems II

8. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II

9. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II
In chapter 2, we solved problems with a single variable.

10. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II
In chapter 2, we solved problems with a single variable. Many
of those problems may be solved using two variables and a
system of equations.

11. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II
In chapter 2, we solved problems with a single variable. Many
of those problems may be solved using two variables and a
system of equations. The advantage of using two variables
instead of one is that it's easier to construct the equations to
form the system.

12. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II
In chapter 2, we solved problems with a single variable. Many
of those problems may be solved using two variables and a
system of equations. The advantage of using two variables
instead of one is that it's easier to construct the equations to
form the system. To do this:
I. identify the unknown quantities and named them as x and y

13. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II
In chapter 2, we solved problems with a single variable. Many
of those problems may be solved using two variables and a
system of equations. The advantage of using two variables
instead of one is that it's easier to construct the equations to
form the system. To do this:
I. identify the unknown quantities and named them as x and y
II. usually there are two numerical relations given, translate
each numerical relation into an equation to make a system

14. Following are key-words translated into mathematic operations.
+: add, sum, plus, total, combine, increased by, # more than ..
–: subtract, difference, minus, decreased by, # less than ..
*: multiply, product, times, “fractions or %” of the amount ..
/: divide, quotient, shared equally, ratio ..
Twice = Double = 2* (amount)
Square = (amount)2
Linear Word-Problems II
In chapter 2, we solved problems with a single variable. Many
of those problems may be solved using two variables and a
system of equations. The advantage of using two variables
instead of one is that it's easier to construct the equations to
form the system. To do this:
I. identify the unknown quantities and named them as x and y
II. usually there are two numerical relations given, translate
each numerical relation into an equation to make a system
III. solve the system.

15. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?

16. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
30 ft

17. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
30 ft
SL

18. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1 30 ft
SL

19. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{ 30 ft
SL

20. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
30 ft
SL

21. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
Substitute L = (2S – 6) into E1, we get
(2S – 6) + S = 30
30 ft
SL

22. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
Substitute L = (2S – 6) into E1, we get
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
30 ft
SL

23. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
Substitute L = (2S – 6) into E1, we get
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
3S = 36
S = 36/3 = 12
30 ft
SL

24. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
Substitute L = (2S – 6) into E1, we get
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
3S = 36
S = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6 = 18.
30 ft
SL

25. Linear Word-Problems II
Example A. A 30-foot rope is cut into two pieces.
The longer one is 6 feet less than twice of the short piece.
How long is each piece?
Let L = length of the long piece
S = length of the short piece
Then L + S = 30 --- E1
L = 2S – 6 --- E2{
Use the substitution method.
Substitute L = (2S – 6) into E1, we get
(2S – 6) + S = 30
2S – 6 + S = 30
3S – 6 = 30
3S = 36
S = 36/3 = 12
Put y = 12 into E2, we get x = 2(12) – 6 = 18.
Hence the rope is cut into 18 and 12 feet.
30 ft
SL

26. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.

27. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.

28. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.
Apple Banana Cake
Maria
Don

29. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.
Apple Banana Cake
Maria 6 4 3
Don

30. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.
Apple Banana Cake
Maria 6 4 3
Don 2 6 10

31. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.
Apple Banana Cake
Maria 6 4 3
Don 2 6 10
b. Beets cost $5/lb and carrots is $3/lb,
Mary bought 8lb of beets and 15Ib of carrots.
Don bought 12Ib of carrots and 9lb of beets .
Beet-cost ($5/lb) Carrot-cost ($3/lb) Total
Maria
Don
Total:

32. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.
Apple Banana Cake
Maria 6 4 3
Don 2 6 10
b. Beets cost $5/lb and carrots is $3/lb,
Mary bought 8lb of beets and 15Ib of carrots.
Don bought 12Ib of carrots and 9lb of beets .
Beet-cost ($5/lb) Carrot-cost ($3/lb) Total
Maria (8×5) 40 (15×3) 45 85
Don
Total:

33. Making Tables
Linear Word-Problems
If the given information are the same types of
for multiple entities, organize the information into a table.
Example B. Put the following information into a table.
a. Maria’s grocery list:
6 apples, 4 bananas, 3 cakes.
Don's grocery list:
10 cakes, 2 apples, 6 bananas.
Apple Banana Cake
Maria 6 4 3
Don 2 6 10
b. Beets cost $5/lb and carrots is $3/lb,
Mary bought 8lb of beets and 15Ib of carrots.
Don bought 12Ib of carrots and 9lb of beets .
Beet-cost ($5/lb) Carrot-cost ($3/lb) Total
Maria (8×5) 40 (15×3) 45 85
Don (9×5) 45 (12×3) 36 81
Total: 166$

34. Linear Word-Problems II
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.

35. Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.

36. Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe
Mary

37. Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary

38. Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

39. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
{
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

40. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method.
Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
{
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

41. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Let x = price of a hamburger, y = price of an order of fries
Linear Word-Problems II
{
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

42. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 362*E1:
Linear Word-Problems II
{
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

43. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

44. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

45. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
–x = –3  x = 3
To find y, put x = 3 into E2,
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

46. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

47. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13
9 + 2y = 13
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

48. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13
9 + 2y = 13
2y = 13 – 9
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

49. 4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13
9 + 2y = 13
2y = 13 – 9
2y = 4 y = 4/2 = 2.
Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

50. Example C. Joe ordered four hamburgers and three fries.
It cost $18. Mary ordered three hamburgers and two fries and
it cost $13. Find the price of each item.
4x + 3y = 18
3x + 2y = 13
The system is
E1
E2
Use the elimination method. LCM of 3y and 2y is 6y.
Multiply E1 by 2 and E2 by 3, subtract
Let x = price of a hamburger, y = price of an order of fries
8x + 6y = 36
) 9x + 6y = 39
2*E1:
3*E2:
Linear Word-Problems II
{
–x = –3  x = 3
To find y, put x = 3 into E2, 3(3) + 2y = 13
9 + 2y = 13
2y = 13 – 9
2y = 4 y = 4/2 = 2.
So a hamburger is $3, and an order of fries is $2.
Berger
Cost $
Fries
Cost $
Total
Cost $
Joe 4x 3y 18
Mary 3x 2y 13

51. Linear Word-Problems II
Following problems use formulas in setting up the equations.

52. The RTD Formula
Linear Word-Problems II
Following problems use formulas in setting up the equations.

53. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
Linear Word-Problems II
Following problems use formulas in setting up the equations.

54. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours,
Linear Word-Problems II
Following problems use formulas in setting up the equations.

55. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6.
Linear Word-Problems II
Following problems use formulas in setting up the equations.

56. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.

57. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems

58. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed)

59. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed) and the current's speed C.

60. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed) and the current's speed C. Going upstream, the
current slows the boat down

61. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed) and the current's speed C. Going upstream, the
current slows the boat down and going downstream it speeds
up the boat.

62. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following problems use formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed) and the current's speed C. Going upstream, the
current slows the boat down and going downstream it speeds
up the boat. In fact, the upstream speed is (R – C),

63. The RTD Formula
Let R = Rate or speed (usually mph)
T = Time (in hours)
D = Distance (in miles)
then RT = D.
For example, if we drove from LA to SF at a rate of 75 mph
and it took 6 hours, it means R = 75, T = 6. Therefore
75(6) = 450 miles is the distance between the two cities.
Linear Word-Problems II
Following formulas in setting up the equations.
Up-and-down-stream-problems
When a boat travels up and down a stream, the speed of the
boat depends on the boat's speed R in still water (cruising
speed) and the current's speed C. Going upstream, the
current slows the boat down and going downstream it speeds
up the boat. In fact, the upstream speed is (R – C), and the
downstream speed is (R + C).

64. Linear Word-Problems II
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current,

65. Linear Word-Problems II
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.

66. Linear Word-Problems II
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles.

67. Linear Word-Problems II
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph

68. Linear Word-Problems II
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

69. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

70. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

71. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
Make a table.
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

72. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
Make a table.
Rate Time (T) Distance (D)
Upstream
Downstream
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

73. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
Make a table.
Rate Time (T) Distance (D)
Upstream R – C
Downstream R + C
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

74. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
Make a table.
Rate Time (T) Distance (D)
Upstream R – C 6
Downstream R + C 3
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

75. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
Make a table.
Rate Time (T) Distance (D)
Upstream R – C 6 24
Downstream R + C 3 24
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

76. Example D. A boat can travel upstream for 24 miles in 6
hours and the same distance downstream in 3 hours.
What is the cruising speed of the boat and the current speed?
Linear Word-Problems II
Let R = cruising speed, C = current speed,
Make a table.
Rate Time (T) Distance (D)
Upstream R – C 6 24
Downstream R + C 3 24
By the formula R*T = D we get two equations.
For example, a boat that travels 3 mph in still water goes
upstream against a 1 mph current, it's upstream rate is 2mph.
It would take 6 hours to go 12 miles. If it travels downstream,
it can travel at 4 mph and it would take 3 hours to go 12 miles.

77. 6(R – C) = 24
3(R + C) = 24{
Linear Word-Problems II
--- E1
--- E2

78. 6(R – C) = 24
3(R + C) = 24{
Linear Word-Problems II
--- E1
--- E2
Simplify each equation, divide E1 by 6 and divide E2 by 3.

79. 6(R – C) = 24
3(R + C) = 24{
R – C = 4 --- E3
R + C = 8 --- E4
Linear Word-Problems II
--- E1
--- E2
{
Simplify each equation, divide E1 by 6 and divide E2 by 3.

80. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+)
R – C = 4 --- E3
R + C = 8 --- E4
Linear Word-Problems II
--- E1
--- E2
{
Simplify each equation, divide E1 by 6 and divide E2 by 3.

81. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R – C = 4 --- E3
R + C = 8 --- E4
Linear Word-Problems II
--- E1
--- E2
{
Simplify each equation, divide E1 by 6 and divide E2 by 3.

82. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R = 6
R – C = 4 --- E3
R + C = 8 --- E4
Linear Word-Problems II
--- E1
--- E2
{
Simplify each equation, divide E1 by 6 and divide E2 by 3.

83. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
R – C = 4 --- E3
R + C = 8 --- E4
Linear Word-Problems II
--- E1
--- E2
{
Simplify each equation, divide E1 by 6 and divide E2 by 3.

84. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
6 + C = 8
C = 2
R – C = 4 --- E3
R + C = 8 --- E4
Linear Word-Problems II
--- E1
--- E2
{
Simplify each equation, divide E1 by 6 and divide E2 by 3.

85. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
6 + C = 8
C = 2
R – C = 4 --- E3
R + C = 8 --- E4
R = cruising speed = 6 mph, C = current speed = 2 mph
Linear Word-Problems II
--- E1
--- E2
{
So,
Simplify each equation, divide E1 by 6 and divide E2 by 3.

86. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
6 + C = 8
C = 2
R – C = 4 --- E3
R + C = 8 --- E4
R = cruising speed = 6 mph, C = current speed = 2 mph
Linear Word-Problems II
--- E1
--- E2
{
So,
Simple Interest Formula
Simplify each equation, divide E1 by 6 and divide E2 by 3.

87. 6(R – C) = 24
3(R + C) = 24{
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
6 + C = 8
C = 2
R – C = 4 --- E3
R + C = 8 --- E4
R = cruising speed = 6 mph, C = current speed = 2 mph
Linear Word-Problems II
--- E1
--- E2
{
So,
Simple Interest Formula
Recall the simple interest of an investment is I = R*P
Simplify each equation, divide E1 by 6 and divide E2 by 3.

88. 6(R – C) = 24
3(R + C) = 24{
Simplify each equation, divide E1 by 6 and divide E2 by 3.
Add the equations to eliminate C
+) 2R = 12
R = 6
Substitute R = 6 into E2,
6 + C = 8
C = 2
R – C = 4 --- E3
R + C = 8 --- E4
R = cruising speed = 6 mph, C = current speed = 2 mph
Linear Word-Problems II
--- E1
--- E2
{
So,
Simple Interest Formula
Recall the simple interest of an investment is I = R*P where
I = amount of interest for one year
R = interest (in %)
P = principal

89. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
Linear Word-Problems II

90. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6%
5%
total
Linear Word-Problems II
Make a table using the interest formula R*P = I

91. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total
Linear Word-Problems II
Make a table using the interest formula R*P = I

92. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total
6
100
5
100 * y
* x
Linear Word-Problems II
Make a table using the interest formula R*P = I

93. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000
6
100
5
100 * y
* x
Linear Word-Problems II
Make a table using the interest formula R*P = I

94. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
Linear Word-Problems II
Make a table using the interest formula R*P = I

95. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
Linear Word-Problems II
Make a table using the interest formula R*P = I

96. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
Linear Word-Problems II
Make a table using the interest formula R*P = I
Multiply E2 by 100 to clear the denominator.

97. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(
Linear Word-Problems II
Make a table using the interest formula R*P = I
Multiply E2 by 100 to clear the denominator.

98. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(
100
Linear Word-Problems II
Make a table using the interest formula R*P = I
Multiply E2 by 100 to clear the denominator.

99. Example E. (Mixed investments)
We have $20,000 saved in two accounts which give 5% and
6% interest respectively. In one year, their combined
interest is $1150. How much is in each account?
rate principal interest
6% x
5% y
total 20,000 1150
6
100
5
100 * y
* x
x + y = 20,000 --- E1
6
100
5
100
y = 1150 --- E2x +{
6
100
5
100
y = 1150)100x +(
100
6x + 5y = 115000 ---E3
Linear Word-Problems II
Make a table using the interest formula R*P = I
Multiply E2 by 100 to clear the denominator.

100. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
Linear Word-Problems II
{

101. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5
-5x – 5y = -100,000
Linear Word-Problems II
{
E1(-5):

102. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+)
Linear Word-Problems II
{
E1(-5):

103. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+) x = 15000
Linear Word-Problems II
{
E1(-5):

104. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+) x = 15000
Sub x=15000 into E1,
15000 + y = 20000
Linear Word-Problems II
{
E1(-5):

105. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+) x = 15000
Sub x=15000 into E1,
15000 + y = 20000
y = 5000
Linear Word-Problems II
{
E1(-5):

106. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+) x = 15000
Sub x=15000 into E1,
15000 + y = 20000
y = 5000
So there are $15,000 in the 6% account,
and $5,000 in the 5% account.
Linear Word-Problems II
{
E1(-5):

107. x + y = 20,000 ---E1
6x + 5y = 115000 ---E3
To eliminate y, multiply E1 by -5 and add to E3.
6x + 5y = 115000
-5x – 5y = -100,000
+) x = 15000
Sub x=15000 into E1,
15000 + y = 20000
y = 5000
So there are $15,000 in the 6% account,
and $5,000 in the 5% account.
Linear Word-Problems II
{
E1(-5):

108. Exercise. A. The problems below are from the word–problems for linear
equations with one variable.
1. A and B are to share $120. A gets $30 more than B, how much does each get?
(Just let A = $ that Mr. A has, and B = $ that Mr. B has.)
2. A and B are to share $120. A gets $30 more than twice of what B gets, how
much does each get?
3. A and B are to share $120. A gets $16 less than three times of what B gets,
how much does each get?
Let x = number of lb of peanuts y = number of lb of cashews (Make tables)
4. Peanuts costs $2/lb, cashews costs $8/lb. How many lbs of each are needed
to get 50 lbs of peanut–cashews–mixture that’s $3/lb?
5. Peanuts costs $3/lb, cashews costs $6/lb. How many lbs of each are needed
to get 12 lbs of peanut–cashews–mixture that‘s $4/lb?
6. We have $3000 more saved at an account that gives 5% interest than at an
account that gives 4% interest. In one year, their combined interest is $600. How
much is in each account?
7. We have saved at a 6% account $2000 less than twice at a 4% account. In
one year, their combined interest is $1680. How much is in each account?
8. The combined saving in two accounts is $12,000. One account gives 5%
interest, the other gives 4% interest. The combined interest is %570 in one year.
How much is in each account?

109. B. Find x = $ of cashews and y = $ of peanuts per lb, solve for x and y with the
following information.
9. Order 1. (in lb) Order 2 (in lb)
Cashews 2 3
Peanuts 3 1
Cost $16 $17
10.
Order 1. (in lb) Order 2 (in lb)
Cashews 2 3
Peanuts 3 1
Cost $17 $15
11. Order 1 consists of 5 lb of cashews and 3 lb of peanuts and it costs $16,
Order 2 consists of 7 lb of cashews and 2 lb of peanuts and it costs $18.
Organize this information into a table and solve foe x and y.
12. Order 1 consists of 3 lb of cashews and 4 lb of peanuts and it costs $38,
Order 2 consists of 2 lb of cashews and 5 lb of peanuts and it costs $30.
Organize this information into a table and solve for x and y.
Note that 9–12 are the same as the hamburger–fries problem.

110. 13. We have $3000 more saved in an account that gives 5% interest than in an
account that gives 4% interest. In one year, their combined interest is $600. How
much is in each account?
14. We have saved in a 6% account $2000 less than twice than in a 4% account. In
one year, their combined interest is $1680. How much is in each account?
15. The combined saving in two accounts is $12,000. One account has 5% interest,
the other has 4% interest. The combined interest is $570 in one year. How much is
in each account?
16. The combined saving in two accounts is $15,000. One account has 6% interest,
the other has 3% interest. The combined interest is $750 in one year. How much is
in each account?
Linear Word-Problems
C. Select the x and y, make a table, then solve.