Solving Simultaneous Linear Solving Simultaneous Linear EquationsEquations

on the Problems of on the Problems of

Relative Motion

Two cars A and B are 140km apart

A B140km

Basic term

Basic term

They travel towards each other

A B

They meet !

Basic term

They travel in the same direction

A B

Car A catches up with Car B

Speed

The speed of a car is 50 km/h.

The speed of a car is 50 km in one hour.

The speed of another car is 100 m/min.

The speed of another car is 100 m in one minute.

Speed Formula: Time

DistanceSpeed

Distance = Speed × Time

Learn how to set up equations

to solve the problems

Question 1 A and B are 21 km apartThey walk towards each other They will meet after 3 hours

What are the speeds of A and B?

A B21 km

They meet after 3 hours: x km

: y km

Let x be A’s speed and y be B’s speed.

After 3 hours, how far will A walk ?

How to equate the distances? 3x + 3y = 21

3x km

3y kmAfter 3 hours, how far will B walk ?

km/h km/h

3x km 3y km

Question 1

A and B are 21 km apart

Walking towards each other

Set up an equation with

2 unknown speeds

A B18 m

A will catch up with B after 4 minutes.

Question 2

• A and B are 18 m apart.

• Walking in the same direction

Set up an equation with 2 unknown speeds.

Choice A

A B18 m

A will catch up with B after 4 minutes.

Question 2

• A and B are 18 m apart.

• Walking in the same direction

Set up an equation with 2 unknown speeds.

Choice B

A B18 m

A will catch up with B after 4 mins. : x m

: y m

Let x m/min be A’s speed and y m/min be B’s speed.

How far will A walk after 4 minutes? 4x m

How far will B walk after 4 minutes? 4y m

4x m

4y m

How to equate the distances? 4x – 4y = 18

Question 2

• A and B are 18 m apart.

• Walking in the same direction

Set up an equation with 2 unknown speeds.

Variables

the speeds

the distance apart

the time

Question 3

A car and a bicycle are a certain distance apart.

Speed of the car : 65km/h

Traveling towards each other, they meet in 2 hours.

the speed of bicycle

the distance apart

Two unknowns:

-- x km

-- y km/h

Do worksheet : Q.3

(a) Draw a diagram to show the situation.

(b) Set up an equation with the unknown distance and speed.

Question 3Speed of the car : 65km/h

Let x km be the distance apart and y km/h be the speed of the bicycle.

Car Bicyclex km

(65 2) km 2y km

65 2 + 2y = x

They meet after 2 hours

Traveling towards each other, they meet after 2 hours

Question 4

Traveling in the same direction,

train N will catch up with train M in 2.5 hours.

Speed of train N : 152 km/h

Two trains M and N are a certain distance apart.

Do worksheet : Q.4

(a) Draw a diagram to show the situation.

(b) Set up an equation with the unknown speed and time.

Question 4

2 trains are a certain distance apart. Speed of train N : 152km/h

Let x km be the distance apart and y km/h be the speed of train M.

Train N Train M

x km

After 2.5 hours152 2.5 km

152 2.5 – 2.5y = x

2.5y km

Train N will catch up with train M in 2.5 hours.

480 m

Towards each other

Question 5

480 m

After 1 minTowards each other

480 m

After 2 minsTowards each other

480 m

Meet in 3 minsTowards each other

480 m

Same direction

480 m

Meet in 3 minsTowards each other

480 m

Same direction After 2 mins

480 m

Meet in 3 minsTowards each other

480 m

Same direction After 4 mins

480 m

Meet in 3 minsTowards each other

480 m

Same direction After 6 mins

480 m

Meet in 3 minsTowards each other

480 m

Same direction The dog catches up with the cat in 8 mins

480 m

Meet in 3 minsTowards each other

Their speeds??

Question 5

A dog and a cat are 480 m apart.

3x + 3y = 480

8x – 8y = 480

Let x m/min be the speed of the dog and y m/min be the speed of the cat.

3y m 3x m

480 m

8x m

8y m480 m

Traveling towards each other, they will meet in 3 minutes.

Traveling in the same direction,the dog will catch up with the cat in 8 mins.

Solve the simultaneous linear equations:

(1)

(2)

The speed of the dog is 110 m/min and the speed of the cat is 50 m/min.

Do worksheet : Q. 6

3y m 3x m

480 m

8x m

8y m480 m

From (1), 3(x + y) = 480 x + y = 160 (3)

From (2), 8(x – y) = 480 x – y = 60 (4)

(3) + (4): 2x = 220 x = 110

(3) – (4): 2y = 100 y = 50

3x + 3y = 480

8x – 8y = 480

Question 6

Ann and Teddy are 60 km apart.

Teddy Ann60 km

1.5y km 1.5x km 1.5x +1.5 y = 60

4y km

4x km

4y – 4x = 60

Cycling towards each other, they will meet in 1.5 hours.

Cycling in the same direction, Teddy will catch up with Ann in 4 hours.

Teddy Ann60 km

Let x km/h be the speed of Ann’s bicycle and y km/h be the speed of Teddy’s bicycle.

Their speeds ??

Solve the simultaneous linear equations:

The speed of Ann’s bicycle is 12.5 km/h and the speed of Teddy’s bicycle is 27.5 km/h.

(1)

(2)

From (1), 1.5(x + y) = 60 x + y = 40 (3)From (2), 4(y – x) = 60 y – x = 15 (4)

(3) + (4): 2y = 55 y = 27.5

(3) – (4): 2x = 25 x = 12.5

1.5x + 1.5y = 60

4y – 4x = 60

Harder Problem 1Harder Problem 1

Kenneth and Betty are 200 km apart.

If driving towards each other, they meet in 2 hours.

If Betty starts driving at noon, andKenneth starts in the same direction at 1 p.m., Kenneth will catch up with Betty at 6:45 p.m.

Set up two equations with two unknown speeds.

Harder ProblemKenneth and Betty are 200 km apart.If driving towards each other, they meet in 2 hours. If Betty starts driving at noon, and Kenneth starts in the same direction at 1 p.m., Kenneth will catch up with Betty at 6:45 p.m. Set up two equations with two unknown speeds.

Let x km/h be the speed of Kenneth’s car and y km/h be the speed of Betty’s bicycle.

2x + 2y = 200

20060

456

60

455 yx

2004

36

4

35 yxor

Susan and Peter are running on a 900m circular track outside the playground. Peter runs faster than Susan. If they start together and run in the same direction, Peter will catch up with Susan 6 minutes later. If they go in opposite directions, they will meet 1.2 minutes later. Set up two equations with two unknown speeds.

Harder Problem 2 Harder Problem 2 (Circular motion)(Circular motion)

Quick review

• The key in setting up equations to solve problems of relative motion:

Equate the distances !

Use of Theory of Learning and Variation

變易理論的運用變 Variant 背景 Background

(Invariant)辨識特徵

Critical feature

Relative Motion (Linear and Circular)

Moving towards each other (meeting problem)

Speed, distance, or timein different units

Speed formula:

Distance = Speed × Time

Using the speed formula to equate the distances traveled to set up equations

Moving in the same direction (catch-up problem)

Both meeting and catch-up problems

http://www.sttss.edu.hk/Mathematics/