Grade 10 Academic Math Chapter 1 – Linear Systems

Grade 10 Academic Math Chapter 1 – Linear

SystemsModelling Word Problems

Days 4 through Days 9

Day 4 Agenda1. Warm-up2. Types of Modelling Problems3. Mixture Problems4. Relative Value Problems5. Practice

Learning GoalBy the end of the lesson… … students will be able to read

and interpret a mixture or relative value word problems and create a pair of linear relation equations, resulting in a linear system

Curriculum Expectations

• Solve problems that arise from realistic situations described in words… by choosing an appropriate algebraic… method

• Ontario Catholic School Graduate Expectations: The graduate is expected to be… a self-directed life long learner who CGE4f applies effective… problem solving… skills

Mathematical Process Expectations

• Connecting – make connections among mathematical concepts and procedures; and relate mathematical ideas to situations or phenomena drawn from other contexts

Modelling Types

• 1. Break-Even Problems• 2. Mixture Problems• 3. Relative Value Problems• 4. Rate Problems

Mixture Problems

• 2 things come together to give a total number or amount

• 2 things come together to form a total cost, weight, points, etc.

• Equations are usually in form Ax + By = C

Mixture Problems

• Ex. 1 Henry sharpens figures skates for $3 a pair and hockey skates for $2.50 per pair. If he earns $240 and sharpens 94 pairs of skates, how many pairs of each type of skate does he sharpen?

Example 1 Mixture (Cont’d)

Mixture ProblemsLet x represent # of figure skates

Let y represent # of hockey skates

x + y = 94 (# of skates eq’n)

3x + 2.5y = 240 (earnings eq’n)

Mixture Problems

• Ex. 2 Joe has 38 loonies and toonies totalling $55. How many of each type of coin does he have?

Example 2 Mixture (Cont’d)

Mixture ProblemsLet l represent # of loonies

Let t represent # of toonies

l + t = 38 (# of coins equation)

l + 2t = 55 (value equation)

Mixture Problems

• Ex. 3 (p.44, #11e)• Benoit invested some money at 8% and

some at 10%. He earned a total of $235 in interest.

Example 3 Mixture (Cont’d)

Mixture ProblemsLet x represent Amount of $

invested at 8%

Let y represent Amount of $ invested at 10%

0.08x + 0.1y = 235 (interest equation)

Note: In order to do a $ invested eq’n, we need the amount invested

Mixture Problems

• Ex. 4, p.51, #4c• The total value of nickels and dimes is 75¢

Example 4 Mixture (Cont’d)

Mixture ProblemsLet n represent # of nickels

Let d represent # of dimes

0.05n + 0.10d = 0.75 (value equation)

Note: In order to do a # of coins eq’n, we need to know the # coins

Relative Value Problems

• Usually 2 unknown numbers, ages, etc.• No set form to the equations• Must follow the directional words such as

more than, less, times, is, twice, sum, difference, etc.

Relative Value Problems

• Ex. 1, p.51, #7• The sum of two numbers is 72. Their

difference is 48. Find the numbers.

Example 1 Relative Value (Cont’d)

Relative Value ProblemsLet x represent the first number

Let y represent the other number

x + y = 72 (sum equation)

x – y = 48 (difference equation)

Relative Value Problems

• Ex. 2, p.51, #8)• A number is four times another number.

Six times the smaller number plus half of the larger number equals 212. Find the numbers.

Example 2 Relative Value (Cont’d)

Relative Value ProblemsLet x represent the first number

Let y represent the other number

x = 4y (multiplication eq’n)

0.5x + 6y = 212 (difference equation)

Relative Value Problems

• Ex. 3, p.24, #7• At the December concert, 209 tickets were

sold. There were 23 more student tickets sold than twice the number of adult tickets. How many of each were sold?

Example 3 Relative Value (Cont’d)

Relative Value ProblemsLet x represent # of student tickets

Let y represent # of adult tickets

x - 23 = 2y (relative # of tickets)

x + y = 209 (# of tickets)

Relative Value Problems

• Ex. 4, p.24, #8• A rectangle with a perimeter of 54cm is 3m

longer than it is wide. What are its length and width?

Relative Value ProblemsLet l represent width of the rect.

Let w represent length of the rect.

2x + 2y = 54 (perimeter eq’n)

l – 3 = w (relative length to width eq’n)

Humour Break

Break-Even Problems

• Usually look for the point at which two things cost the same

• Can refer to the point at which cost and number of things are equal

• Equations usually take the form of y = mx + b

Break-Even Problems

• Ex. 1. Barney’s Banquet Hall charges $500 to rent the room, plus $15 for each meal and Patrick’s Party Palace charges $400 for the hall plus $18 for each meal. When will both places cost the same amount?

Example 1 Break-Even (Cont’d)

Break-Even ProblemsLet x represent # meals

Let y represent the cost

y = 15x + 500 (Barney’s BH)

y = 18x + 400 (Patrick’s PP)

Break-Even Problems

• Ex. 2. The Millennium Wheelchair Co. has just started its business. It costs them $125 to make each wheelchair plus $15,000 in start-up costs. They plan to sell the chairs for $500 each. How many chairs do they have to sell in order to break even?

Example 2 Break-Even (Cont’d)

Break-Even ProblemsLet x represent # of wheelchairs

Let y represent cost or revenue

y = 125x + 15000 (Cost eq’n)

y = 500x (Revenue eq’n)

Break-Even Problems

• Ex. 3. p.44, #11c • It costs $135 to rent the car, based on $25

per day, plus $0.15/km

Example 3 Break-Even (Cont’d)

Break-Even ProblemsLet x represent # of days

Let y represent # of km driven

25x + 0.15y = 135 (Cost eq’n)Note: This is not a usual example. Usually if you are dealing with car rental, you have an eq’n like

y = 0.15x + 25

Humour Break

Rate (Speed Distance Time) Problems (Copy)

• Usually looking for time, speed or distance• Distance = Speed x Time (from science –

can be rearranged for speed and time also)

• Easiest to use a chart to help develop the equations

Rate (Speed Distance Time) Problems

• But first, we have the

Distance = Speed x Time (equation)

Or...

D = S x T

Rate (Speed Distance Time) Problems

• We can also rearrange this eq’n to solve for speed...

Speed = Distance ------------ TimeOr...

Rate (Speed Distance Time) Problems

• We can also rearrange this eq’n to solve for Time...

Time = Distance ------------ Speed

Rate (Speed Distance Time) Problems

• Ex. 1 Fred travelled 95 km by car and train. The car averaged 60 km/h and the train averaged 90 km/hr. If the trip took 1.5 hours, how long did he travel by car?

• Let’s use a speed distance time chart to organize our information...

Example 1 Rate (Cont’d)

Rate (Speed Distance Time) Problems

Let x represent the time in the car

Let y represent the time on the train

Distance (km)

Speed (kph)

Time (h)

Car

Train

Total

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

Car 60

Train

Total

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

Car 60

Train 90

Total

Rate (Speed Distance Time) Problems

Distance (km) (recall D = S x T...

Speed (kph)

Time (h)

Car 60x 60 x

Train 90

Total

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

Car 60x 60 x

Train 90y 90 y

Total

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

Car 60x 60 x

Train 90y 90 y

Total 95

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

Car 60x 60 x

Train 90y 90 y

Total 95 1.5

Rate (Speed Distance Time) Problems

x + y = 1.5 (total travelling time)

60x + 90y = 95 (total distance travelled)

Rate (Speed Distance Time) Problems

• Ex. 2 (text p.137, #6) A traffic helicopter pilot finds that with a tailwind, her 120km trip away from the airport takes 30 minutes. On her return trip to the airport, into the wind, she finds that her trip is 10 minutes longer. What is the speed of the helicopter? What is the speed of the wind?

Example 2 Rate (Cont’d)

Rate (Speed Distance Time) Problems

Let h represent the speed of the helicopter

Let w represent the speed of the wind

Distance (km)

Speed (kph)

Time (h)

With tail windWith headwindTotal

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

With tailwind

120

With headwind

120

Total

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

With tailwind

120 h + w

With headwind

120

Total

Rate (Speed Distance Time) Problems

Distance (km)

Speed (kph)

Time (h)

With tailwind

120 h + w ½

With headwind

120 h - w 2/3 (keep as a fraction)

Total

Rate (Speed Distance Time) Problems

• Recall that

Speed = Distance ------------ Time

Rate (Speed Distance Time) Problems

h + w = 120/0.5 (with tailwind... )

h – w = 120/(2/3) (with headwind)

h + w = 240 (simplified)

h – w = 120 x (3/2) (flip 3/2 and x’s)

h – w = 180 (simplified)

Humour Break