Lesson 21 – Conservation Laws (Continued)

Algebra Textbook Page 329

Exhaust Gas Problem - Carbon monoxide is a gas that is 43% carbon. Carbon dioxide is only 27% carbon. Suppose that the Environmental Protection Agency (EPA) analyzes exhaust from cars, a mixture of carbon monoxide and carbon dioxide.

Carbon monoxide gas: 43% carbon


Carbon dioxide gas: 27% carbon


Problem

i. Define variables from the number of milligrams (mg) of each gas in a sample. Then write an expression for the number of mg of carbon in the sample.

x = amount of CO
y = amount of CO2

Conservation Law and the Conservation of Carbon: The total amount of carbon in the sample equals the amount of carbon in the CO plus the amount of carbon in CO2.

Now use your dictionary above to translate this into algebra.

Carbon in sample = 43% of x + 27% of y

Amount of carbon in mixture = 0.43x + 0.27y

Conservation Law and the Conservation of Gas: amount of gas in the sample = amount of CO gas + amount of CO2 gas

ii. Evaluate the expression in part (a) if one mixes 200 mg of carbon monoxide and 3000 mg of carbon dioxide

x = 2,000 mg; y = 3,000 mg

Use the equation above and plug in your variables

Amount of carbon in the mix = 0.43 • 2000 + 0.27 • 3000
→ 860 + 810
→ 1,670 mg

iii. What is the percent of carbon in the mixture of part (b)? Percentage equals the amount of carbon in the mix divided by the amount of gas

Total amount = 5,000 mg
Amount of Carbon = 1,670 mg

1670
5000

iv. If the EPA finds that a 1600-mg sample of exhaust gas has 32% carbon, how many mg of the sample were carbon monoxide, and how many were carbon dioxide?

Amount of mixture = 1,600 mg with 32% carbon

Conservation of Carbon: 32% of 1600 = 43% of x + 27% of y
Conservation of Gas: 1600 mg = x + y

0.32 • 1,600 = 0.43x + 0.27y
1600 = x + y

512 = 0.43x + 0.27y
x + y = 1600
x= 1600 – y

Simplify
512 = 0.43(1,600 – y) + 0.27y

Add like terms
512 = 688 – 0.43y + 0.27y
512 = 688 – 0.16y
-176 = -0.16y
y = 1,100

Plug in the value for y in the equation
1,600 = x + y
x = 500

Go back to your original definitions and translate the algebra back into English language. There are 1,100 mg of CO2 and 500 mg of CO in the sample.

Slushie Sugar Problem – The sugar content of a slushie can be though of as the percent of the amount of sugar in the slushie. For instance 89 sugar slushie would be 89% sugar and 11% something else. Suppose that a wholesaler has 84 sugar slushie and 91 sugar slushie.

  1. The wholesaler mixes x gallons of 84% sugar slushie and y gallons of 91% sugar slushie. Write and expression for the number of gallons of sugar in the mixture.

    Amount of sugar in the mix = 84% of x + 91% of y
    Amount of sugar in the mix = 0.84x + 0.91y

  2. Evaluate the expression in part (a) if 1000 gallons of 84-sugar slushie is mixed with 2000 gallons of 91-sugar slushie.

    x = 1,100 gallons of slushie
    y = 2,000 gallons of slushie

    Amount of sugar in the mix = 0.84 • 1000 + 0.91 • 2000
    → 840 + 1820
    → 2660 gallons of sugar in the mix

  3. What will the sugar content of the mixture in part (b) be? In other words, what is the sugar number?

    Sugar number = total amount of sugar in mix
                                    total amount of mix

    Remember the law of conservation of matter– the total amount of matter after the mixing is equal to the amount of matter before the mixing

    Amount of mix = x + y = 1000 + 2000 = 3000

    Sugar amount (from part b): 2,660
    Mix amount (from above) : 3,000

    The sugar number is 89. (This answer has no algebraic terms or expressions. Just plain English. Does this answer make sense? Yes, because it is between 84 and 91. Is it closer to 84 or 91? Does this make sense?

  4. If the wholesaler receives an order for 15,000 gallons of 89.7-sugar slushie, how much of each kind should be mixed to fill the order?

Key

Amount in the mix: 15,000
Sugar number : 89.7

Equations

89.7% of 15,000 gallons = 0.84x + 0.91y
x + y = 15,000 or x = 15,000 – y

Translate into algebra
0.897 • 15,000 = 0.84x + 0.91y

Simplify
13,455 = 0.84x + 0.91y

Solve for x using substitution and the other equation
13,455 = 0.84(15,000 – y) + 0.91y

Distribute
13,455 = 12,600 – 0.84y + 0.91y

Simplify
13,455 = 12,600 + 0.07y

Subtract 12,600 from both sides
855 = 0.07y

Divide both sides by 0.07
12,214 = y

Substitute 12,214 for y in equation # 2 from above
15,000 = x + 12,214

Solve for x by dividing both sides by 12,214
x = 2,786

Now translate the problem back into everyday English.


To fill the order, he must mix 2,786 gallons of 84 sugar slushie and 12,214 gallons of 91 sugar slushie.

Displacement      =     velocity     •     time elapsed

Δ x      =     v      •     Δ t

(delta x     =     velocity      •     delta t)


The Greek letter Δ is delta and represents “change in”

The change in position equals the velocity multiplied by the change in time.

In other words, the distance you travel equals how fast you go times how long you go for.

Example 1

Fasty is driving at 72 miles per hour on the interstate.

a.)   How far does she drive in 2 hours and 45 minutes?

Make a key for the equation so that you can translate it into algebraic terms.

Key

displacement = Δx = unknown
velocity = 72 miles per hour
time elapsed = Δt = 2 hours and 45 minutes

Note: you must translate the 45 minutes into algebraic terms as well. This can be easily done by dividing the number of minutes by 60.

Since one hour is equal to 60 minutes, you can multiply 45 minutes by the formula
45 minutes • 1 hour
                     60 minutes

The minutes cancel out and you are left with 45 hours, which equals 3 hours
                                                                       60                                  4

Since you are looking for the distance she traveled, you want to know ∆x. Use the equation: Δx = v • Δt

Plug in your values for each variable
Δx = 72 • 2.75

Simplify
Δx = 198 miles

b.) How long would it take Fasty to drive from Urbana to St. Louis, 250 miles away?

Key

Δx = 250 miles
v = 72 miles per hour
Δt = unknown

Use the equation: Δx = v • Δt
250 = 72 ^ Δt

Divide each side by 72 47.     = Δt

Are we finished? No! We have to translate the answer back into everyday English. Since we have a value with a decimal point, we have to convert that 3.47 into hours and minutes.

This time use the equation 60 minutes • .47 hours
                                               1 hour

Multiply 60 by 0.47 and the hour value cancels out
You are left with 28 minutes.
3 hours 28 minutes

It takes Fasty 3 hours and 38 minutes to drive 250 miles.

c.) Like all of us, Fasty is sometimes in a hurry. If she made it to St. Louis in 3 hours, how fast was she driving?

Key

Δx = 250
v = unknown
Δ t = 3 hours

Use the equation: Δx = v • Δt
250 = v • 3

Divide both sides by 3
83 = v

Fasty drove 83 miles per hour!

d.) If Fasty begins her journey at 3pm in Springfield (which lies 50 miles closer to St. Louis), when will she arrive if she drives 72 miles per hour?

Δx = position of end point minus position of starting point
Position of end point is 250 miles away.
Position of starting point is 50 miles closer.
250 – 50 = 200 miles
Δt = arrival time minus departure time
     = T (arrival time) – 3pm
velocity = 72 miles per hour

The equation looks like this
200 = 72 • (T – 3)

Distribute 72
200 = 72T – 216

Add 216 to each side
416 = 72T

Divide both sides by 72
T = 5.7 pm

Change 0.7 to minutes
0.7 hours • 60 minutes = 42 minutes
                     1 hour
= 5 hours 42 minutes

What was the question asking? It asked at what time did she arrive.
Fasty arrived at 5:42pm.

Swimming Pool Problem – Page 508 in Algebra Book

Topic – Rates are additive!

Problem

A water pipe can fill an empty swimming pool in 5 hours. With the water pipe and the garden hose both running, it takes only 3 hours to fill the empty pool. How long would it take to fill the pool with just the garden hose?

The first equation we need is a rate equation. It is similar to calculating distance.

Amount = rate • time

In this equation, “amount” can also stand for amount of work done.

Translate the problem into statements
A pipe fills a pool in 5 hours
What work does the pipe do?

Work = filling 1 pool           therefore, 1 pool = 5 hours • rate
Time = 5 hours

To find the rate, divide 1 pool by 5 hours
1 pool = rate = 0.2 pools per hour
    5

Key

rate = work done    and    rate = displacement = velocity
              time                                      time

In order to answer the question, we need to know the rate at which the hose is filling the pool. To keep the information straight, it helps to draw a picture.

If we know the rate and the amount, then we can compute the time.

If both the hose and the pipe are on, then the rates are additive!
The rate of the pipe plus the rate of the hose.

Work = 1 pool
Time = 3 hours
Rate = rhose + rpipe
1 pool = (rhose + rpool) • 3 hours
1 = (rhose + 0.2) • 3

Distribute
1 = 3rhose + 0.6

Subtract 0.6 from both sides
0.4 = 3rhose

Divide both sides by 3
rhose = 0.1333 pools per hour

Plug 0.1333 into the equation
1 pool = 0.13 pools per hour • time

Divide both sides by 0.13
t = 7.69 hours

To get 0.69 in minutes, you must multiply by 60 minutes
0.69 hour • 60 minutes = 41 minutes
                      1 hour

Write back in everyday English
It would take 7 hours and 41 minutes to fill the pool with just the hose.

When you prepare the pool for your girly spring party, you notice that the pool started to leak and empties in 2 hours and 37 minutes. You know that you can fill the pool in 3 hours if it is not leaking at all. If you start filling the pool 4 hours before the party starts, will the pool be full?

rhose = 1 pool = 1
             3 hr       3

rleak =   - 1 pool      =     - 1  
             2 hr 37 min     2.616

rate = rhose + rleak

rate =  1   +   - 1   = 0.33 + -0.38
           3        2.16

rate = -0.05 pools per hour

1 pool = rate • time

1 = -0.05 • t

Divide each side by -0.05
t = -20 hours

This means that if the pool is full and you have the hose and pipe on, you have only 20 hours until the pool is empty. This problem shows that rates can be subtractive!

Problem

Janet has just skied down to the bottom of a hill. Her skis made a trail 360 feet long, from the top of the hill to the bottom. She is going to climb back to the top of the hill, along the trail that her skis made when ski skied down.
     Every minute that Janet climbs up the hill, she moves forward 3 yards during the first 30 seconds, but slides back 2 yards while she rests during the next 30 seconds.
     How long will it take her to reach the top of the hill?i

Consecutive Integers
Word problems that involve more than one unknown.

Problem
Find two consecutive integers, whose product is 56.

An example would be 2 • 3 = 6. But this is not correct. Perhaps you can do this problem in your head through trial and error, but algebra can be very useful when the problem involves larger numbers.

To begin, make a dictionary to translate from everyday English to algebra.
x = small integer
y = larger integer

Consecutive integer means larger integer equals the smaller integer plus one.
y = x + 1       this is the first equation

Given: the product of the two consecutive integers is 56
x • y = 56       this is the second equation

Now there are two equations for two unknowns. This is algebra.
y = x + 1
x • y = 56

Use substitution for y in the second equation because y is already by itself
x • (x + 1) = 56

Distribute
x2 + x = 56

Subtract 56 from both sides to get a quadratic equation
x2 + x – 56 = 0
a = 1, b = 1, c = -56

Plug the numbers into the equation
x = -1 ± √12 – 4 • 1 • (-56)
                     2 • 1

=-1 ±√225
         2

= -1 ± 15
         2

= -1 + 15 = 14 = 7   or   = -1 – 15 = -16 = -8
         2             2                     2              2

x = 7 or -8

Is this okay? Do integers have negative numbers?

Counting numbers are numbers such as 1, 2, 3, …
Whole numbers are numbers such as 0, 1, 2, 3, …
Integers are numbers such as 0, 1, 2, 3, … and their opposites -3,-2,-1,0,1,2,3

Yes, negative numbers are considered integers and correct for this problem.

The problem is not finished yet. Using the other equation x • y = 56
7 • y = 56
y = 8

or

8 • y = 56
y = -7

So the solution is [7,8] or [-8,-7].

Problem 13 - Find two consecutive integers whose product is 72

Let x = smaller integer, y = larger integer

Find two equations
y = x + 1
x • y = 72

Plug in y
x (x + 1) = 72

Distribute
x2 + x = 72

Subtract 72 from both sides to set up a quadratic equation
x2 + x – 72 = 0
a = 1, b = 1, c = -72

Plug in the numbers
x = -1 ± √12 – 4 • 1 • (-72)
                       2 • 1

=-1 ±√289
         2

= -1 + 17   or   -1 – 17
         2                  2

x = 8 or -9
y = 9 or -8

Problem 14 – Find two consecutive integers whose product equals 90.

Perhaps you can do this one in your head using trial and error. When the problems become larger, however, trial and error gets more difficult.

Problem 16 – Find 2 consecutive integers whose product is 9,312.

While it will take too long to do trial and error, there is a short cut that uses trial and error and algebra. This is called an educated guess.

Start with two equations
y = x + 1
x • y = 9,312

Since the number 1 is such a small number, it is almost insignificant. We can ignore it for a moment and just take the square root of 9,312
x = √9,312
x = 96.49

Round 96.49 up to 97. Multiply 96 by 97. What do you get?

Problem 17 – Find two consecutive integers whose sum of their squares equals 3,445.
This is too hard to do with trial and error. Let’s use algebra. Begin with two equations.

y = x + 1
x2 + y2 = 3,445

Substitute for y
x2 + (x + 1)2 = 3,445       Note: (x + 1)2 is equivalent to (x + 1)(x + 1)

Distribute
x2 + x2 + x + x + 1 = 3,445

Simplify
2x2 + 2x + 1 = 3,445

Subtract 3,445 from both sides to set up a quadratic equation
2x2 + 2x =3,444
a = 2, b = 2, c = 3,444

x = -2 ± √22 – 4 • 2 • -3,444
                    2 • 2

Simplify
x = -2 ± √ 27,556
              4

x = -2 – 166   or    x = -2 + 166
          4                           4

x = -42 or 41
y = -41 or 42

Let’s use algebra with a short cut to make an educated guess.
y = x + 1
x2 + y2 = 3,445

Substitute for y
x2 + (x2 + 1)2 = 3,445

Approximate that x2 + x2 = 3,445
2x2 = 3,445

Simplify by dividing by 2
2x2 = 3,445
2           2

x2 = 1,722.5

x = √1,722.5

x = 41.5

Round up to 42 and add the squares of 41 and 42
412 + 422 = 3,445

Homework Problem

Use the educated guess method to find two consecutive integers whose sum of their squares equals 1,513.

Begin with 2 equations
y = x + 1
x2 + y2 = 1,513

Educated guess method
x2 + x2 = 1,513

Simplify and divide both sides by 2
2x2 = 1,513
2           2

Take the square root of each side
√x2 = √756.5
x = 27.5

Substitute 27 for x
y = 27 + 1
y = 28

Substitute for x and y in the other equation to check
272 + 282 = 1,513