3 6 equations amp problem solving

7/28/2019 3 6 Equations Amp Problem Solving

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3-6 Equations & Problem Solving


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The first step is to define the variables

There is no magic formula for doing this

because each problem is a little different.

You need to identify the variable based on the problem.

Usually you will need to identify one variable in terms of

another.

Ex.The perimeter of a rectangle is 24 in.

One side is 6 in. longer than the other side.

The variables are side #1 and side #2.

Identify one of them as s,

If Side #1 = s, then Side #2 = s + 6


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There are 5 typical types of problems:

Perimeter problems

Consecutive integer problems

Combined money problemAge/Comparison problems

Distance problems

Lets do an example of each!


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The perimeter of a rectangle is 24 in.

One side is 6 in. longer than the otherside. What are the dimensions?

Perimeter problems:

These are usually pretty easy! You can always draw it

The variables are side #1 and side #2

Identify one of them as s,

If Side #1 = s, then Side #2 = s + 6Since the perimeter is 24add up all sides

And set them equal to 24.

s

s + 6

s + s + (s + 6) + (s + 6) = 24

4s + 12 = 24

4s = 12

s = 3 This isnt the complete answer!

The answer is 3 units by 9 units


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The sum of 3 consecutive integers is 258, find them!

Consecutive integer problems:

These are usually pretty easy as well!

The variables are integer #1, integer #2, and integer #3Identify one of them as x,

The next one would be 1 more, or x + 1

The next one would be 2 more or x + 2

Remember that their sum is 258.

x + (x + 1) + (x + 2) = 258

3x + 3 = 2583x = 255

x = 85 This isnt the complete answer!

The answer is 85, 86 & 87


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Tony has twice as much money as Alicia. She has $16 less than

Ralph. Together they have $200. How much money does each

person have?

Combined money problems:

The variables are Tonys$, Alicias $, and Ralphs $.

The key to these problems is picking the right variable!

So, let Alicias $ = a

That means Tonys $ = 2a and Ralphs $ = a + 16

And together they have a sum of 200so the equation would be:

a + 2a + (a + 16) = 200

Look at the problem, and identify which one is somehow related to

the other two. I know the relationship b/t Tonys& AliciasAlicias & Ralphs

Since Alicia is connected to both, make her $ the variable.


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Age or comparison problems:

These are very similar to the combined money problems.

The key to these problems is picking the right variable!

State College has 620 students. There are 20 more women thanmen. How many men are there? How many women are there?

Ifw= the number of women, then the number of men = w20

Alternative method

Ifm = the number of men, then the number of women = m + 20Its a good idea to avoid subtraction if possible, but either

method will work!

The number of men plus the number of women = 620

m + (m + 20) = 620

2m + 20 = 620

2m = 600

m = 300

This isnt the complete answer!

There are 300 men and 320 women


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Distance problems tend to be the

toughest.

The problem is usually translating them from English into

Algebra.


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Start with the formula for speed (aka rate)

Rate =distance

time

This formula can be manipulated..

times


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Start with the formula for speed (aka rate)

Rate =distance

time

This formula can be manipulated..


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You end up with 3 different (similar)

formulas: d = rt

r =

t =

d

t

d

r


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There are also three types ofdistance problems:

Motion in opposite directionsMotion in the same directionRoundtrip


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Each type of problem is solved a little

differentlydepending on what information

you have.

Motion in opposite directions problems usually use

d1 + d2 = total dMotion in the same direction usually use

d1 = d2Roundtrip problems usually use

d1 = d2


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No matter which type of problem it is, you

should set up a chart like this:

distance timerate


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Ex. Motion in opposite directions problems usually use

d1 + d2 = total dJane and Peter leave their home traveling in opposite

directions on a straight road. Peter drives 15 mph faster

than Jane. After 3 hours, they are 225 miles apart. Find

Peters rate and Janes rate.Jane Peter

225 miles

What should the variable be?Look at the questionfind the ratetheres the clue!


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Ex.

Jane and Peter leave their home traveling in opposite directions

on a straight road. Peter drives 15 mph faster than Jane. After 3

hours, they are 225 miles apart. Find Peters rate and Janes rate.If Janes rate = r, what is Peters rate?

Peters rate = r + 15The time and the distance are both given!

t= 3 hours and d = 225 miles

How do I put this all together using a formula?

Think about the distance travelledits their TOTAL distanceI have

to add Janes distance plus Peters distance to get 225.

distance timerate

Jane

Peter

r

r + 15

3

3


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Janes distance + Peters distance = 225 miles

3r + 3(r + 15) = 2253r + 3r + 45 = 225

6r + 45 = 225

6r = 180

r = 30

Look back at how you defined the variables.

Janes rate = 30 mph and Peters rate = 45 mph

Janes rate = r

Janes time = 3So Janes distance = 3r

Peters rate = r + 15

Peters time = 3So Peters distance = 3(r+ 15)

Peters distance = Peters rate * Peters time

Janes distance = Janes rate * Janes time

Janes distance + Peters distance = 225 miles, so:


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Motion in the same direction problems usually use

d1 = d2Ex. Motion in opposite directionsJane and Peter leave their home traveling in the same direction.

Peter drives 55 mph, Jane drives 40mph. Jane leaves at noon, Peter

leaves at 1 P.M. When will Peter catch up to Jane?

distance timerate

Jane

Peter

40

55

What should the variable be?

Look at the question..when..theres the cluefind the time!NB! time = time spent traveling, NOT time of day!

Since she left 1 hour

earlier, her time is

t + 1

t + 1

t

If Peters time = t

What is Janes time?

h d bl ll


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Motion in the same direction problems usually use

d1 = d2Jane and Peter leave their home traveling in the same direction.

Peter drives 55 mph, Jane drives 40mph. Jane leaves at noon, Peter

leaves at 1 P.M. When will Peter catch up to Jane?distance timerate

Jane

Peter

40

55

t + 1

t

Use the formula d = r * t

40(t+1)

55t

40(t+1) = 55t

40t + 40= 55t

40 = 15tt = 2

t = 2

15

10

3

2

What does this mean? When Peter has been traveling 2 hours and

Jane has been traveling 3 , they will have travelled the same

distanceor to put it another way, Peter will have caught up to Jane.

The time will be 3:40 P.M.

3

2 3

2

Janes d = Peters d

d bl ll d d


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Ex. Round trip problemJane goes to the mall. She drives 35mph. On the way home, there is

lots of traffic. She averages 15 mph. Her total travel time was 2 hours.How long did it take her to get to the mall?

distance timerate

there

back

35

15

What should the variable be?

Look at the questionhow long...theres the cluefind the time!

If Janes time there = t

What is her time

home?

Since her total time is 2and it takes her t hours

to get there, her time

home is 2t

t

2 - t

Roundtrip problems usually use d1 = d2

You cant avoid subtraction on this one.

J h ll Sh d i 35 h O h h h i


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Jane goes to the mall. She drives 35mph. On the way home, there is

lots of traffic. She averages 15 mph. Her total travel time was 2 hours.

How long did it take her to get to the mall?distance timerate

there

back

35

15

t

2 - t

Use the formula d = r * td there = d back

35t = 15(2-t)

35t = 30 15t

50t = 30

t =53

35t

15(2-t)

What does this mean? It took Jane of an hour to getto the mall and it took her 1 hours to get home.

If you wanted to convert this to minutes, just multiply

by 60.

5

3

5

2

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