unit 2 – solving distance word problems and linear equations

Avery Point Academic Center

Author: Thomas Geisler

Last updated: June 24, 2013

Avery Point Academic Center

Unit 2

Solving Distance Word Problems

and Linear Equations

EXAMPLES: DISTANCE PROBLEMS

AND LINEAR EQUATIONS Following are examples of using these steps to solve word

problems using fairly simple math (i.e., distance traveled at various speeds for various periods of time and 2-variable pairs of linear equations), plus exercises for you in the same topics.

Thereafter, several Module Units are included which apply the steps to problems in various other subject matters, including interest problems, more complex linear equation problems (supply/demand), projectile motion problems, and gas law problems. Practice exercises are provided in each Module Unit.

The best way to learn how to solve problems is by practicing solving problems.

Example: Speed/Distance Formulas -I

Let’s say you have been studying the

relationship between speed (velocity) and

distance traveled. You have learned the

following two formulas.

Formula 1: The total distance traveled at a

constant speed over a period of time is:

d = vt

(d is distance; v is velocity; t is time)

d = vt

The graph above depicts the formula d = vt, where the x axis is t, the time period of movement at v, the speed. The y-axis shows the total distance traveled. (Here, each unit of x is one hour and each unit of y is 100 miles; the v depicted is 60 miles per hours.

Example: Speed/Distance Formulas -II

Formula 2: If two objects, a and b, start from the

same point, at same time, and in the same

direction at the respective speeds, va and vb,

(where va is greater than vb), the distance

between a and b at time t is:

d = vat − vbt

d = t(va − vb)

(d is the distance between a and b; t is the time

elapsed; va is a’s speed and vb is b’s speed.)

Problem 1- Solution I Solve the following problem based on these two formulas:

Problem 1: “Charlie gets into his new convertible and drives west along Route 95 at 60 miles an hour, starting in Mystic, for three hours. How far does he go?”

Solution:

1. First, read and reread the problem until you understand it.

2. Organize the data: Charlie drives West

on Route 95

at 60 miles per hour

starting from Mystic

for three hours.

Some of this data appears irrelevant to solving the problem: the direction, the highway, and the starting location.

We can write the rest of the data using symbols common in distance formulas:

v = 60 miles/hour

t = 3 hours

Problem 1- Solution II

3. The problem asks how far Charlie traveled, i.e., the

distance he goes moving at that speed and for that

period of time. The unknown is the total distance

traveled, which we call d since both formulas use d

to represent distance.

4. Since the problem is that of a single moving object

(Charlie) and since we are given v and t and asked to

find d, Formula 1 seems the right mathematical

model, or formula, to solve this problem:

d = vt

Problem 1- Solution III

5. Substitute the data into the formula:

d = vt d = (60 mi./hr.)∙(3 hrs.)

6. Then solve mathematically

d = 180 miles

7. We multiplied the units of the terms together:

(miles/hour)(hours) = miles, so the answer is

in the proper unit of measurement.

Answer: Charlie went a distance of 180 miles.

Problem 2 – Solution I

Problem 2: Arnold and Bill

begin driving east on

Route 95 from New

London at the same time.

Arnold’s luxury

convertible travels at 65

miles an hour. Bill’s car,

an old jalopy, travels at of

50 miles an hour. After

they have driven for 90

minutes, how far is

Arnold ahead of Bill?

Problem 2 - II

1. First, read and reread the problem until you

understand the problem, what it tells you and

what it asks.

2.-7. YOUR TURN.

Using the steps for turning word problems into

math problems to solve problem 2.

What is the correct answer?

Problem 2 - III

The correct answer is 22.5 miles.

If you got the correct answer, congratulations!

If not, the following analysis shows how to use

the steps to solve the problem.

Problem 2 - IV 1. Read the problem until you understand it.

2. Organize the given data:

Arnold’s speed is 65 mph

Bill’s speed is 50 mph;

each drives 90 minutes

they start at the same point

they drive in the same direction.

The starting point [New London], the condition of the cars, and which road they travel don’t seem relevant.

The relevant data is stated as follows (with subscript a for Arnold’s speed and b for Bill’s speed):

va = 65 mi./hr.

vb = 50 mi./hr.

t = 90 minutes

Problem 2 - V

3. The problem asks how far apart Arnold and Bill will be after 90 minutes of driving? The distance between them is designated the unknown d, since both distance formulas use d for distance.

4. Of the two mathematical formulas, Formula 2 seems a better fit, since it involves two moving objects (here, Arnold and Bill). The problem data fits Formula 2, and the distance between the two objects (the d in Formula 2, is precisely what is asked for:

d = t(va − vb)

Problem 2 - VI

5. We substitute the data into the formula:

A. However, before we do that, we notice that the

two speeds are given in hours, but the period of

travel is given in minutes.

B. The units of time must be consistent, i.e.,

expressed in the same units. Therefore, we

translate 90 minutes into 1.5 hours (using what

we all know: 1 hour = 60 minutes). We are then

ready to substitute the data into Formula 2:

Problem 2 - VI 5. So now we can go ahead and actually substitute the

data into the formula:

d = t(va − vb)

d = 1.5(65 − 50)

6. We then solve the problem mathematically:

d = 1.5(15)

d = 22.5

7. Multiply the units together as we did the numbers:

(hours)(miles/hour) = miles

Answer: The final answer is 22.5 miles.

Additional Exercises

Other problems involving these motion/distance

formulas are supplied in the accompanying

written materials for you to try. If you have

difficulty solving them, guidance is available

from tutors or in written form.

Linear Equation Problem - 1 Problem: “Phil wants to buy a soft drink in a

store. He has a total of thirteen coins on him: some are dimes and the others are quarters. The soft drink costs exactly $2.50. He counts up the value of his coins and finds he has exactly $2.50, so he can buy the soft drink. How many dimes and how many quarters did Phil have?”

This kind of problem does not involve a formula, as in the distance problems, but instead involves a different mathematical model (i.e., how to solve a pair of linear equations).

Linear Equation Problem - 2

Solving a pair of linear equations involves four

steps:

1. put the equations in standard form (the y alone

on the left-hand side of each equation);

2. equate the two right-hand sides of the standard

form equations (which contain the x-terms);

3. solve for x.

4. substitute that value for x in one of the equations

and solve for y.


We can use the basic steps in translating math

word problems in this situation as well.

1. First, read and reread the problem until you

understand what it is telling you and what it

is asking you.

2. Organize the data:

The problem involves Phil’s coins, which are of

two types: dimes and quarters.

The total number of the coins equals 13.

The total value of the coins is $2.50.


The quantities involved (number of coins, value of coins, number of dimes, number of quarters) do not seem related to any formula we have studied.

Therefore, we cannot assign specific variables from formulas to the known quantities.

3. The two unknowns which we are asked to determine are the number of dimes and the number of quarters Phil has. Since they do not seem related to any formula we have learned, we assign them the generic variables, x and y.

Linear Equation Problem - 5 4. The problem gives us two different relationships

involving the dimes and quarters:

The mathematical model does not seem to be a formula.

But we have studied how to solve pairs of linear

equations in two variables, and the information the

problem tells us seems to fit into two separate equations:

The coins total 13 coins.

The total value of the coins is $2.50

We can express these relationships can be expressed as

two linear equations. The mathematical model we use to

solve this problem is the 4-step method of solving a pair

of linear equations in two variables.


5. We create from the problem information the

following specific two linear equations (substituting

the data the problem gives us for the generic

coefficients and constant of a linear equation

(ax + bx = c):

x + y = 13

10x + 25y = 250

(Each dime is 10 cents and each quarter 25 cents.)

6. Now we can solve the problem using the four-step

linear equation method, as follows:


Four-Step Solution:

1. Turn both linear equations into standard

form:

y = −x + 13

y = −(2/5)x + 25

2. Equate the right-hand side of both standard form

equations:

−x + 13 = −(2/5)x + 10


3. Solve for x:

−x + 13 = −(2/5)x + 10

−x + (2/5)x = 10 − 13

(−3/5)x = −3

(−5/3)(−3/5)x = (−5/3)(−3)

x = 5


4. Now we have a value for x, namely, x = 5, so we

substitute 5 in place of x in either of the original two

equations (preferably in standard form). Here, the

first equation seems simpler:

x + y = 13

5 + y = 13

y = 8

Solution: Phil had 5 dimes and 8 quarters.

(In this case, 7., is taken care of because the two

variables were simply the numbers of each type of

coin; no complicated units were involved.)

Additional Linear Equation Problems

Other, more complicated linear equation

problems are contained in the separate unit

going into more detail as to linear equations.

Now let’s practice grammar,

punctuation and

diagramming sentences!!!

Or not. . . .

unit 2 – solving distance word problems and linear equations

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