Author: Thomas Geisler
Last updated: June 24, 2013
Unit 2
Solving Distance Word Problems
and Linear Equations
Avery Point Academic Center
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 2variable 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 = v a t − v bt
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.
Linear Equation Problem - 3
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.
Linear Equation Problem - 4


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.
Linear Equation Problem - 6
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:
Linear Equation Problem - 7
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
Linear Equation Problem - 8
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
Linear Equation Problem - 9
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. . . .