Translating Problems into Equations (Level
2) In this video we are going to continue translating
problems into equations, this time around we will go over more challenging examples.
Making sure we use the 3 step plan introduced in the previous video. Alright let’s go
over the first example. Translate each problem into an equation.
The length of a rectangle is one meter more than its width. The perimeter of the rectangle
is 60 m. In this example we are dealing with a rectangle
so let’s go ahead and draw a rectangle to help us visualize and analyze this problem.
Recall that the first step, is to identify the unknown variables, in this example the
unknown quantities are the length and width of the rectangle. Having identified our unknowns
we now have to assign variables to them. As As a rule of thumb we first want to assign the
variable to the quantity for which we have the least amount of information. In this problem
notice that we have a description that relates the length with the width, but we don’t
have a description that relates the width with the length. This means that we should
assign the variable to the width so let’s assign the letter x to it. Now that the width
is represented by the variable x, the length of this rectangle would be denoted as x + 1
since the length is one meter more than the width. Having assigned variables to both unknown
quantities we are ready to translate the word problem into an equation. Notice that we are
given that the perimeter of this rectangle is equal to 60 meters. We are going to use
this fact in combination with the variable expressions and relate them to form an equation. In
order to accomplish this, we need to make use of a formula specifically the formula
for the perimeter of a rectangle. The perimeter is just going to be equal to the sum of all
the sides of this rectangle in other words twice the length plus twice the width. In
this problem we know that the perimeter is equal to 60, in addition we also know that
the width is represented by the variable x, and the length is represented by the variable
expression x + 1, substituting these expressions into the formula results in the following
equation, 60=2x + 2(x+1) and this is our equation. Keep in mind that you might have
to use formulas in order to successfully translate a word problem into an equation. It all depends
on the type of word problem that you are face with. Alright let’s try the next example. A triangle has two equal sides and a third
side that is 15 cm long. The perimeter of the triangle is 50 cm. In this example we are dealing with a triangle
so let’s go ahead and draw a triangle. Lets first identify the unknowns in this problem,
notice that we are given the measurement of one side of the triangle but have no idea
what the measurement of the other two sides are. For this problem the unknowns are the measurements
of these two sides of the triangle. Next let’s assign variables to these unknowns, from the
first sentence we are told that this triangle has two equal sides so we are actually dealing
with an isosceles triangle, a triangle that has two of its sides with the same measurement
this means that we can assign the same variable to both sides since they are equivalent, so
let’s assign each side with the variable x as follows. Now it’s just a matter of translating
these expressions into an equation. We first need to figure out how to relate all of these
expressions with one another. Notice that we are told that the perimeter is equal to
50 cm. We can relate all the expressions by using the formula for the perimeter of a triangle.
We essentially need to add all the sides of the triangle, so our equation will be equal
to: x + x + 15=50, and this is our final answer. Alright up to this point we have been
going over examples that contain two unknowns, Now let’s try some examples that contain
3 facts about 3 unknown. Luis is three times as heavy as his luggage.
His luggage is 20 lbs heavier than his backpack. The weight of Luis, his luggage and back pack
total 170 lbs. In this example we actually have 3 unknowns
as oppose to 2 unknowns like in the previous problems. Regardless of the number of unknowns
the procedure is essentially the same we first need to identify our unknowns. After reading
the problem it is clear that we are dealing with three unknowns, the first unknown is
Luis’s weight, the second unknown is the weight of the luggage, and the third unknown is
the weight of the back pack. Having identified the unknowns we now need to choose a variable
and represent the rest of the unknowns with this variable. First, let’s figure out which
unknown we absolutely have no information about. We know that Luis’s weight is 3 times
the weight of his luggage, we also know that his luggage is 20 lbs heavier than his back
pack, but we have no idea how the backpack is related to the other unknowns, so let’s
go ahead and represent the weight of his back pack with the variable x, this way the weight
of the luggage can be expressed as x + 20, and Luis’s weight can be expressed as 3
times the weight of the luggage or using our expressions as 3(x + 20), now we need to translate
these expressions into an equation. In the last sentence we are told that the combined
weight of the luggage, back pack and Luis’s weight is equal to 170 lbs. So we can express
this fact as follows, x + (x+20) + 3(x+30)=170 and this is our equation. When dealing
with multiple unknowns the procedure is essentially the same, just be a bit more careful and make
sure you relate the correct unknowns with one another. Alright let’s go over the next
example. Maribel, Dave, and Henry have \$180 together.
Maribel has three times more money than Dave. Dave has two times more money than Henry. Similar to the previous problem we first need
to identify all of the unknown quantities. In this particular problem the unknown quantities
are the amount of money that each person has. In this example we have three unknown quantities.
Once we have determined the unknown quantities we are ready to assign variables to them.
Let’s first determine which unknown quantity we have the least information on. We know
that Maribel has 3 times more money than Dave. We also know that Dave has two times more
money than Henry, but we have no information of how Henry’s amount is related to the
other two amounts. Hence we will assign the variable x to Henry’s amount, this means
that Dave’s amount will be represented as: 2x, and lastly Maribel’s amount will be
represented as: 6x. Having assigned a variable to each unknown quantity we are ready to translate
the word problem into an equation. We know that if we were to add the amount of all three
individuals it would add up to 180 dollars, so we can denote this relation as: 6x + 2x
+ x=180, and this is our equation. Alright let’s go over the final example.
A pipe ten feet long is cut into three pieces. One piece is one foot longer than the shortest
piece and two feet shorter than the longest piece.
Alright this word problem is perhaps the most challenging problem in this video. Let’s
use everything that we learned to tackle it. Let’s first make a drawing of our pipe broken
into 3 pieces. Let’s first start by identifying the unknowns, in this example we have no idea
how much each of the pieces measure. So the length of each of the pieces will be our unknowns.
Next let’s go ahead and assign variables to each of the unknown lengths. Because we
have three distinct pieces we will label them as the shortest, medium and longest piece
so we can easily distinguish them. From the second sentence we are told that one piece
is one foot longer then the shortest piece. So let’s go ahead and assign a variable
to the shortest piece, let’s assign the variable x, with this assignment the medium piece will
be denoted as x + 1. Finally we are also given that this same piece (the medium piece) is
two feet shorter than the longest piece. Let’s think about this, if the medium piece is two
feet shorter than the longest piece that means that the longer piece is 2 feet longer than
the medium piece, in other words it can be denoted as x + 1 the length of the medium
piece + 2 since it is 2 feet longer. Having assigned the variables to all three unknowns
we are ready to translate the word problem into an equation. From the first sentence
we know that the original pipe length was equal 10. So if we were to add the individual
pieces they should add up to 10. So our final equation is equal to: x + (x+1) + [(x+1)+2]
=10. Alright make sure you read the word problem as many times as you need to. You
really need to understand the problem before you start identifying unknowns and assigning
variables to them. Ok in our next video we will expand the skills developed so far and
start solving word problems over a given domain.