Algebra I: Translating Words Into Symbols (Level 2 of 2) | Simple Phrases, Formulas


Translating Words Into Symbols (level 2)
Alright we are going to work on various examples. It is very important that you understand how
to translate phrases into variable expressions; this is the first step in solving more complicated
word problems that you will encounter in your studies of algebra and other related science
classes such as physics and chemistry. Alright let’s try the following examples:
Translate each phrase into a variable expression our first phrase is:
“10 more than a number” This phrase is describing an addition between
two numbers because of the key word, more so this phrase translates to 10 + n or n + 10
either form represents the same translation of the phrase.
Let’s try this one: “Five times a number”
This phrase is describing a product because of the key word times; this phrase translates
to five times n or 5n. Let’s try the next one:
“A number decreased by 11” This phrase represents a subtraction between
2 numbers because of the key word decreased, so this phrase translates to n – 11. Many students
have a hard time with phrases that involve subtraction because they are not sure which
numbers are being subtracted from each other. For example the variable expression 11 – n
is read “11 decreased by a number” These two expressions are completely different because
for the most part they yield different results when you substitute the variable n with a
random number. So be extra careful when translating phrases involving subtraction make sure you
are subtracting the numbers in the correct order.
Alright let’s try the next one: “The difference between a number and 2”
Once again we have a subtraction because of the key word difference, so this phrase translates
to n – 2. Alright let’s go to the next phrase: “The quotient of 15 and d”
Here we have a division because of the key word quotient, this quotient involves the
number 15 and the variable d so the phrase translates to 15 divided by d or 15 over d
both answers are acceptable but once again we usually want the answer with the fraction
bar. Let’s try a slightly more complicated one:
“6 more than twice y” This phrase has two key words, more which
means addition and twice which means multiplication so its 6 more than 2y or 2y + 6. Notice that
when we are dealing with addition the order in which the numbers are added is not important
a word of caution this is not the case for subtraction so be careful with phrases
involving the subtraction operator the order in which the numbers are being subtracted
matters. Let’s try the next example:
“10 times the sum of a number and 8” Here we have a multiplication because of the
word times and we have an addition because of the word sum. Also, notice that the product
involves the number 10 and the sum of a number and 8 we are going to need grouping symbols
for this one, so the phrase translates to 10 times the quantity n + 8. Whenever you
have multiplication its always between 2 numbers. In this case one number is 10 and the other
was the sum of a number and 8. So we need to use grouping symbols to group the expression
formed by the sum which is then multiplied by 10.
Alright let’s try the next one: “The difference between three times a number
and one” So here we have a subtraction because of the
key word difference, the two expressions that are being subtracted are 3n and 1 so the final
answer is 3n – 1 Let’s try the following phrase:
“Twice a number, increased by 3” Here we have a sum because of the key word
increased, so this phrase translates to 2n + 3 or 3 + 2n, whenever you have phrases such
as twice, tripled, half, a third, these phrases denote a multiplication.
Alright so now let’s go over examples that involve translating phrases that are slightly
more practical. Answer the following with a variable expression.
Bob is 2 ft taller than Joe. If Joe’s height is J ft, how tall is Bob?
It’s important to read the question as many times as it takes you to understand it completely.
The question is describing the heights of Bob and Joe. The question is asking how tall
Bob is given the fact that Bob is 2 ft taller than Joe, we are also told that Joe’s height
is represented by the variable J. So Bob’s height is equal to Joe’s height + 2 or written
algebraically j + 2 ft and this is our final answer. Let’s try the next one: Maria is 10 in. shorter than Claudia. If Claudia’s
height is C inches, how tall is Maria? Once again re-read the problem over and over
until you understand it. The question is asking how tall Maria is given the fact that Maria
is 10 inches shorter than Claudia whose height is represented by the variable C. So since
Maria is 10 inches shorter than Claudia we take Claudia’s height C and subtract 10
from that. This results in a final expression C – 10 inches. Which represents Maria’s
height. Let’s try a slightly more challenging one:
Sam is 3 feet taller than Dan. If Dan’s height is D ft, how tall is Sam? Also,
If Sam’s height is S ft, How tall is Dan? Let’s answer the first question; we know
that Sam is taller than Dan by 3 feet. We also know that Dan’s height is D feet so
to find Sam’s height we take Dan’s height and add 3. This results in a final expression
of D + 3 feet. Notice that this expression is from the reference point of Sam, imagine
that you are Sam and you glance at Dan you see that you are taller than Dan. So this
expression makes sense. The second question tells us that Sam’s height is S feet and
is asking us how tall is Dan. So if you were Dan and look at Sam you would notice that
you were shorter than Sam so your height would be equal to Sam’s height minus 3 this results
in a final expression of S – 3 feet. The first variable expression we obtained is from the
reference point of Sam and the second variable expression is from the reference point of
Dan. Both contain the same constrain the fact that Sam is 3 feet taller than Dan. The variable
expressions are different depending on whose point of view we focus on. At times it’s
useful to put yourself on the characters shoes and think about their perspectives relative
to the other. Alright let’s work on questions that require
the use of formulas Answer the following with a variable expression.
Matthew drove for H + 6 hours at a constant rate of R + 5 mph. How far did he go?
This problem is a typical distance equals rate times time problem so we use the distance
traveled formula at a constant rate. D=rt In this problem the rate is equal to the quantity
R + 5 so we use grouping symbols and substitute this expression for r on our formula and in
the same manner the time interval traveled is equal to the quantity h + 6 so we substitute
this expression using grouping symbols for the variable t our final variable expression
is equal to the quantity R + 5 times the quantity H + 6 Miles.
Alright lets try the next one: Pencils cost W cents each and note books cost
N cents each. How much will 3 pencils and 2 notebooks cost?
This problem is an application of the cost formula C=n times P, but first notice that
we have 2 distinct items in this case pencils and notebooks so to find the total cost we
need to add the total cost of the pencils and the total cost of the notebooks. The total
cost of the pencils is equal to 3w and total cost of the note books is equal to 2N so the
total cost is equal to 3W + 2N cents. This expression represents the total cost of 3
pencils and 2 notebooks. Notice that the formula we used was a slightly modified version of
the cost formula because of the fact that we had 2 different types of items in this
case pencils and notebooks. Make sure that you understand the problem before you blindly
use a formula. Alright let’s work on the final example.
A rectangle has a length of 5R yd and width of 2S yd. What is the perimeter and area of
the rectangle? To find the perimeter of the rectangle we
use the formula P=2L plus 2W or just add all the sides of the rectangle so using the
formula and simplifying the expression we have that the perimeter is equal to 10R + 4s.
In the same manner the area of the rectangle can be found by using the formula length times
width so substituting the expressions into the formula and simplifying we have that the
final variable expression for the area of the rectangle is equal to 10 times R times S or 10RS squared yards, and that’s our final answer. Alright it’s important that you develop
this important skill of translating phrases into variable expressions and start thinking
algebraically by working solely with variable expressions as oppose to numerical expressions.
Okay in our next video we will expand on this skill of translating phrases into variable
expressions by taking sentences and translating them into equations.

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4 thoughts on “Algebra I: Translating Words Into Symbols (Level 2 of 2) | Simple Phrases, Formulas

  1. This is so helpful! All through high school I never could figure out word problems but after watching this multiple times I think I have it! (But I’m going to watch s few more times to make sure) thank you!!

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