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WEEK 2
SECTION 2.4, 2.5, 2.6 AND
SECTION 2.7
SRABASTI DUTTA
EQUATIONS WITH MANY VARIABLES
Sometimes an equation might have more than one variables.
Example: 2x + 3y = 4. As we learned in the other PPT files, x and y are variables.
The question might ask us to solve either for x or y.
Whenever you are asked to solve for a particular mean, you have to make that
variable stand alone.
Example: Solve for x for 2x + 3y = 4. Here x is not standing alone. First of all it has 2
attached (multiplied) to it. Then 3y is added to it. So, first get rid of 3y: 2x + 3y – 3y
= 4 – 3y  2x = 4 – 3y. Then, divide both sides by 2: 2x/2 = (4 – 3y)/2  x = (4 –
3y)/2 and that’s the solution for x (since it is now standing alone).
Solve for y: 2x + 3y – 2x = 4 – 2x  3y = 4 – 2x  3y/3 = (4 – 2x)/3  y = (4 –
2x)/3 and thus that’s the solution since y is now standing alone.
FUNCTIONS
In the previous slide, we had x = (4 – 3y)/2 and y = (4 – 2x)/3.
So, we expressed x in terms of y in x = (4 – 3y)/2. Mathematically, we say that x
is a function of y.
We expressed y in terms of x in y = (4 – 2x)/3. Mathematically, y is a function
of x.
A function is a rule for determining uniquely values of one variable from values
of one or more variable. When we say that x is a function of y, we mean that we
can determine the values of x from the values of y. Thus, if y = 1, then we will
have x = (4 – 3*1)/2 = (4 – 3)/2 = ½. If y = 2, then, x = (4 – 3*2)/2 = (4 – 6)/2 =
-2/2 = -1.
In such situations, we say that y is the independent variable; x is the
dependent variable (as its values depend on that of y’s).
OUTLINES
•In sections 2.4 to 2.7, you will be learning about solving word problems.
•Many of the word problems will involve concepts such as decimal numbers,
rounding off numbers, area, perimeter, rates, discount, percentage, simple
interest and other.
•For many of such concepts, the book might not have included definitions.
•So, the next few slides will explain many such concepts.
DECIMAL NUMBERS
•We have learned about whole numbers and fractions.
•Decimals are another kind of numbers.
•Decimal numbers are denoted using points, known as decimal points, like 12.5, 181.894,
2222.01, etc.
Decimal Point
•The first number after the decimal is known to be in tenth place; the second number is known
to be in hundredth place; the third number in thousandth place and so on. Thus, 5 in 12.5 is in
tenth place; 9 in 181.894 is in hundredth place; 4 in 181.894 is in thousandth place.
•For a long decimal number, you might be asked to round it of to hundredth place or two
decimal places. That means, there should be only two rounded-off numbers after the decimal
place; if you are asked to round off to thousandth place or three decimal places, that means
there should be three rounded off numbers after the decimal place.
DECIMAL NUMBER PLACE VALUE
Here is the place-value for decimal numbers. Note that the place-values move
from left to right (instead of whole number place-values which moved from
right to left).
The place values start after the decimal point.
Decimal
Point (.)
1/10 = 0.1
1/100 = 0.01
1/1000 =
0.001
1/10,000 =
0.0001
1/100,000 =
0.00001
Places
Tenths
Hundredths
Thousandths
Tenthousandths
Hundredthousandths
Places
First (one)
Second (two) Third (three)
Fourth (four)
Fifth (five)
Example: 2.5 has one decimal place because it has just one number after the
decimal point
Example: 5.871 has three decimal places because it has three numbers after the
decimal point.
CONVERTING DECIMAL TO FRACTION
•Each decimal number can be converted into fraction. To do so, count the
number of decimal places a number has and divide the number by 10, 100,
1,000, etc. according to the number of places
5
•Example: 0.5 has one decimal place. Thus, 0.5 = 10
287
•Example: 2.87 has two decimal places. Thus, 2.87 =
100
NOTE 1: When writing the number in fraction, we do not include the decimal
point.
NOTE 2: The numbers before and after the decimal place are written together as
one whole number in the numerator. The denominator has 10, 100, 1000 based
on number of decimal places and the place-value table (refer to the previous
slide).
ROUNDING OFF DECIMAL NUMBERS
Steps for rounding numbers
 Locate the digit just to the right of the place you want to round off to
 If that digit is less than 5, then omit it and all other remaining numbers to
its. The digit to its left remain as it is.
 If that digit is greater than 5 or equal to 5, then omit it and all other
remaining numbers to its right, and add 1 to the digit to its left.
Example: Round off 5.982317 to two decimal places. 8 is in the second
decimal place. The number to the right of 8 is 2. It is less than 5. Thus all the
numbers from 2 to right is omitted. The final solution is 5.98.
Example: Round off 5.982317 to five decimal places. 1 is in the fifth decimal
place. The number to its right is 7. It is greater than 5. Thus, 7 gets omitted and
1 (the number immediately to the left of 7) is increase by 1. The final solution
is 5.98232.
PERIMETER
•Perimeter of a geometric figure is the sum of lengths of all its sides.
•Example: Suppose we have a triangle with one side being of length 3 inches,
the other side being of length 4 inches and the last side being of length 2 inches.
Find the perimeter of the triangle.
•Solution: add the lengths = 3 + 4 + 2 = 9 inches.
•Perimeter of a rectangle is 2L + 2W, where L and W represents length and
width.
•Perimeter of a square is 4L where L is the length of any one side.
AREA
Area is the total place (surface) enclosed within a figure.
Area of the above rectangle is the whole yellow-region.
Formulas: Area of a rectangle = LW where L is the length and W is the width
Area of a square = L2 where L is the length of one side of the square
Area of a triangle = 1 bh , where b is the base of the triangle and h is the
2
height.
RATES
Rate is a measurement of one quantity against another quantity.
It is written as a ratio
Example: A car travels 100 miles on 16 gallons of gas. Then mile to gas ratio
is a rate: 100 miles to 16 gallons =
100miles
10 *10
5*5 25



16 gallons 2 * 2 * 2 * 2 2 * 2 4
Or
1
6
4
= 6.25 miles/gallon
PROPORTIONS
•When two rates are equal, it is known as proportion.
•If a/b = c/d are two equal ratios, then the statement
a c

b d
is known as a proportion.
PROPERTY OF PROPORTIONS
a c
If  , then ad = bc, that is product of numerator of first ratio and
b d
denominator of second ratio is equal to the product of numerator of second ratio
and denominator of first ratio.
Example: 2  4  2*5  3* 4  10  12
3 5
USE OF PROPERTY OF PROPORTIONS
We use the property of proportions to solve simple equations.
Example 1 Solve for x:
x 3

5 4
Step 1: Apply the property: 4*x = 3*5  4x = 15 (multiplying both sides by 4
and then multiplying both sides by 5).
Step 2: Multiply both sides by reciprocal of 4:
1
1
15
4 x  15  x 
4
4
4
3
Step 3: Write down the answer: x = 15/4 or 3 or 3.75
4
APPLICATION OF PROPORTIONS
A woman drives her car 235 miles in 5 hours. At this rate, how
far will she travel in 7 hours?
Step 1: Convert the word-problem into a proportion:
235 miles to 5 hours = how far to 7 hours
235 how  far


5
7
Step 2: Represent the unknown quantity with a variable x: 235  x
5
Step 3: Solve for x  235*7 = 5*x  1645 = 5x
Multiplying both sides by 1/5 (which is same as dividing both
sides by 5):1645  5 x  329 = x
5
5
Step 4: The woman will travel 329 miles in 7 hours
7
PERCENT
•During the recent election, you must have heard statements like “turnout in
midterm elections is far lower, peaking at 48.7% .”
•48.7% is read as 48 point seven percent
•Percent means per hundred.
•So, 48.7% means 48.7 out of 100
CONVERTING PERCENT TO FRACTIONS
•To convert percent into fractions, write the percent out of 100 and then reduce
the fraction into lowest form
•Example 1: 57% = 57
100
1
1
 
•Example 2: 25% = 25  5*5


100
10*10
 
2* 2
4
CONVERTING PERCENT TO DECIMAL
•Step 1: Write the percent out of 100
•Step 2: Reduce the fraction into lowest term
•Step 3: Write the lowest-term fraction as decimal
•Example: 30%
•Step 1: 30% is 30 out of 100
•Step 2: 30
3*10
3
100

10*10

•Step 3: 3/10 = 0.3

10
SHORT-CUT METHOD FOR CONVERTING PERCENT INTO
DECIMALS
•To change percent to a decimal, drop the % symbol and move the decimal point
two places to the left.
•Example 1: 42% = .42 (note we have shifted two places to the left and put down
the decimal point).
•Example 2: 201% = 2.01
We have shifted two numbers to the left. The two numbers are 01. Thus
put a decimal point before 0 to get 2.01
•Example 3: 80.1% = .801
•Example 4. 0.5% = ._ _ 5(the decimal point was before 5. Shift it two places to
the left; the two places are indicated by the blanks. In the blanks we put zeros.
Thus, the final answer is 0.5% = 0.005
CHANGING DECIMAL TO PERCENT
•There are two ways to convert a decimal to percent.
•Method 1 (short-cut method): move the decimal point two places to the right
and use the % symbol.
•Method 2: Write the decimal as a fraction of lowest term and then multiply the
top by 100; use the % symbol.
EXAMPLE 1
Method 1: Convert 0.45 into percent
Move the decimal two places to the right and use the % symbol.
Thus, 0.45 = 45.% = 45%
45
Method 2: Write 0.45 as a fraction: 100
Multiply the top by 100: 45  100
100
Cancel the 100 from top and bottom and use the % symbol to get
45 100
 45%
100
EXAMPLE 2
•Convert 0.09 into percent
•Step 1: Move the decimal point two places to the right:
.09  09.
•Step 2: 09 is nothing but 9
•Step 3: using % symbol we have 9% as the solution
CHANGING FRACTIONS TO PERCENT
•To change a fraction into percentage, we can follow either of the two methods:
•Method 1: Convert the fraction into a decimal and change the decimal into
percent
•Method 2: Multiply the fraction by 100 and simplify (that is make sure
everything is in lowest term).
EXAMPLE
Convert ¾ into percent
Method 1: Step 1: ¾ = 0.75 in terms of decimal.
Step 2: Shift the decimal point two places to the right to get
75%
3
300
 75%
Method 2: Multiply ¾ by 100 and simplify: 100 
4
4
BASIC PERCENT PROBLEMS
•The basic percent problems can be of three types:
1. What number is 20% of 52?
2. What percent of 52 is 27?
3. 30 is 50% of what number?
To solve any of the above questions, you first have to convert
the word-problem into equation and then solve. We will only
look at number 1.
WHAT NUMBER IS 20% OF 52?
The number is unknown.
We know that the number = 20% of 52
Represent the number as n
Thus, n = 20% of 52. “Of “is nothing but multiplication.
Thus, we have the equation n = 20%*52
 n = 20  52  10.4
100
COMMISSION
•Commission is the earning that a salesperson earn based on a
percentage of amount the salesperson sell.
•The commission rate is also always represented in percent.
•Commission problems are nothing but percent application
problem.
•To find the commission, you convert the word problem into an
equation and solve the equation.
EXAMPLE -- COMMISSION
A real estate agent has a commission rate of 3%. If a piece of
property sells for $94,000, what is her commission?
Commission = percent of earning.
Percent = 3%
Earning = $94,000
Therefore, commission = 3% of $94,000 =
Thus, the commission is $2,820
DISCOUNT
Discounts are also percentage problems as discounts are always represented in
percentage.
During a clearance sale, a three-piece suit that usually sells for $300 is marked
15% off. What is the discount? What’s the sales price?
Discount = 15% of $300 =
15
 $300  $45
100
Sales price = Original price – discount = $300 - $45 = $255
SIMPLE INTEREST
•Whenever we borrow money, we have to pay interest on the amount of money
borrowed.
•Whenever we deposit money in any account, the money earns interest.
•There are two types of interest – simple interest and compound interest
•In this course, we will only learn how to solve problems involving simple
interest
•A simple interest is nothing but the interest that is paid on the principal alone.
FORMULA
•Principal – the amount that is invested.
•Interest rate -- the percent of principal that is earned (in case of investment) or
the percent of principal that needs to be paid (in case of borrowing money)
•Interest – the amount of money earned
•Therefore, the simple interest formula is I = PRT, where I = interest, P =
principal, R = rate, T = time
EXAMPLE
•A farmer borrows $8,000 from his local bank at 7%. How much
does he pay back to the bank at the end of the year when he pays
off the loan?
•Step 1: Write down the values of P, R and T: P = $8,000; R =
7%; T = 1
•Step 2: Convert percent into decimal: 7% = 0.07
•Step 3: Apply the formula I = PRT  I = $8,000*0.07*1 =
$560
•So, he pays back $8,000 + $560 = $8,560 at the end of the year.
ANGLES
•Complimentary angles: if two angles add up to 90◦, we call them
complimentary angles
•Supplementary angles: if two angles add up to 180◦, we call them
supplementary angles.
Examples:
1. Angle 43◦ and angle 47◦ are complimentary angles because 43 + 47 = 90
degree
2. Angle 102◦ and 78◦ are supplementary angles because 102 + 78 = 180
degree.
3. Find x such that x and 51◦ are (a). complimentary angles, (b).
Supplementary angles.
(a). X + 51 = 90  x = 90 – 51 = 39 degree
(b). X + 51 = 180  x = 180 – 51 = 129 degree
STRATEGIES FOR SOLVING WORD PROBLEMS
In summary, here are the rules you should follow in solving any word problems:
1. Read the problem and underline all the information that are given in the
problem.
2. Write down what’s being asked to solve in the problem.
3. Whatever is asked being solved or whatever the unknowns are, represent it
(them) with variables (x, y, z, w, etc).
4. Use formulas if applicable
5. Draw pictures to represent the problem. Often time it helps.
6. Write equations/algebraic expressions to represent the unknown variable in
terms of the known variables/value/information given in the problem.