Math Notes-Homework Helper.docx

Natural and Whole Numbers, Integers, Rational and Irrational Numbers

If drawn carefully, the number line has 0 in the middle, with the positive numbers increasing on the right, and the negative numbers increasing to the left.

On this number line we note that we can see the set of all natural numbers {1, 2, 3, 4, 5….}

We also see that we have the set of whole numbers, which is the same as the natural numbers with zero added. {0, 1, 2, 3, 4……}

Also, on this number line, we notice that we have the set of integers {….-4, -3, -2, -1, 0, +1, +2, +3, +4….}

The dots (….) indicate that the numbers continue to increase.

Besides these numbers, we have some other numbers that can be placed on the line that are not necessarily included in the numbers we have introduced so far.

We have the set of rational numbers which might also be called fractions, because they are defined as numbers that can be written as a/b where a and b are both integers and b is not zero.

By this definition, the following are rational numbers: ½ 1/3 2/5 5 -2 0

But the following are not rational numbers:

4/0 π

√

−

1

The first is an “undefined” number since division by zero is impossible. The second is called an irrational number, but it is a very useful one. The approximate value of pi is:

(Some people learned that pi = 22/7. This is inaccurate. 22/7 is an approximate value for pi but not its exact value)

√

−

1

√

that you wish to find what number multiplied by itself will give you the number under the

√

Also, we can build what’s called a Venn Diagram that shows symbolically how these kinds of numbers relate to each other.

We can start with the natural numbers. The circle represents all the natural numbers. 1,2,3,4,5… etc.

\

Included in the set of whole numbers is one extra number, zero, that is not part of the set of

natural numbers. So, we can make the drawing anew:

\

We say that the set of natural numbers is a subset of the set of whole numbers, because every natural number is included in the set of whole numbers. The opposite in not true.

Then we add the set of integers. Integers are both positive and negative whole numbers, and zero. So now, we have:

\

1,2,3,4…

1,2,3,4… 0

1,2,3,4… Natural

0

If numbers are real numbers (not imaginary), and they are not integers, then they must either be rational numbers (can be written as a/b where b  0) or they are irrational numbers.

Note that there is no intersection between the set of rational numbers and the set of irrational numbers.

1,2,3,4… Natural Numbers

0

…-4, -3, -2, -1

Whole Numbers Integers Rational numbers ¼, - ½, - ¾, + ½ Real

Numbers

Order of Numbers on the Number Line and Absolute Value

Now, let’s return to the number line.

There are a couple of important concepts to introduce at this point.

1) The number farther to the right on the number line is always larger than any number to its left.

So…

+5 > +3 (easy) +3 > 0 (also easy) 0 > -1 (not so easy) -1 > -2 (counterintuitive)

Of course, the symbol > means “greater than” Reversed, the symbol < means “less than” So,

5>3 is read “5 is greater than 3” while 3<5 is read “3 is less than 5”

2) The absolute value of a number is always positive and represents its distance from zero on the number line.

| +3 | = 3 | 0 | = 0 | -2 | = 2

By convention, the absolute value is written without a sign, but is understood to be positive.

Fractions and Decimals

Now, lets get more deeply into looking at fractions and decimal numbers.

Let’s consider adding two fractions with the same denominator

2/7 + 3/7

If two fractions have the same denominator, addition is simple. Just add the numerators. 2/7 + 3/7 = 5/7

However, if two fractions have different denominators, then we must find a common denominator before we can do the addition.

2/7 + 3/5 = ?

One sure way to find a common denominator is to multiply the two denominators together. 7x5 = 35

So, how do we get 2/7 and 3/5 as fractions with a denominator of 35?

We take both the numerator and the denominator of the fraction, and multiply it by the same number.

2 x 5 = 10

-

--7 x 5 = 35

3 x 7 = 21

-

---5 x 7 = 35

More complicated is an addition of three fractions

2/7 + 3/5 + 1/4

There are two ways to proceed here:

Notice that we have already done the sum of 2/7 + 3/5 above, so we can do the addition like ( 2/7 + 3/5 ) + 1/4

(10/35 + 21/35) + 1/4 (31/35) + 1/4

Now we have the problem 31/35 + 1/4 to solve

A common denominator between 35 and 4 is 35 x 4 = 140 31 x 4 124

-- = ---35 x 4 140

1 x 35 35

- =

----4 x 35 140

The second way we can solve this problem is to recognize that we’re going to need a common denominator between 5, 7, and 4 to begin with, so we multiply all the denominators together to get a common denominator.

Note: if you have a string of fractions to add or subtract, multiplying all the denominators together will always give you a common denominator, though not necessarily the lowest common denominator.

5 x 7 x 4 = 140 So we want the denominator of all these fractions to be 140

2 x 5 x 4 40

- =

---7 x 5 x 4 140

3 x 7 x 4 84

- =

---5 x 7 x 4 140

1 x 5 x 7 35

- =

---4 x 5 x 7 140

Adding, Subtracting, Multiplying, and Dividing Positive and Negative Numbers

Adding

There are just two simple rules for adding positive and negative numbers.

1. If the numbers have the same sign, add their absolute values, and keep the sign. (+5) + (+2) = +7

(-4) + (-5) = - 9

2. If the numbers have different signs, subtract their absolute values, and keep the sign of the number with the biggest absolute value.

(+5) + (-3) = +2 (+5) + (-7) = -2 (-6) + (+2) = -4 (-4) + (+10) = +6

Subtracting

You can convert all subtractions to additions. What we need to understand is the concept of the

negative of a number or the opposite of a number.

Consider - (+3) can be converted to -3 - (-5) becomes +5

Here’s the rule: If there is a negative sign in front of parentheses, remove the parentheses and change the sign of whatever is inside.

So, the following expression:

(+3) – (- 4) + (- 2) – (+5) – (- 1)

…after removing the parentheses, becomes +3 + 4 - 2 - 5 +1

I like having only one sign attached to a number. As long as I follow the rules, then I can be confident that all I need to do now is add all the numbers that remain. (using the rules for addition)

From left to right: + 3 + 4 – 2 – 5 + 1 = + 7 - 2 – 5 + 1 = +5 - 5 + 1 = 0 + 1 = +1

Multiplication and Division:

Multiplication and Division of positive and negative numbers is actually much simpler than Addition and Subtraction. Here are the rules:

1. If multiplying or dividing two numbers with the same sign, the result is positive.

(+5) x (+2) = + 10

(-3) x (-4) = + 12

(-10) / (-2) = + 5

2. If multiplying or dividing two numbers with different signs, the result is negative.

Putting all the operations together

When combining addition, subtraction, multiplication and division with negative and positive numbers you now have all the rules you need except for the order of operations.

Rule for the Order of Operations

1. Do operations between parentheses or brackets first, starting with the innermost. 2. Then, from left to right, do the multiplications or divisions.

3. Then, from left to right, do the additions or subtractions.

Example:

{ (+10) x [(+3) + (-5)] } / { (-7) + [ (-22)/(-2) ] }

= { (+10) x [ - 2] } / { (-7) + [ + 11]}

= (-20) / (+4)

Further Operations with Fractions:

The Rules

1. When multiplying a fraction by a whole number, multiply it by the numerator only.

Ex: 1/3 x 5 = 5/3

2. When multiplying two fractionstogether, multiply the numerators together, and the denominators together.

Ex: 2 3 6

- x -- =

--3 7 21

3. When dividing two fractions, take the second fraction, flip it, and multiply.

Ex: 2 5

--  -- =

3 8

2 8 16

-- x -- =

--3 5 15

Knowing these rules how would you divide the following ½ 5

Converting Fractions to Decimals

We can convert fractions to decimals by doing the long division.

3/8 =

3/8 =

It’s just like normal long division, we just have to keep track of the decimal point. For the number under the division sign (3), we just keep adding zeros to the right of the decimal point until the division ends up with a zero remainder…

Notice here that we finished after we received zero as a remainder, but sometimes, we never get there…

3 8

3.0 2 4 6 0 5 6 4 0 4 0 0 8

1/3 =

Note that as we divide 1 by 3, we keep getting 3 times with a remainder of 1. So, the most accurate representation of 1/3 in decimal form is 0.333

where the bar over the 3s indicates that it repeats forever…

We can also have a whole string of numbers to repeat. Consider 2/7

2/7 =

1.0 9 1 0 9 1 0 9 1 3

0.3 3 3

2.0 1 4 6 0 5 6 4 0 3 5 5 0 4 9 1 0 7 3 0 2 8 2 0 7

0.2 8 5 7 1 4

Converting decimals to fractions:

Converting decimals to fractions is much, much easier. Here are the rules.

1. If there is only one number to the right of the decimal point (.), write it over 10. Ex: 0.4 = 4/10

3.2 = 32/10

2. If there are two numbers to the right of the decimal point (.), write it over 100.

Ex: 0.47 = 47/100 1.32 = 132/100

3. For each digit after the decimal place, add one more zero to the denominator Ex: 0.473 = 473/1000

1.3321 = 13321 / 10000 4. Reduce the fraction if possible

Ex: 0.5 = 5/10 = ½ 0.25 = 25/100 = ¼

Fractions are reducible if both the numerator and denominator are divisible by the same number.

Ex: 2/6 is reducible to 1/3 because the numerator and denominator are both divisible by 2

Rules for Divisibility

2. A number is divisible by 2 if the last digit is 0, 2, 4, 6, or 8 Ex: 16654 is divisible by 2

16653 is not divisible by 2

3. A number is divisible by 3 if the sum of the digits is divisible by 3

Ex: 1236 is divisible by 3 because 1+2+3+6 = 12 which is divisible by 3 1234 is not divisible by 3 because 1+2+3+4=10 which is not divisible by 3 4. A number is divisible by 4 if the last two digits are divisible by 4

Ex: 12453940 is divisible by 4 because the last two digits, 40, is divisible by 4

12453938 is not divisible by 4 because the last two digits, 38, is not divisible by 4 5. A number is divisible by 5 if the last digit is 0 or 5

Ex: 12459605 is divisible by 5 12459606 is not divisible by 5

6. A number is divisible by 6 if it is divisible by both 2 and 3 Ex: 123 is divisible by 3 but not 2

124 is divisible by 2 but not 3 126 is divisible by 6

9. A number is divisible by 9 if the sum of the digits is divisible by 9 Ex: 123 is not divisible by 9

Adding and Subtracting Decimal Numbers

The Rules

1. Make sure that all the numbers have the same number of digits after the decimal point. (You can always add zeros to the right of the decimal point)

2. Add or subtract as normal, keeping the decimal point in place.

Ex: 0.2 + 0.13 + 0.255 =

0.200 0.130 0.255 ---0.585

Ex 1.2 - 0.355 =

1.200

- 0.355

Multiplying Decimals

The Rules

1. Multiply as you would any numbers, disregarding the decimal point 2. Add the number of digits after the decimal point in both numbers 3. Put the decimal point that far to the left after the answer

Ex: 0.2 x 0.5 2x5 = 10, and

There are two digits to the right of the decimal point, so the answer is 0.10

Ex: 0.25 x 0.25

25 x 25 = 625, and

Equations in one variable (usually represented by X)

An equation is a different way of stating something. We can say, for example that 2 times some number is 12, or we can write

2X = 12

This states the same thing more succinctly.

If we say that two times a number minus 5 is 13, then we may write 2X - 5 = 13

How would I state the following equations in word terms? 3X = 21

5X + 3 = 23 X2 _{= 36}

The reason we use the symbology in mathematics instead of the sentences is because it is much easier to find the missing numbers once you know the rules for manipulating the equations.

If I told you I was thinking of a number that if I multiply it by 2 and add 5 to it I get 27, that might be an easy problem to solve with a little trial and error, but it works out easily if we use

algebra.

2X + 5 = 27 2X = 22

X = 11

Equations in One Variable: The Key

The MAIN Rule for Dealing with Equations:

A Balance is the key to solving algebraic equations.

Just like with a balance, if I add something to one side, I have to do it to the other in order to keep it in balance, so it is with an equation.

The other Key to solving equations is to get the variables alone by themselves.

The Objective: Get the Variable by Itself:

We do this by looking at the operation that was done, and doing the opposite operation to get rid of the numbers and get the variables by themselves.

So, consider the following:

X + 5 = 11

What is the operation that was done that stands in the way of getting X by itself? Addition.

So, we’ll do the opposite operation - SUBTRACTION – to get rid of the 5.

But, remember the BALANCE, I’ll have to do the same thing on BOTH SIDES of the equation.

X + 5 - 5 = 11 - 5

X + 0 = 6

Now, consider the following:

X - 4 = 10

What operation was done? Subtraction. Do the opposite to get the 4 away.

X - 4 + 4 = 10 + 4

X + 0 = 14

X = 14

Opposites:

So, consider the following:

3 X = 33

This equation says that 3 multiplied by some unknown X equals 33. You can probably guess what X must be, but lets do it by our procedures.

1. Identify the operation(s) that is done to keep X from being alone.

2. Do the opposite operation to both sides of the equation. (Remember: ) 3. Simplify to find the result.

3 X = 33

3 X 33

--- =

---3 3

Cancel the 3 in the numerator with 3 in the denominator. Do the division on the right side of the equation, to obtain:

Before going too much further, let’s review some rules of algebraic arithmetic.

Remember to Keep your and Straight !

1. Rules for Variable Addition, Subtraction:

Just like apples and oranges, we can’t add or subtract things that don’t have the same label. We CAN add and subtract things that DO have the same label.

Ex: 2X + 3Y cannot be written any other way without knowing X or Y But: 2X + 3X = 5X

And 4X - X = 3X (Why? Because X means 1X)

This rule works for things like X2 _{and X}3_{. These things are treated like apples and}

oranges. You can only add/subtract things having the exact same label. Ex: 3 X2 _{+ 4 X}2 _{= 7 X}2

But: 3 X2 _{+ 4 X}3 _{cannot be written any other way without knowing the value of X.}

NOTE: The label stays the same. We don’t change the exponents ! (But this will happen in multiplication, so be careful !!!)

2. Rules for multiplication of numbers and fractions with variables.

When you multiply a variable by a number (whether a whole number or a fraction), you just write the number next to the variable, keeping in mind that there is a multiplication. If it’s a fraction, you multiply the variable only by the numerator.

So:

3 x X = 3 X (where x represents multiplication, and X is the variable) ¼ x X = 1X / 4 or simply X/4

3. When multiplying a variable expression (such as 3X) by another number, multiply the numbers only, leaving the variable alone.

4. When dividing a variable or variable expression by another number, divide the numbers only, and leave the variable alone (in the numerator):

Ex: 12X  4 = 3X

Solving more difficult equations in one variable

Now let’s consider something a little more complicated:

5X - 4 = 3X - 10

Notice that Xs are on both sides of this equation, as are numbers.

Whenever we encounter problems like this, follow the following procedure.

1. Get all the numbers (this means -4 and -10 in this example) to the right side of the equal sign.

2. Get all the variables (this means 5X and 3X in this example) to the left side of the equal sign.

3. Don’t forget the “Equation is a Balance Rule” when doing the above. Gotta do the same thing to both sides of an equation.

So, we do the following

5X - 4 + 4 = 3X - 10 + 4 [To get +4 from the left to the right side] 5X + 0 = 3X - 6 [Remember the rule for adding -10 and +4]

5X = 3X - 6 [Now, let’s get 3X to the left side]

5X - 3X = 3X - 3X - 6 [This will do it!]

2X = -6 [Now, divide both sides by 2]

2X -6

--- =

---2 2

1X = -3

or

How to do it in a less complicated way.

Here are the rules:

1 If something is added or subtracted on one side of the equal sign, and you want to move it to the other side – pick it up and move it over, and change the sign.

Ex. 3X - 4 = 11

3X - 4 = 11

3X = 11 + 4

3X = 15

X = 5

Once you get used to it, this procedure will make solving equations much faster and easier!

2 After moving all the things that are added or subtracted, think of both sides of the equation as having a numerator and a denominator. Some equations may have a denominator. For those that don’t, the denominator is 1 .

3X = 15 means

3X 15

--- =

1 1

3 Now, with the equation set up like that, get rid of all the coefficients of X (here it is 3)

by moving them from the numerator of one side of the equation to the denominator of the other side.

Ex.

3X 15

---3 X 15

--- =

1 1

This gives

X 15

---- =

1 3 x 1

…where the little x means multiplication (3 x 1 = 3), so.

X = 15

---3

X = 5

This will work in every case where you have multiplication in the numerator on one side of the equation. You can move it to the other side, and multiply it by the denominator.

This will also work the other way around, from denominator to numerator.

Consider the following problem: X/2 = 11

X 11

--- =

---2 1

X 11 x 2

--- =

1 1

X = 22

Some more complicated examples !!!

Consider:

5X 3X 1

--- - 5 = - +

7 4 3

This is a combination of everything we’ve learned so far. Positive and negative numbers, fractions, and equations in one variable. To solve it, do the following

1 Get the numbers to the right side, the X expressions to the left side.

5X 3X 1

--- - 5 = - +

7 4 3

5X 3X 1

--- - --- = + --- + 5 7 4 3

2 Find a common denominator for the left side expressions, and for the right side

5X x 4 3X x 7 1 5 x 3 --- --- - --- --- = + - + 7 4 4 7 3 1 3

20X 21X 1 15

3 Now, do the fraction arithmetic on both sides, since you have common denominators.

- 1X + 16 ---- =

28 3

[Always write the sign with the numerator, this will make it less confusing !]

4 Move all the denominators across the equal sign to be multiplied by the numerators.

- 1X + 16 ---- =

28 3

-1X x 3 = +16 x 28 [I’m always using “x” to mean multiplication]

-3X = +448

X = - 448

3

Solving Problems with Algebra: Rectangles, Circles, Trianges, Cylinders etc.

The AREA of a Rectangle (shaded) is given by the equation: A = L x W

The PERIMETER (distace around) of a Rectangle is given by the equation:

P = 2xL + 2xW

Often we write the multiplication without the sign,

P = 2L + 2W

AREA is useful when you want to solve problems that have to do with things like:

1. How much seed would I need to put on a field that has length of 30 ft, width of 10 feet, where you need 2 lb of seed / square foot of land.

2. A landowner wanted to convert his lot from rectangular to square. His land currently is 40 feet long and 10 feet wide. What would the dimensions of an equivalent square field be?

PERIMETER is useful in solving problems like:

1. How many feet would you have to walk to walk around a field 300 ft long and 100 ft Length = L

Circles

For Circles, we also have AREAS and PERIMETERS. For the PERIMETER (distance around) we have a different term - CIRCUMFERENCE.

It turns out that the AREA and CIRCUMFERENCE of a CIRCLE are related to a value called PI (π), whose approximate value is 3.14. In the problems we do involving π, you can leave it as a symbol. You don’t have to replace it by its value.

The AREA (Shaded) of a circle is given by

A = π x R

Or, dispensing with the multiplication symbol:

A = πR

So, for a circle of Radius R=5 inches , the Area is A = π (52_{) = 25 π square inches}

Or inches2

The units for Area will be the square of the units for Diameter or Radius If Radius is given in inches, then Area is square inches

If Radius is given in feet, Then Area is in square feet…and so on

The RADIUS (R) of a circle is the distance from the Center of the Circle to the circle.

The DIAMETER (D) of a circle is the distance from one side to the other thru the center.

Circumference of a Circle

Now, let’s consider another property of circles – the CIRCUMFERENCE.

Again, it is analogous to the Perimeter in rectangles – it is the distance around the figure.

The formula for the Circumference C is : C = 2 x π x R

Or: C = 2 π R

Since the diameter is twice the radius (D = 2R) then: C = π D

So, for a circle of Radius of 5 inches (D = 10) C = π x 10 = 10 π inches

Triangles

The Triangle shown here is a Right Triangle, meaning that it has a right angle (= 90o_).

For a right triangle, the length of the hypoteneuse (the side opposite the right angle) is

Length of Hypoteneuse =

√

(

Base

)

2

+

(

Height

)

2

So, if the Base = 3 inches, and the Height = 4 inches, then

Hypoteneuse =

√

(

3

)

2

+

(

4

)

2

=

√

(

9

)

(

16

)

=

√

25

Base = B

This side is called the Hypoteneuse

About Square Roots:

The Square Root symbol

√

Perimeter of a Right Triangle

So, for the triangle shown here, the Perimeter = 3 + 4 + 5 = 12

Since we now know how to find the length of the hypoteneuse of a right triangle, we can now find the Perimeter (distance around it).

In general, for a right triangle, the Perimeter is

PΔ = Base + Height + Hypoteneuse

Replace the Hypoteneuse with our formula for finding its length…

PΔ = Base + Height

√

(

Base

)

2

+

(

Height

)

Area of a Triangle

The Area of a triangle is much easier. The Area is given by

AΔ = ½ x Base x Height

We usually abbreviate this as:

AΔ = ½ b h

So, for the triangle whose base is 3 inches and whose height is 4 inches

Cylinders

A Cylinder is a three dimensional figure that is like a cup. It is formed from a circle that has a height.

The Volume (how much liquid it will hold, for example) of a three dimensional figure is given in cubic feet, cubic inches, cubic centimeters, etc…the cube of whatever the Height and Radius dimension is.

If the radius is given in inches, the Volume is cubic inches (like your car’s engine) If the radius is given in feet, the Volume is cubic feet (or ft3_).

So, for a Cylinder where R = 2 inches , and H = 10 inches, the Volume is given by

Vcyl = π R 2 H

Vcyl = π x 22 x 10 = 40π cubic inches or in3 Vcyl = 120.56 cubic inches (approximately)

Question: If your car has 4 cylinders, all of which have the same dimensions given above, what is the size of your engine in cubic inches?

Question 2: If your car’s engine has 6 cylinders, what is its size?

Circle with Radius = R Height

Practical Problems with Geometric Figures and Algebra

Consider a field that you want to water with a sprinkler. The field is 100 feet long, 100 feet wide (what is the area of the field?)

The sprinkler is placed in the middle of the field, and it has a range of 50 feet. How many square feet of the field will go unwatered?

The problem is as shown in the diagram. The unwatered section will be the area between the square and the circle.

A

=

Area of square section - Area of circular section

A

=

A

- A

= L x W - (π x R2₎

= (100x100) - (502 _{x π)}

= 10,000 sq ft - 2500 π sq ft

If you want a numerical answer, replace π by its approximate value (3.14)

A

Rectangle and ½ Circle

The key is recognizing this as a Perimeter problem.

For the perimeter of the rectangle we have P = 2L + 2W = 2 x 12 + 2 x 8 = 24 + 16 = 40

But this is not the whole story, you also need to construct the ½ circle. For this you need ½ the circumference of the circle. The Radius of this ½ circle is 4 ft (why?).

C(1/2 Circle) = ½ x 2πR = ½ x 2 x π x 4

= 4 π

So, the total length of tape needed is

P+ C(1/2 Circle) = 40 + 4 π = 40 + 12.56 (approx.) = 52.56 feet

Suppose you are constructing a basketball court on your paved driveway. You are using tape to construct the part of the court that contains the free throw line and the 3 second area as shown. If the length of the rectangle is 12 feet, and the width is 8 feet, what is the length of tape needed to construct the figure as shown? 12 ft

Parts of a circle

Consider the parts of a circle shown. What is the area and perimeter?

(Don’t forget the “straight line” section !!!)

Area = π R2 _{= π (2)}2 _{= 4 π (square inches)}

Circumference = 2 π R = 2 x π x 2 = 4 π (inches) = 12.56 in (approx) R = 2 inches

for all figures

Area = ½ x π R2

= ½ x π x 22

= ½ x 4 π

= 2 π square inches

= 6.28 square inches (approx) Perimeter = ½ circle + diameter

= ½ x 2 π R + 4 = ½ x 2 x π x 2 + 4 = 2 π + 4 = 6.28 (approx) + 4 = 10.28 inches (approx)

Area = ¼ x π R2

= ¼ x π x 22

= ¼ x 4 π

= 1 π square inches

= 3.14 square inches (approx) Perimeter = ¼ circle + 2 x Radius

= ¼ x 2 π R + 2R = ¼ x 2 x π x 2 + 4 = 1 π + 4

Area = ¾ x π R2

= ¾ x π x 22

= ¾ x 4 π

= 3 π square inches

Consider a triangle formed from the division of a rectangle by a diagonal. What is the area of the triangle formed?

Therefore, if you know the dimensions of the rectangle, you could figure out the dimensions (and AREA) of the triangle.

What if you knew the AREA of the Rectangle, and one of its sides?

If Arectangle = L x W

Then Lrectangle = Arectangle / W

So, if the area is 400, and the width is 10, what is the length? L = 400 / 10 = 40

What is the area of the triangle formed by the diagonal? A = ½ b h

Swimming Pool

Consider a swimming pool with a diameter of 20 feet (Radius = ?) and a height of 4 feet.

What volume will the pool hold?

Vcyl = π R 2 H

Vcyl = π x 102 x 4

Vcyl = π x 100 x 4

Vcyl = 400π cu ft (= 1256 cu ft approx)

If the pool is already ¼ full, how much more water is needed to fill the pool

Vneeded to fill = Vcyl - V1/4 cyl

Vneeded to fill = 400 π - π x 102 x 1 ft

Vneeded to fill = 400 π - 100 π

3 ft

Fences, etc.

How many fence sections 10’ long are necessary to build a fence around a field that is 40 feet long, 20 feet wide?

Perimeter of rectangle = 2 L + 2 W = 2(40) + 2 (20) = 100 feet Fence Sections needed = 100 ft needed / 10 feet per fence section Fence Sections needed = 100/10 = 10 sections

You can probably solve this one in your head or with a simple diagram, but as problems become more complex, the algebra helps.

How much wood is needed to build all the fences if the boards of the fence are 4 feet high, and 10 feet long as shown in the diagram?

Total wood needed = 10 sections x (6 boards x 4’ high + 2 boards x 10’ long) Total wood needed = 10 x (24 + 20) = 10 x 44 = 440 board feet of wood BUT…

What happens if the boards are 10’ long, you have 2’ of waste for every 2 4’ sections you cut. What if the boards are 12’ long? You have 2’ of waste for the 10’ boards. Which would you rather purchase - 10’ boards or 12’ boards and why?

Fence sections = 10’ long

4’ high

40 feet

Inequalities

Let’s bring our number line back for the discussion of “inequalities.”

If we solved an equation in one variable like 2X = 10 to find X=5, then we may “plot” it on the number line as shown below:

The thing is, we may have an equation that involves inequalities such as: 2X – 10 > 0

2X – 10 < 0 - 2X +10 ≤ 0 2X – 10 ≤ 0

or something like that. And we can solve these equations (really, we should say “in-equations”) much in the same way as equalities, with only a slight variation.

Here’s the process:

1. Treat an inequality just like an equation, you can add, subtract, multiply or divide – as long as you do the same thing to both sides. Keep the sign the same.

2. ONE EXCEPTION: If, in the process of solving for X, you multiply or divide by a negative number – then you have to REVERSE the sign. ( < will become > and vice versa; ≥ will become ≤ and vice versa)

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Application. Let’s do the first one above. 2X – 10 > 0

Add 10 to both sides 2X – 10 + 10 > 0 +10

2X > 10

Divide both sides by 2 (positive 2 – so we don’t have to reverse the sign) 2X/2 > 10/2

X > 5

Typically we graph this on a number line with a solid line to the right of 5 (greater than 5), and either an open circle at 5, or an open parenthesis ( .

or

Notice that the “solution set” for this inequality includes ALL numbers on the number line greater than +5.

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Now, let’s do one with a ≥ or a ≤ sign.

- 2X + 4 ≥ 10 (with the negative sign in front of the 2X, this indicates that we may have to divide both sides by a -2 after getting rid of the 4, and this will cause us to reverse the ≥ sign !!!)

- 2X + 4 - 4 ≥ 10 - 4 - 2X ≥ 6

Now divide both sides by -2, being careful to reverse the ≥ sign when you do so.

-2X 6

--- ≤

-2 -2

So:

X ≤ -3

X is less than or equal to -3. How does this look on the graph?

Or

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The short cuts we learned for equalities also work for inequalities – the only thing to be careful of is when you move a negative number across the sign from the numerator to the denominator or vice versa.

Example 1:

2X + 5 ≥ 15

2X ≥ 15 - 5

2X ≥ 10

X ≥ 10/2

X ≥ 5

Example 2:

- 5 X - 2 < 13

- 5 X < 13 + 2

- 5 X < 15

When moving the -5 from the numerator to the denominator, remember to reverse the < sign.

X

>

X > -3

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Dealing with Inequalities and Absolute Value

When dealing with inequalities and absolute values such as:

| X - 5 | > 3 or

| X + 2 | ≤ 6

… there are basically only two procedures to follow. Each procedure has two parts.

Procedure 1: To be used for < and ≤signs:

PART 1:

1. Write the expression that is between the absolute value signs | | without that sign 2. Write the sign used (either < or ≤ )

3. Write the number that is on the other side of the sign 4. Solve that inequality.

5. This will tell you that X < something or X ≤ something. Graph it.

PART 2

6. Then, write the same expression between the absolute value signs | | without that sign once again.

7. If you had the < sign before, write > now. If ≤ write ≥.

8. Write the opposite of the number that is on the other side (e.g. if 6 write -6) 9. Solve that inequality.

Example: | X + 2 | ≤ 6

1. X + 2 2. X + 2 ≤ 3. X + 2 ≤ 6 4. X ≤ 4

5. So, X is less than or equal to 4 6. X + 2

7. X + 2 ≥ 8. X + 2 ≥ -6 9. X ≥ -8

Now, graph both, and choose the region in common.

X ≤ 4

X ≥ -8

Region in common: X simultaneously is ≥ - 8 and X ≤ 4

We can write this as:

-8 ≤ X ≤ 4

and graph it as follows:

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- 8 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7

Procedure 2: To be used for > and ≥ signs:

PART 1:

1. Write the expression that is between the absolute value signs | | without that sign 2. Write the sign used (either > or ≥ )

3. Write the number that is on the other side of the sign 4. Solve that inequality.

5. This will tell you that X >something or X ≥something. Graph it.

PART 2

6. Then, write the same expression between the absolute value signs | | without that sign once again.

7. If you had the >sign before, write < now. If ≥ write ≤.

8. Write the opposite of the number that is on the other side (e.g. if 6 write -6) 9. Solve that inequality.

Example:

| X - 5 | > 3 1. X – 5 2. X – 5 > 3. X – 5 > 3 4. X > 8

5. So, X is greater than 8 6. X – 5

7. X – 5 < 8. X – 5 < - 3

9. X < + 2

10. So, X is smaller than + 2

X > 8

X < + 2

The solution can be written as: X > 8 AND X < + 2

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Cartesian Coordinates or the X-Y Plane

Don’t be afraid of this. All we’re gonna do here is draw some nice pictures.

This is the Cartesian Coordinate Plane (or the Rectangular Coordinate Plane or X-Y Coordinate System.

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An “ordered pair” means two numbers, in order. Usually they are given in the form (X, Y) where X and Y are two numbers.

Starting from the intersection of the two axes – the origin – let’s say we wanted to plot the point (+5, +3), we’d go +5 to the right, and then up +3, and put a point there.

(+5, +3)

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Now, let’s plot the ordered pair (-2, +5)

The first number in the ordered pair always represents the position on the horizontal axis. So, we start at zero and go two to the left.

Then, since the second number in the ordered pair is positive, we go up five.

Notice that in this region, the X value is negative, while the Y value is positive. (-2, +5)

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Now let’s plot the point (-5, -3). This means go left five units, then go down three units and put a point there.

Notice that in this region, both the X value and Y value are negative.

(-5, -3)

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Now, let’s plot the point (+2, - 4). This means go to the right two units, then down, four units, and put a point there.

Notice that in this region, the X value is positive, while the Y value is negative.

(+2, -4)

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Notice that the intersection of the X and Y axes (the

+

QUADRANT I X is Positive Y is Positive QUADRANT II

X is Negative Y is Positive

QUADRANT IV X is Positive Y is Negative QUADRANT III

X is Negative Y is Negative

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Now, lets look at some lines. VERTICAL and HORIZONTAL LINES are easy. Consider first the HORIZONTAL LINE Y = + 3

Count UP three from the origin, and draw a HORIZONTAL LINE thru that point.

Any equation of the form Y = something (where the something is a NUMBER) is a HORIZONTAL LINE !

The HORIZONTAL Line Y = +3

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The expression X = something means a VERTICAL LINE.

So, for example, to draw the line X = - 4 , start at the origin, and GO LEFT four units, then draw a VERTICAL LINE thru that point.

The VERTICAL Line

X = - 4

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Now, lets consider a special line that will help us later when we are drawing other lines. This is the special equation of a line:

Y = X

What this equation means is that when X is 0, then Y is also 0. When X is 1, then Y is also 1, etc.

So, we can see that we’ll have the following points as part of this line: (0, 0) (1, 1) (2, 2) (3, 3)

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Now that we have these points we can draw a line through them. (you only need two to draw a line, but it’s good to draw a few extra just to check yourself)

This is the line

Y = X. (or Y=1X) Notice that for all points on the line, the Y value is equal to the X value.

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Another way to draw the line Y = X is to start at the origin (0, 0), move over one, and up one (like the ladder shown in the diagram below) and draw the line between these points.

For the line Y = X, start at the origin ( 0 , 0 ) and go up one and over one, draw a point, and continue in this manner until you can draw a line thru the points.

The line should be a 45o

angle line thru the origin. At every point on this line, the Y value equals the X value.

(When X=0, Y=0; When X=1, Y=1; When X=2, Y=2; And so on)

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Notice that for every increase of 1 on the Y scale (vertical), there is also an increase of 1 on the X scale (horizontal). This is known as the SLOPE, and it is defined as:

Change in Y Coordinate Δ Y + 1

--- = ---- = ---- (in this example)

Change in X Coordinate Δ X + 1

What about situations where the Y Coordinate changes faster (or slower) than the X Coordinate?

If you can calculate the slope, and if the line goes thru the origin (0,0), then writing the equation of the line is easy.

For example, this line goes thru the origin (0,0) and the slope is ΔY/ ΔX = 3/1, so the equation of the line is:

Y = 3X

The slope is the COEFFICIENT of X in the equation of the line Y = mX where m is the slope and the line goes thru the origin.

Is the slope of the line > 1 or < 1 ?

( Slope = ΔY/ ΔX )

ΔY = 3

ΔX = 1 Δ X = 1

Δ Y = 1

Slope of the line = 1/1 = +1

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Usually, however, our lines have the form Y = (slope) X + something. We write the general form of this line as:

Y = mX + b …where m represents the slope, and

…where b is the place on the AXIS where the line crosses. b is also called the Y-Intercept.

Over 4, Up 2. Over 2, Up 1. What is the slope?

2 1

What is the equation of this line that goes thru the origin?

Slope = ΔY/ ΔX = ½ Equation of the line: Y = ½ X

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Now you are armed with enough information to draw any line that is in Slope-Intercept Form.

The Slope-Intercept form of a line is:

What is the “Y-Intercept” ? 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Let’s draw Y = 3 X + 2

Y Intercept = +2

ΔY = 3

ΔX = 1

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Draw Y = 2 X - 5

Notice that we have a -5 as the Y-Intercept. Slope = +2

Y-Intercept = -5

ΔY =2

ΔX = 1

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Now, try Y = - 2 X

The Slope = -2, Y Intercept is zero (it goes thru the origin).

ΔX = +1

ΔY = -2

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So, when you go upstairs, the slope is positive

And when you go downstairs, the slope is negative

Line with a Positive Slope

Now, draw the line Y = - 3X - 4

Δ Y = - 3 Δ X = + 1 Y Intercept = - 4

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Manipulating the Forms of a Line

Now, you are armed with the ability to draw a line if it is written in the form Y = m X + b

…where both m and b are numbers, that represent the slope and Y-intercept of the line.

But, suppose you are given an equation like: 6 X + 2 Y - 10 = 0

This is an equation of a line, but it is not in the form we expect.

We want to do the following to the equation:

6 X + 2 Y – 10 = 0

Remember the rules that we learned for moving anything that is added or subtracted to the other side?

6 X + 2 Y - 10 = 0 Keep the Ys on

the Left side of the equals sign, and move everything else to the right of the sign.

6 X becomes - 6X

Now, we have to divide both sides by 2 to get rid of the coefficient of Y. 2 Y = - 6 X + 10

Y = - 3 X + 5

Notice that we have to divide everything in the right side by 2 (everything that is added or subtracted, that is).

How to find a line being given a POINT and a SLOPE, or TWO POINTS.

POINT and SLOPE

Suppose you are given the information that the slope of a line is +2, and that the line passes thru the point (3, 3). How can you find the equation of the line?

Well, what is the definition of SLOPE again?

Change in the Y Coordinate Δ Y SLOPE = --- or ---Change in the X Coordinate Δ X

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We have an equation that is called the POINT SLOPE Form of a line, and it is written like this:

The M is the SLOPE of the line that we are given in the problem set up.

The Y and X are variables. They don’t change. They stay in the problem as letters.

However, the X1 and Y1 are actual numbers. We take the value given for X, and the value

given for Y in the problem setup, and plug them in for X1 and Y1.

In the problem given, both X1 and Y1 are 3. But, don’t forget, the first number in an order pair is

the X value, the second number is the Y value. So, for (3, 3), X and Y are both 3. Now we just plug the numbers that we know into the form

Y - 3 + 2 = ---X - 3

Don’t forget that we have to subtract X1 and Y1. (If X1 or Y1 were negative, what would

happen?)

So, now we have an equation that we can manipulate to the form: Y = something x X + something else

Or, in the stricter language of math Y = mX + b

Where m is the slope, b is the y-intercept.

(Y – 3) If + 2 =

---(X – 3)

Y - Y1

To manipulate, multiply both sides of the equation by (X – 3)

+ 2 (X – 3) = Y – 3

Now, +2 must be multiplied by everything between parentheses (“distributive property”)

+ 2 X - 6 = Y - 3

Let’s just switch the order of the equals sign (If b = a, then a = b)

Y – 3 = +2X - 6

And now bring the – 3 over to the other side (becomes + 3) Y = +2X – 6 + 3

Simplifying

Y = +2X – 3

TW O POINTS

Suppose you are given two points such as (2, 3) and (4, 4) and are asked to write the equation of the line between them.

What is the slope of a line (Change in Y/ Change in X or ΔY/ΔX)

If we know two points, we can find ΔY/ΔX by subtracting the two Ys and the two Xs.

Here’s an equation to find the SLOPE.

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What this equation says is that you can find the SLOPE of a line if you know TWO POINTS, by subtracting the Y values, and then dividing by the difference in the X values.

You can start with either point, but you have to stay consistent.

4 – 3 1

Here M = --- =

--4 – 2 2 because X1 is 2, X2 is 4, Y1 is 3, and Y2 is 4.

where (X1, Y1) is (2, 3)

and (X2, Y2) is (4, 4)

SO, of course now you have a POINT and a SLOPE, and you can use the process we learned earlier to get the SLOPE-INTERCEPT form from it. (Pages 61-63)

INEQUALITIES ON THE CARTESIAN PLANE

Consider the INEQUALITY Y < 2X - 1

How do you graph this?

For a Linear Inequality, the answer must be the REGION on ONE SIDE OF THE LINE.

Which side? Pick a point and test it!

Y = 2X - 1

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So, the entire region to the right of the line is the region Y < 2X - 1

Y = 2X - 1

For the point (4,1) is Y < 2X - 1 or is Y > 2X - 1

Plug in values for X and Y 1 < 2(4) - 1 ?

1 < 8 - 1 ? 1 < 7 YES

So, the region on this side of the line is the right region, where Y < 2X - 1

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Here’s some grids for you to use for practice:

In all of this region: Y < 2X - 1

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Solving Simultaneous Equations:

One of the things we do a lot of in algebra is solving two equations (or sometimes 3 or more) to get a specific point that satisfies both equations.

Geometrically, the solution represents the intersection of two lines on a graph.

“Simultaneous solution” means the “one and only” point that is in common between the two equations or two lines.

The graph shows that the two lines that represent these functions intersect at the point (1,3), and therefore, that is the one point in common between the two.

But we can do the same thing with algebra that we can do graphically. The point (1,3) represents

the simultaneous solution of the equations:

Y = x/2 + 2.5 and Y = x + 3

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Here are our two equations:

Y = X/2 + 2.5 Y = X + 2

One way to solve simultaneous equations is to get them to equal the same thing – then,if they equal the same thing, they must be equal to each other.

This has already been done if we write the equations in slope-intercept form as shown here. So, X/2 + 2.5 = X + 2

Now we have an equation in 1 variable, which we recall how to solve. (Get X by itself on the left, get the numbers on the right).

X/2 - X = 2 – 2.5 - ½ X = - ½

Therefore X = 1

What is Y when X = 1?

One way to be extra careful that you solved the simultaneous equations correctly for X is to plug the value for X back into both equations. You should get the same value for Y.

Y = X/2 + 2.5 = ½ + 2.5 = 3 Y = X + 2 = 1 + 2 = 3

Another Way to Solve Simultaneous Equations:

Consider the following equations: 2X + Y = 20

3X - 2Y = 9

We could manipulate the two equations to obtain the slope-intercept form, or we could proceed another way.

Remember that I can multiply or divide and equation by any number without changing the equation, as long as I do exactly the same thing to both sides of the equation.

Notice if you multiply the first equation by 2, you can get the opposite coefficient to Y in the second equation. That means we can add the two equations together and eliminate Y. That is our purpose.

Try it.

2 x ( 2X + Y ) = 2 x 20

4X + 2Y = 40 1st _{equation after modifications}

3X - 2Y = 9 2nd _{equation, unchanged}

Both sides are equal, so we can basically add the two equations together…. 7X + 0Y = 49

Therefore 7X = 49 and X = 49/7 = 7 Plug back in the value of X = 7 in both equations.

1st _{Equation 2}nd _Equation

2X + Y = 20 3X – 2Y = 9

2(7) + Y = 20 3(7) – 2Y = 9

14 + Y = 20 21 – 2Y = 9

Y = 6 2Y = 21 – 9

2Y = 12 Y = 6

Applications

Now let’s look at some applications.

We can recall that the Fahrenheit Temperature is related to the Celsius Temperature by the following equation:

F =

9 5C+32

Therefore, if the F value is the Y axis, and the C value is the X axis, we can draw the line that represents the conversion from C to F as follows: (next page)