Math 7 ( Period 4)

Solving Inequalities Using Multiplication or Division 9.7

Solving an inequality is similar to solving an equation. You can use the four arithmetic operations to isolate the variable on one side of the inequality There is ONE important difference.

DAY TWO of this lesson (January 6)  

When you multiply an inequality by a negative number, you must reverse (switch) the inequality symbol!

5 > -4 but if you multiply both sides by -2

-10 < 8

Divide  by -3:    -6 < 3 becomes 2 > -1

Properties of Inequalities

1. Adding or subtracting the same number on each side of the inequality produces an equivalent inequality.

2. Multiplying or dividing each of an inequality by the same positive number produces an equivalent inequality.

3. Multiplying or dividing each of an inequality by the same negative number AND reversing the direction of the inequality symbol produces an equivalent inequality.

-2x < 6

Divide each side by -2 and as you do that rewrite the inequality reversing the inequality symbol

-2x/-2 > 6/-2

x > -3

-1/2(x) ≥ 6

This time you need to multiply by -2 and as you do that rewrite the inequality reversing the symbol

(-2)(-1/2)x ≤ 6(-2)

x ≤ -12

What about 6x > -42

You are not multiplying or dividing by a negative number—so just divide both sides

6x/6 > -42/6

x> -7

Algebra Honors (Periods 6 & 7)

Complex Fractions Chapter 6 Extra 

A complex fraction is a fraction whose numerator or denominator  contains one or more fractions. To express a complex fraction as a simple faction use one of the following methods:

Method 1:  Simply the numerator and the denominator.  Express the fraction as a quotient using the division symbol  ÷ .   Multiply by the reciprocal of the divisor. 

Method 2: Find the LCD of ALL the simple fractions in the complex    fraction. Multiply the numerator and the denominator of the complex  fraction by the LCD 

Using Method 1

Method 1:  Simply the numerator and the denominator.  Express the fraction as a quotient using the division symbol  ÷ .   Multiply by the reciprocal of the divisor. 

Simplify above to get

Using Method 2

Method 2: Find the LCD of ALL the simple fractions in the complex    fraction. Multiply the numerator and the denominator of the complex  fraction by the LCD 

   

Same problem as above but this time Find the LCD of all the simple fractions in the complex fraction... The LCD would be 2ab. Multiply each simple fraction by 2ab

Which simplifies to

Then you can factor:

Hopefully you realize you have been able to obtain the same simplified fraction using EITHER method. Use whichever method works BEST for you!

Math 6A ( Periods 1 & 2)

Equations: All 4 Op’s

Review Sections 8-2 & 8-3

x = 48 + 15

There isn’t any transformations necessary

We just need to add the numbers on the right hadn side

x = 63

4( 3-1) + x = 34

4(2) + x = 34

8 + x = 34

subtract 8 from both sides

x = 26

Make sure to box your answers~

(Include the variable in your BOX!

n/16 = 6

We need to multiply both sides by 16

n = 96

17n = 289

Wait a minute… you should look at this and say… I know this one!

divide both sides by 17

n = 17

You know your squares and square roots!

YAY

Math 6A (Periods 1 & 2)

Equations: Decimals  8-4

You can use transformations to solve equations which involve decimals
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5

1.6n = 3.6
1.6n/1.6 = 3.6/1.6
Carefully do the math and you find that
n = 2,25
If you need to-- practice with decimals! You need to be able to divide and multiply accurately!


Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.8


n + 0.519 = 0.597
we need to subtract 0.519 from both sides
n = 0.078

x - 0.323 = .873
we need to add 0.323 to both sides
x = 1.196


Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.)

Also remember that the understood (invisible) coefficient of x in the equation x = c is 1.

[Can you picture the poster in the front of the room?]

Algebra Honors (Periods 6 & 7)

Adding & Subtracting Fractions  6-5

In chapter 2 Section 9 we reviewed adding and subtracting fractions.

 

With that we found we could do the reverse….

To add or subtract fractions with the same denominator, you add or subtract their numerators and write that results over the common denominator.

To simplify an expression involving fractions, you write it as a single fraction in simplest form

3c/16 + 5c/16 = 8c/16 = c/2

Be careful when you distribute the subtraction sign to every term in the 2^nd  fraction.

The following really seems simple but many students try to simplify it… BE CAREFUL

 

STOP! You cannot simplify any farther!

Notice that 3-x = -( x - 3) , the LCD is x – 3

Simplify

You must determine the LCD first. Rewrite the fractions using the LCD of 36

Simplify:

  

Simplify:

  

Factor completely FIRST

You realize that the LCD is a(a-2)(a+2)

Algebra Honors ( Periods 6 & 7)

Adding & Subtracting Fractions:  
The Least Common Denominator 6-4

We know that we can write a fraction in simpler form by dividing its numerator and denominator by the same nonzero number.

and the reverse is true as well

You can write a fraction in a different form by multiplying the numerator and denominator by the same nonzero number.

3/7 = ?/35

You realize that you multiply 7 by 5 to obtain 35 so you would multiply 3 by 5 to obtain the correct number for the numerator.

3/7 = 15/35

the same applies to fractions with variables…

8/3a = ?/18a²

What do you multiply 3a by to obtain 18a²?   6a

so you must multiply both by 6a

  Complete:  

You notice that you need to multiply the denominator by (x +1) so you must do the same to the numerator.

When you add or subtract fractions with different denominators, you will find that using the Least Common Denominator (LCD)  will simplify your work

Find the LCD of: 

First factor each denominator completely. Factor Integers into primes!

6x-30 = 6(x - 5)= 2·3(x - 5)

9x -45 = 9(x - 5) = 3∙3(x - 5)

Form the product of the greatest power of each
(Remember that the LCM is the product of every factor to its greatest power)

2·3²(x - 5) = 18(x - 5)

Therefore the LCD is 18( x – 5)

Re write the following with their LCD

x²- 8x +16 = (x - 4)²

and

x²- 7x + 12 = (x - 4)(x - 3)

the LCD is (x -4)²(x -3)

Now rewrite each fraction using the LCD

Math 7 (Period 4)

Problem Solving Strategies 4.6

WORD PROBLEMS! YAY!

Often there are several ways to solve a real-life problem It can be helpful to use more than one method and then compare results…

We turned to the textbook Page 193 and looked at the Example 1 problem…

A logging company is making a reforestation plan. The company has 90 square miles of logged land and is logging 15 square miles more each year. The company plants 30 square miles f new trees each year.

Predict how long it will take for the logged land to be reforested.

Method 1

You can use a table and look for a pattern For the logged land start with 90 square miles and add 15 square miles each year. For the reforested land, start at 0 (because they hadn’t started this program, yet) and add 30 square miles each year.

See the chart:

Year	0 (NOW)	1	2	3	4	5	6
Logged land ( in square miles)	90	105	120	135	150	165	180
Reforested (in square miles)	0	30	60	90	120	150	180

From the table you can see that will take 6 years to reforest all the and that has been logged.

Method 2

Please turn to page 193 and look at the coordinate graph to visualize the same data as in the table above. That graph shows that the amount of reforested land steadily approaches the amount of logged land. It also shows that it will take 6 years to reforest the logged land.

Method 3

Use an equation to solve this problem

Please look at Page 194 for the Verbal Model

   

We know:                                                                    

Land already logged = 90

Land logged per year = 15

Number of years= THAT”S WHAT WE DON”T KNOW!

Let n = the number of years

Land reforested per year = 30

90 + 15n = 30n

NOW we can solve the equation

subtract 15n from both sides

90 = 15n

90/15 = 15n/15

6 = n

So it will take 6 years to reforest all the logged land!

MOVIES

Suppose the cost of seeing a movie at a theater is $7. Buying a DVD player costs  $360 and renting a movie to use on the DVD player is $4.

Write an expression to represent the cost of seeing y movies at a theatre… and write an expression to represent the cost of renting the same number of movies to use on your DVD player ( include the cost of the DVD player)

7y = the cost at the theater

360 + 4y = the cost at home

How many moves must you see so that the  two costs are the same? Write an equation that models that situation and solve for y- the number of movies watched.

7y = 360 + 4y

3y = 360

y =120

You would need to watch 120 movies to have the cost the SAME!

Suppose you watched more than 120 movies… which would cost less at that point… going to the theatre or renting DVD movies?

Renting DVD movies. The cost is only $4 for a DVD  and it is $7 at the theater.

Aesop’s Fables

In one of Aesop’s Fables, a tortoise and a hare are in a race. The hare is far ahead and SURE to win, so it takes a nap. When the hare wakes up, it sees that the tortoise is about to cross the finish line.

Suppose the hare runs 600 inches per second and the tortoise runs 3 inches per second. When the hare wakes up, the tortoise is 1000 feet (12,000 inches) ahead. Which equation represents when the hare will catch up with the tortoise? (Hint: If t represents time in seconds, then 3t is the distance in inches the tortoise can  travel in t seconds…)

A.  3t = 600t  +  12,000            B. 600t = 3t + 12,000

C. 3t – 12000 = 600t                 D. 600t = 12,000

We had a class discussion on Wednesday about the correct equation—which is B.

Using 600t = 3t + 12, 000, we solved for t

and discussed what our solution meant.

597t = 12,000

597t/597 = 12,000/597

t =20.1

20.1 seconds is the time when the hare will catch up with the tortoise.

If the tortoise is 5 feet (60 inches) from the finish line when the hare wakes up, who will win the race? Explain…