Monday, March 16, 2020

Classwork for College Readiness Math March 16, 2020

Welcome! Here is where I will upload documents we use in class and videos of content we have done in class. We also have a class Remind101 to send texts about what is going on in class. Please make sure you join. Just text @mrsberk20 to 81010. I also have some extra practice problems on DeltaMath. My teacher code is 399031. 

Here is the link to your work:http://ashleyberkeley.weebly.com/college-readiness-math.html

Student Handbook
https://www.georgiastandards.org/Georgia-Standards/Frameworks/Course-Materials-College-Readiness-Mathematics.pdf
Over the next few weeks that school will be closed, the expectations are as follows:

You will continue to look over Unit 5 material. 
Work on the worksheet over solving systems of inequalities.
Work on the review sheet for Unit 5.
You will look over the material for Unit 6 (Quadratics)
Work on the worksheets posted below.
Work on the review sheet for Unit 6.

Upon returning to class, we will have one amnesty day to turn in everything you have done over the break. We will then have a review day for Unit 5 the following day. We will test over Unit 5 the day after that. The following day will be a review day for Unit 6. We will test over Unit 6 the next day. The days that follow we will pick up with Unit 7 and continue instruction as normal according to the pacing guide. Please let me know if you have any questions. 

I will have online office hours from 10:00 AM until 11:30 AM where I will be working/posting materials for you. I will be online for tutoring/virtual instruction from 12:00 PM until 3:00 PM. 


Classwork Algebra 1 , Monday, March 16, 2020


Algebra 1, March 16, 2020

Great morning scholars! I hope that everyone is healthy and happy on this beautiful day! Please see today's virtual learning tasks below:

1. Check out Pebblebrook's Algebra 1 blog for an overview of what we will be learning each day this week: 
http://pebblebrookalgebra1.weebly.com/lesson-316---320.html

Next week's lessons can be found here: 
http://pebblebrookalgebra1.weebly.com/lesson-323--327.html.

2. View today's Powerpoint on the blog if needed (we previously learned Completing the Square). I have attached the PowerPoint here in case you cannot access it from the blog.
3. Join the video conference if you need assistance . The site to join the conference is https://zoom.us/j/6134333209 and the meeting code is 6134333209. The conference is scheduled to begin at 10:30 a.m. - 11 a.m. on Mondays , Wednesdays and Fridays. Here, I can answer your questions directly.


4. If you cannot join the the zoom video conference , you can emailing me for assistance from 9 a.m. to 12 noon.

5. If you do not have internet access, here is a link to an article about Comcast offering 2 months of internet for free to low-income customers.  

Stay safe!

Sunday, March 15, 2020

Algebra 1 Week of March 16, 2020

Click on the link for this week's assignments:

pebblebrookalgebra1.weebly.com/lesson-316---320.html

Monday, March 9, 2020

Week of March 9, 2020 College Readiness Math

Systems of Linear Equations


linear
Linear Equation is an equation for a line.
A linear equation is not always in the form y = 3.5 − 0.5x,
It can also be like y = 0.5(7 − x)
Or like y + 0.5x = 3.5
Or like y + 0.5x − 3.5 = 0 and more.
(Note: those are all the same linear equation!)

System of Linear Equations is when we have two or more linear equations working together.

Example: Here are two linear equations:

2x+y=5
−x+y=2
Together they are a system of linear equations.
Can you discover the values of x and y yourself? (Just have a go, play with them a bit.)
Let's try to build and solve a real world example:

Example: You versus Horse

horse
It's a race!
You can run 0.2 km every minute.
The Horse can run 0.5 km every minute. But it takes 6 minutes to saddle the horse.
How far can you get before the horse catches you?

We can make two equations (d=distance in km, t=time in minutes)
  • You run at 0.2km every minute, so d = 0.2t
  • The horse runs at 0.5 km per minute, but we take 6 off its time: d = 0.5(t−6)

So we have a system of equations (that are linear):
  • d = 0.2t
  • d = 0.5(t−6)
We can solve it on a graph:
you vs horse graph
Do you see how the horse starts at 6 minutes, but then runs faster?
It seems you get caught after 10 minutes ... you only got 2 km away.
Run faster next time.
So now you know what a System of Linear Equations is.
Let us continue to find out more about them ....

Solving

There can be many ways to solve linear equations!
Let us see another example:

Example: Solve these two equations:

system linear equations graph
  • x + y = 6
  • −3x + y = 2
The two equations are shown on this graph:
Our task is to find where the two lines cross.
Well, we can see where they cross, so it is already solved graphically.
But now let's solve it using Algebra!

Hmmm ... how to solve this? There can be many ways! In this case both equations have "y" so let's try subtracting the whole second equation from the first:
x + y − (−3x + y) = 6 − 2
Now let us simplify it:
x + y + 3x − y = 6 − 2
4x = 4
x = 1
So now we know the lines cross at x=1.
And we can find the matching value of y using either of the two original equations (because we know they have the same value at x=1). Let's use the first one (you can try the second one yourself):
x + y = 6
1 + y = 6
y = 5
And the solution is:
x = 1 and y = 5
And the graph shows us we are right!

Linear Equations

Only simple variables are allowed in linear equations. No x2, y3, √x, etc:
linear vs nonlinear
Linear vs non-linear

Dimensions

Linear Equation can be in 2 dimensions ...
(such as x and y)
 2D Line
... or in 3 dimensions ...
(it makes a plane)
 3D Plane
... or 4 dimensions ...  
... or more!  

Common Variables

For the equations to "work together" they share one or more variables:
A System of Equations has two or more equations in one or more variables

Many Variables

So a System of Equations could have many equations and many variables.

Example: 3 equations in 3 variables

2x+y2z=3
xyz=0
x+y+3z=12
There can be any combination:
  • 2 equations in 3 variables,
  • 6 equations in 4 variables,
  • 9,000 equations in 567 variables,
  • etc.

Solutions

When the number of equations is the same as the number of variables there is likely to be a solution. Not guaranteed, but likely.
In fact there are only three possible cases:
  • No solution
  • One solution
  • Infinitely many solutions
When there is no solution the equations are called "inconsistent".
One or infinitely many solutions are called "consistent"
Here is a diagram for 2 equations in 2 variables:
system of linear equations types: no solution, one solution, infinite solutions

Independent

"Independent" means that each equation gives new information.
Otherwise they are "Dependent".
Also called "Linear Independence" and "Linear Dependence"

Example:

  • x + y = 3
  • 2x + 2y = 6
Those equations are "Dependent", because they are really the same equation, just multiplied by 2.
So the second equation gave no new information.

Where the Equations are True

The trick is to find where all equations are true at the same time.
True? What does that mean?

Example: You versus Horse

you vs horse graph
The "you" line is true all along its length (but nowhere else).
Anywhere on that line d is equal to 0.2t
  • at t=5 and d=1, the equation is true (Is d = 0.2t? Yes, as 1 = 0.2×5 is true)
  • at t=5 and d=3, the equation is not true (Is d = 0.2t? No, as 3 = 0.2×5 is not true)
Likewise the "horse" line is also true all along its length (but nowhere else).
But only at the point where they cross (at t=10, d=2) are they both true.
So they have to be true simultaneously ...
... that is why some people call them "Simultaneous Linear Equations"

Solve Using Algebra

It is common to use Algebra to solve them.
Here is the "Horse" example solved using Algebra:

Example: You versus Horse

The system of equations is:
  • d = 0.2t
  • d = 0.5(t−6)
In this case it seems easiest to set them equal to each other:
d = 0.2t = 0.5(t−6)

Start with:0.2t = 0.5(t − 6)
Expand 0.5(t−6):0.2t = 0.5t − 3
Subtract 0.5t from both sides:−0.3t = −3
Divide both sides by −0.3:t = −3/−0.3 = 10 minutes
Now we know when you get caught!
Knowing t we can calculate d:d = 0.2t = 0.2×10 = 2 km

And our solution is:
t = 10 minutes and d = 2 km

Algebra vs Graphs

Why use Algebra when graphs are so easy? Because:
More than 2 variables can't be solved by a simple graph.
So Algebra comes to the rescue with two popular methods:
  • Solving By Substitution
  • Solving By Elimination
We will see each one, with examples in 2 variables, and in 3 variables. Here goes ...

Solving By Substitution

These are the steps:
  • Write one of the equations so it is in the style "variable = ..."
  • Replace (i.e. substitute) that variable in the other equation(s).
  • Solve the other equation(s)
  • (Repeat as necessary)
Here is an example with 2 equations in 2 variables:

Example:

  • 3x + 2y = 19
  • x + y = 8
We can start with any equation and any variable.
Let's use the second equation and the variable "y" (it looks the simplest equation).

Write one of the equations so it is in the style "variable = ...":
We can subtract x from both sides of x + y = 8 to get y = 8 − x. Now our equations look like this:
  • 3x + 2y = 19
  • y = 8 − x

Now replace "y" with "8 − x" in the other equation:
  • 3x + 2(8 − x) = 19
  • y = 8 − x

Solve using the usual algebra methods:
Expand 2(8−x):
  • 3x + 16 − 2x = 19
  • y = 8 − x
Then 3x−2x = x:
  • x + 16 = 19
  • y = 8 − x
And lastly 19−16=3
  • x = 3
  • y = 8 − x

Now we know what x is, we can put it in the y = 8 − x equation:
  • x = 3
  • y = 8 − 3 = 5
And the answer is:
x = 3
y = 5

Note: because there is a solution the equations are "consistent"

Check: why don't you check to see if x = 3 and y = 5 works in both equations?

Solving By Substitution: 3 equations in 3 variables

OK! Let's move to a longer example: 3 equations in 3 variables.
This is not hard to do... it just takes a long time!

Example:

  • x + z = 6
  • z − 3y = 7
  • 2x + y + 3z = 15
We should line up the variables neatly, or we may lose track of what we are doing:

x  +z=6   
 3y+z=7   
2x+y+3z=15   

WeI can start with any equation and any variable. Let's use the first equation and the variable "x".

Write one of the equations so it is in the style "variable = ...":
x    =6 − z  
 3y+z=7   
2x+y+3z=15   

Now replace "x" with "6 − z" in the other equations:
(Luckily there is only one other equation with x in it)
 x    =6 − z  
  3y+z=7   
2(6−z)+y+3z=15   

Solve using the usual algebra methods:
2(6−z) + y + 3z = 15 simplifies to y + z = 3:
x    =6 − z  
 3y+z=7   
  y+z=3   
Good. We have made some progress, but not there yet.

Now repeat the process, but just for the last 2 equations.

Write one of the equations so it is in the style "variable = ...":
Let's choose the last equation and the variable z:
x    =6 − z  
 3y+z=7   
    z=3 − y  

Now replace "z" with "3 − y" in the other equation:
x    =6 − z  
 3y+3 − y=7   
    z=3 − y  

Solve using the usual algebra methods:
−3y + (3−y) = 7 simplifies to −4y = 4, or in other words y = −1
x    =6 − z  
  y  =−1   
    z=3 − y  
Almost Done!

Knowing that y = −1 we can calculate that z = 3−y = 4:
x    =6 − z  
  y  =−1   
    z=4   
And knowing that z = 4 we can calculate that x = 6−z = 2:
x    =2   
  y  =−1   
    z=4   

And the answer is:
x = 2
y = −1
z = 4

Check: please check this yourself.
We can use this method for 4 or more equations and variables... just do the same steps again and again until it is solved.
Conclusion: Substitution works nicely, but does take a long time to do.

Solving By Elimination

Elimination can be faster ... but needs to be kept neat.
"Eliminate" means to remove: this method works by removing variables until there is just one left.
The idea is that we can safely:
  • multiply an equation by a constant (except zero),
  • add (or subtract) an equation on to another equation
Like in these examples:
elimination methods

WHY can we add equations to each other?

Imagine two really simple equations:
x − 5 = 3
5 = 5
We can add the "5 = 5" to "x − 5 = 3":
x − 5 + 5 = 3 + 5
x = 8
Try that yourself but use 5 = 3+2 as the 2nd equation
It will still work just fine, because both sides are equal (that is what the = is for!)

We can also swap equations around, so the 1st could become the 2nd, etc, if that helps.

OK, time for a full example. Let's use the 2 equations in 2 variables example from before:

Example:

  • 3x + 2y = 19
  • x + y = 8
Very important to keep things neat:
3x+2y=19   
x+y=8   

Now ... our aim is to eliminate a variable from an equation.
First we see there is a "2y" and a "y", so let's work on that.
Multiply the second equation by 2:
3x+2y=19   
2x+2y=16   
Subtract the second equation from the first equation:
x  =3   
2x+2y=16   
Yay! Now we know what x is!

Next we see the 2nd equation has "2x", so let's halve it, and then subtract "x":
Multiply the second equation by ½ (i.e. divide by 2):
x  =3   
x+y=8   
Subtract the first equation from the second equation:
x  =3   
  y=5   
Done!
And the answer is:
x = 3 and y = 5

And here is the graph:
Graph of (19-3x)/2 vs 8-x
The blue line is where 3x + 2y = 19 is true
The red line is where x + y = 8 is true
At x=3, y=5 (where the lines cross) they are both true. That is the answer.
Here is another example:

Example:

  • 2x − y = 4
  • 6x − 3y = 3
Lay it out neatly:
2xy=4   
6x3y=3   
Multiply the first equation by 3:
6x3y=12   
6x3y=3   
Subtract the second equation from the first equation:
00=9   
6x3y=3   
0 − 0 = 9 ???
What is going on here?

Quite simply, there is no solution.

They are actually parallel lines: graph of two parallel lines
And lastly:

Example:

  • 2x − y = 4
  • 6x − 3y = 12
Neatly:
2xy=4   
6x3y=12   
Multiply the first equation by 3:
6x3y=12   
6x3y=12   
Subtract the second equation from the first equation:
00=0   
6x3y=3   
0 − 0 = 0
Well, that is actually TRUE! Zero does equal zero ...

... that is because they are really the same equation ...

... so there are an Infinite Number of Solutions
They are the same line: graph of two lines superimposed
And so now we have seen an example of each of the three possible cases:
  • No solution
  • One solution
  • Infinitely many solutions

Solving By Elimination: 3 equations in 3 variables

Before we start on the next example, let's look at an improved way to do things.
Follow this method and we are less likely to make a mistake.
First of all, eliminate the variables in order:
  • Eliminate xs first (from equation 2 and 3, in order)
  • then eliminate y (from equation 3)
So this is how we eliminate them:
elimination methods
We then have this "triangle shape":
elimination methods
Now start at the bottom and work back up (called "Back-Substitution")
(put in z to find y, then and y to find x):
elimination methods
And we are solved:
elimination methods
ALSO, we will find it is easier to do some of the calculations in our head, or on scratch paper, rather than always working within the set of equations:

Example:

  • x + y + z = 6
  • 2y + 5z = −4
  • 2x + 5y − z = 27
Written neatly:
x+y+z=6   
  2y+5z=−4   
2x+5yz=27   

First, eliminate x from 2nd and 3rd equation.
There is no x in the 2nd equation ... move on to the 3rd equation:
Subtract 2 times the 1st equation from the 3rd equation (just do this in your head or on scratch paper):
elimination methods
And we get:
x+y+z=6   
  2y+5z=−4   
  3y3z=15   

Next, eliminate y from 3rd equation.
We could subtract 1½ times the 2nd equation from the 3rd equation (because 1½ times 2 is 3) ...
... but we can avoid fractions if we:
  • multiply the 3rd equation by 2 and
  • multiply the 2nd equation by 3
and then do the subtraction ... like this:
elimination methods
And we end up with:
x+y+z=6   
  2y+5z=−4   
    z=−2   
We now have that "triangle shape"!

Now go back up again "back-substituting":
We know z, so 2y+5z=−4 becomes 2y−10=−4, then 2y=6, so y=3:
x+y+z=6   
  y  =3   
    z=−2   
Then x+y+z=6 becomes x+3−2=6, so x=6−3+2=5
x    =5   
  y  =3   
    z=−2   

And the answer is:
x = 5
y = 3
z = −2

Check: please check for yourself.

General Advice

Once you get used to the Elimination Method it becomes easier than Substitution, because you just follow the steps and the answers appear.
But sometimes Substitution can give a quicker result.
  • Substitution is often easier for small cases (like 2 equations, or sometimes 3 equations)
  • Elimination is easier for larger cases

Week of March 9, 2020 Algebra 1

Greatest Common Factor

The highest number that divides exactly into two or more numbers.
It is the "greatest" thing for simplifying fractions!

Let's start with an Example ... 

greatest common factor

Greatest Common Factor of 12 and 16

  1. Find all the Factors of each number,
  2. Circle the Common factors,
  3. Choose the Greatest of those

So ... what is a "Factor" ?

Factors are numbers we can multiply together to get another number:
factors
A number can have many factors:
Factors of 12 are 1, 2, 3, 4, 6 and 12 ...

... because 2 × 6 = 12, or 4 × 3 = 12, or 1 × 12 = 12.
(Read how to find All the Factors of a Number. In our case we don't need the negative ones.)

What is a "Common Factor" ?

Say we have worked out the factors of two numbers:

Example: Factors of 12 and 30

Factors of 12 are 1, 2, 3, 4, 6 and 12
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30
Then the common factors are those that are found in both lists:
  • Notice that 1, 2, 3 and 6 appear in both lists?
  • So, the common factors of 12 and 30 are: 1, 2, 3 and 6
It is a common factor when it is a factor of two (or more) numbers.

Here is another example with three numbers:

Example: The common factors of 15, 30 and 105

Factors of 15 are 1, 3, 5, and 15
Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30
Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105
The factors that are common to all three numbers are 1, 3, 5 and 15
In other words, the common factors of 15, 30 and 105 are 1, 3, 5 and 15

What is the "Greatest Common Factor" ?

It is simply the largest of the common factors.
In our previous example, the largest of the common factors is 15, so the Greatest Common Factor of 15, 30 and 105 is 15
The "Greatest Common Factor" is the largest of the common factors (of two or more numbers)

Why is this Useful?

One of the most useful things is when we want to simplify a fraction:

Example: How can we simplify 1230 ?

Earlier we found that the Common Factors of 12 and 30 are 1, 2, 3 and 6, and so the Greatest Common Factor is 6.
So the largest number we can divide both 12 and 30 exactly by is 6, like this:
 ÷ 6 
right over arrow 
1230 = 25
right under arrow 
 ÷ 6 
The Greatest Common Factor of 12 and 30 is 6.
And so 1230 can be simplified to 25

Finding the Greatest Common Factor

Here are three ways:
1. We can:
  • find all factors of both numbers (use the All Factors Calculator),
  • then find the ones that are common to both, and
  • then choose the greatest.
Example:
Two NumbersFactorsCommon FactorsGreatest
Common Factor
Example Simplified
Fraction
9 and 12 9: 1,3,9
12: 1,2,3,4,6,12
1,33912 = 34
And another example:
Two NumbersFactorsCommon FactorsGreatest
Common Factor
Example Simplified
Fraction
6 and 18 6: 1,2,3,6
18: 1,2,3,6,9,18
1,2,3,66618 = 13

2. Or we can find the prime factors and combine the common ones together:
Two NumbersThinking ...Greatest
Common Factor
Example Simplified
Fraction
24 and 1082 × 2 × 2 × 3 = 24, and
2 × 2 × 3 × 3 × 3 = 108
2 × 2 × 3 = 1224108 = 29

3. Or sometimes we can just play around with the factors until we discover it:
Two NumbersThinking ...Greatest
Common Factor
Example Simplified
Fraction
9 and 123 × 3 = 9 and 3 × 4 = 123912 = 34
But in that case we must check that we have found the greatest common factor.

Greatest Common Factor Calculator

OK, there is also a really easy method: we can use the Greatest Common Factor Calculator to find it automatically.