Daily Scribe

Friday in class we continued to work on how to solve equations. Mostly, we focused on equations that have negative numbers in them, or you have to use a negative number in order to solve it. For example:

7+3x=5x+13

When you are solving and equation like this you have to remember that whatever you do to one

side, you have to do to the other in order to maintain balance.

For example:

7 + 3x = 5x + 13

-7 -7 Subtract & from each side

_____________

3x = 5x + 6

Now, you have to make the equation so one side is equal to x and the other to a number.

7 + 3x = 5x + 13

-7 -7 Subtract 7 from each side

______________

3x = 5x + 6

-5 -5 subtract 5 from each side

_________________

-2x = 6

/ -2 / -2 divide by -2 on each side

Now you have a negative number one one side and a positive on the other. You still have to get one side to x so you would divide -2 by -2 and 6 by -2 in order to get the answer.

__________________

x= -3

Now that you have solved the equation you want to check your answer. You do that by plugging in the number you figured out that x was, anywhere in the equation where you see there is an x. Then you follow the order of operations in order to solve the equation. In the end, if the answer is correct both numbers on each side of the equal sign should be the same.

For example:

The equation was: 7+3x=5x+13

The answer we got is: x=-3

*reminder- the order of operations is PE

MD

AS

or Parentheses Exponent Multiplication Division Addition Subtraction

Now check to see if the answer is correct.

7+3(-3)=5(-3)+13

7+ (-9)= (-15)+13

-2 = -2

Moving Straight Ahead Problem 3.3 hw launch 3.4

Moving Straight Ahead Problem 3.3 hw launch 3.4

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Moving Straight Ahead Problem 3.3 bt

Moving Straight Ahead Problem 3.3 bt

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Daily Scribe Tues. April 27

Today we learned how to find an answer to an equation using coins and pouches. With these problems, you find how many coins are in a pouch. For example...

C= coins P=pouches

cccccccccc=pppcccc


So first you put
3p (3 pouches)+4 (the 4 coins on the side it was with) =10 (total coins on the other side)

Then you do
3p+4=10
-4 -4
__________
3p=6

So you subtract 4 from 4, or in any problem, the number of coins on the pouches side. Then you'd subtract 4 from 10, or 4 from the 10 coins on the other side.

You see after you do the first steps you see 3p=6

So then you'd do

3p=6
________
3 3
_______
1p= 2

So 3 divided by 3 is 1, so you have 1 pouch. Then you do 6 divided by 3 and get 2, so there's 2 coins per pouch.

This is what we learned in class.

Daily Scribe for 4/26/10

Today in class, we began to focus, among other things, upon mantaining an equivalent value between two things.
In other words, we were learning how to keep a balance.

We did a labsheet involving "gold coins" and "diplomatic pouches". The idea was that each pouch contained a certain number of gold coins. On one side, there was a certain amount of gold coins, and perhaps a pouch or two. On the other side was a diferent amonut of coins, and possibly pouches. By eliminating the same number of coins and pouches on each side, we were eventually able to come up with one pouch, and however many gold coins.


For pouches, I am going to use X's. For coins, I am going to use Y's.

************************************************************************

A) y y y y y y y y y y = x x x y y y y

10 y's = 3 x's plus 4 y's. First, you can eliminate 4 y's from each side. The new equation will be:

y y y y y y = x x x

So, 6 y's = 3 x's. This means 1 x = 2 y's.

Therefore, if x means pouches and y means coins, there are 2 coins per pouch.



B) x x x y y y = y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y

3 x's plus 3 y's = 30 y's, so you can eliminate 3 y's from each side.

x x x = y y y y y y y y y y y y y y y y y y y y y y y y y y y

3 x's = 27 y's, so 1 x = 9 y's. Therefore, there are 9 coins per pouch.



B2) x x y y y y = y y y y y y y y y y y y

2 x's and 4 y's = 12 y's. This means that 4 y's can be taken from each side.

x x = y y y y y y y y

2 x's = 8 y's. This means that 1 x = 4 y's. In other words, there are 4 coins per pouch.



B3) x x x = x x y y y y y y y y y y y y y

3 x's = 2 x's and 12 y's, so this time 2 x's can be eliminated from each side.

x = y y y y y y y y y y y y

x = 12 y's, so there are 12 coins per pouch.



B4) x x x y y y = x x y y y y y y y y y y y y

3 x's and 3 y's equals 2 x's and 12 y's. First take 2 x's away from each side.

x y y y = y y y y y y y y y y y y

1 x and 3 y's = 12 y's, so now 3 y's can be taken away from both sides as well.

x = y y y y y y y y y

x = 9 y's. Therefore, there are 9 coins per pouch.



B5) x x y y y y y y y y y y y y y y y y y y y y y = x x x x x y y y

2 x's and 21 y's = 5 x's and 3 y's. First take away 2 x's from each side.

y y y y y y y y y y y y y y y y y y y y y = x x x y y y

21 y's = 3 x's and 3 y's, so now eliminate 3 y's on each side.

y y y y y y y y y y y y y y y y y y = x x x

18 y's = 3 x's, so 6 y's = 1 x. There are 6 coins per pouch.

****************************************************************

That was a run-through of pretty much everything we covered in class today. Hopefully, you have now gained a clearer understanding of equivalent balances!

Signed,

Grace T :)

Moving Straight Ahead Problem 3.2

Moving Straight Ahead Problem 3.2

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Math Reflection 1

1. In a linear relationship, the dependent variable changes as the independent variable changes. A linear relationship is when the variables go up at a steady rate. For example, if someone starts off with $100 at the beginning of the week, then is left with $80 the next and $60 the next and so on, then the money (dependent variable) is decreased at a steady rate, as the days (independent variable) is also decreasing at a steady rate. The rest of the week would look like this:
Number of Days – Money Left
0 - 100
1 - 80
2 - 60
3 - 40
4 - 20
5 - 0


2. The pattern of change for a linear relationship shows up in a table if the numbers change by decreasing or increasing at a certain rate. For example, if the table:

Miles walked-Time in minutes
2 - 20
3 - 30
4 - 40
5 - 50
The number of miles increases by 1 mile every 10 minutes, so the pattern of change would be 1 mile

The pattern of change for a linear relationship shows up in a graph is the data points are connected in a straight line. For example,

In this graph, the data points are lined up in a straight line, so this shows that the pattern of change in the graph is linear. If the data points were scattered in different places on the graph, then it would show that the points would not be lined up correctly, so there would not be a linear relationship.

The pattern of changes for a linear relationship shows in an equation if the variables are being multplied of divided. Equations are usually used for linear relationships, so if data is not linear, then there usually no equation. For example, in the equation m=20w (w=weeks) (m-money left), since the variable is multiplied, it would be a linear relationship.