Separated Variables
- Thread starter 2unique
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- Apr 12, 2005
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Standard dogma: As long as you do exacly the same thing to both sides of the equation, you're good.
Here's is the one you can do.
12 - 5x = 7x
ADD 5x to both sides:
12 - 5x + 5x = 7x + 5x
12 = 12x
Here's the one you submitted:
5m = 18 + 2m
SUBTRACT 2m from both sides
5m - 2m = 18 + 2m - 2m
3m = 18
Why is that different?
Really, if these are significantly different to you, I blame your teacher or your text book. It is EXACTLY the same thing. Put them together in your mind.
Note: The "Standard Dogma" isn't really a good idea, but it will do for now. In your studies, you will see how this idea is formalized.
Here's is the one you can do.
12 - 5x = 7x
ADD 5x to both sides:
12 - 5x + 5x = 7x + 5x
12 = 12x
Here's the one you submitted:
5m = 18 + 2m
SUBTRACT 2m from both sides
5m - 2m = 18 + 2m - 2m
3m = 18
Why is that different?
Really, if these are significantly different to you, I blame your teacher or your text book. It is EXACTLY the same thing. Put them together in your mind.
Note: The "Standard Dogma" isn't really a good idea, but it will do for now. In your studies, you will see how this idea is formalized.
J
JeffM
Guest
2unique said:
The thrust of this site seems to be Socratic in nature, which means that most questions are answered by questions so that you learn the answer on your own. So, in that spirit, try answering these simpler questions.
Is it true or false that (7 + 4) = (4 + 7)?
If addition of numbers is commutative, meaning that (7 + 4) = (4 + 7), then is it true that (18 + 2m) = (2m + 18)?
If it is true that (18 + 2m,) = (2m + 18), does it make any difference whether the equation is written as 5m = (18 + 2m) or 5m = (2m + 18)?
Ok, but it says the answer to the problem I gave you is x=1
( I am studying for the ged test) and they give you online worksheets and instructions how to do the problems and an example problem.
Here is the example of how they are saying to do this.Let’s try the example from above: 6y = 2y +12
Step 1: Get the variable to ONE side of the equation.
You can do that in the same way that you did above: do the inverse operation on BOTH sides of the equation
6y = 2y +12
-2y -2y____
4y = 12
Step 2: Now solve the equation like you did above
4y = 12
4 4_
y = 3
Still very confused. Sorry....
( I am studying for the ged test) and they give you online worksheets and instructions how to do the problems and an example problem.
Here is the example of how they are saying to do this.Let’s try the example from above: 6y = 2y +12
Step 1: Get the variable to ONE side of the equation.
You can do that in the same way that you did above: do the inverse operation on BOTH sides of the equation
6y = 2y +12
-2y -2y____
4y = 12
Step 2: Now solve the equation like you did above
4y = 12
4 4_
y = 3
Still very confused. Sorry....
- Joined
- Apr 12, 2005
- Messages
- 11,339
You must learn to manipulate the sumbols.
2x + 4x = 6x -- It's like counting things.
2 dogs + 4 dogs = 6 dogs.
Don't get TOO hung up on that, since 'x' is a little more abstract than "dog".
"use the inverse operation" is a bit tricky. Don't rely on it.
1) What do you wish to accomplish? Generally, it is get all the variables on one side and all the constants on the other.
2) How do you do that? Generally, collect all the variables are remove all the constants.
3) How do you remove all the constants? Generally, decide how they are attached to the variables and find the operation that un-does the attachment:
Example:
6y = 2y + 12
We aren't collected, so let's collect. Subtract 2y from each side.
6y - 2y = 2y + 12 - 2y
Simplify -- 6y - 2y = 4y, so this leaves
4y = 2y + 12 - 2y
The right side is a little trickier. You need those awesome rules that let you move things around. In this case a+b = b+a. Addition doesn't care what order you use, right? Thsi lets us more the 12 and a 2y around, giving.
4y = 2y - 2y + 12
Simplify -- 2y - 2y = 0, so this leaves
4y = 0 + 12
Or just
4y = 12
We're almost done. That '4' is attached to the 'y' by multiplication? How do you un-do multiplication? Division!! Divide bith sides by 4.
4y/4 = 12/4
4/4 = 1, so the left side is just 'y', leaving
y = 12/4
12/4 = 3 on the right side and we are left with the solution
y = 3
Really, you just have to work a few hundred. You'll get the hang of it.
2x + 4x = 6x -- It's like counting things.
2 dogs + 4 dogs = 6 dogs.
Don't get TOO hung up on that, since 'x' is a little more abstract than "dog".
"use the inverse operation" is a bit tricky. Don't rely on it.
1) What do you wish to accomplish? Generally, it is get all the variables on one side and all the constants on the other.
2) How do you do that? Generally, collect all the variables are remove all the constants.
3) How do you remove all the constants? Generally, decide how they are attached to the variables and find the operation that un-does the attachment:
Example:
6y = 2y + 12
We aren't collected, so let's collect. Subtract 2y from each side.
6y - 2y = 2y + 12 - 2y
Simplify -- 6y - 2y = 4y, so this leaves
4y = 2y + 12 - 2y
The right side is a little trickier. You need those awesome rules that let you move things around. In this case a+b = b+a. Addition doesn't care what order you use, right? Thsi lets us more the 12 and a 2y around, giving.
4y = 2y - 2y + 12
Simplify -- 2y - 2y = 0, so this leaves
4y = 0 + 12
Or just
4y = 12
We're almost done. That '4' is attached to the 'y' by multiplication? How do you un-do multiplication? Division!! Divide bith sides by 4.
4y/4 = 12/4
4/4 = 1, so the left side is just 'y', leaving
y = 12/4
12/4 = 3 on the right side and we are left with the solution
y = 3
Really, you just have to work a few hundred. You'll get the hang of it.
J
JeffM
Guest
No duh about it.
Sometimes it does make a difference what order you do things; sometimes it does not.
With addition of numbers, it does not.
tkhenny is absolutely right: to learn even a seemingly obvious principle in math, you need to do lots of examples so the principle becomes a habit.
Go get that ged
Sometimes it does make a difference what order you do things; sometimes it does not.
With addition of numbers, it does not.
tkhenny is absolutely right: to learn even a seemingly obvious principle in math, you need to do lots of examples so the principle becomes a habit.
Go get that ged
I have another question. This may be redundant but here goes.
How would you work a problem like this? I have yet to come across a problem with 2 constants.
Do you work them separately ex: 5x+4 then 3x+20 then subtract the answers and that gives you x?
5x + 4 = 3x + 20
How would you work a problem like this? I have yet to come across a problem with 2 constants.
Do you work them separately ex: 5x+4 then 3x+20 then subtract the answers and that gives you x?
5x + 4 = 3x + 20
J
JeffM
Guest
2unique said:
I am not sure I understand the question. Presuming that the question is how do you attack an equation of the form (5x + 4) = (3x + 20), then let's get Socratic again.
In the algebra that you are studying, x is just a number, right?
As a general principle (in math as well as life), it is usually a good idea to try to simplify a problem before trying to solve it.
Let's simplify the two terms containing x. Think of the equation as a balance. So, to keep things in balance, you need to take the same number of x (remember x is just a symbol for a number not yet known) off both sides of the balance. How do you do that to get a an equation with x on just one side of the equation? You already know how to do that, right?
OK Now you have a new equation with a term in the unknown on just one side of the equation but knowns on both sides. The knowns are still numbers though, right?
So now do the same trick of simplifying to get knowns on just one side of the equation, except you want to get the known amount on the other side of the equation from the unknown. You know how to do that, right?
OK Now you have an equation with an unknown on one side and a known on the other. You know how to solve that, right?
If this seems obscure, try it a step at a time, and let me see you answer for each step. OK?
You can use google to find helpful sites...like:
http://www.mathleague.com/help/algebra/algebra.htm
http://www.mathleague.com/help/algebra/algebra.htm