bluestreak711

09-10-2007, 10:06 AM

Is there a forum with homework help or can someone help me here?:confused:

(Please see last page on forum.)

(Please see last page on forum.)

View Full Version : Algebra Yikes!

bluestreak711

09-10-2007, 10:06 AM

Is there a forum with homework help or can someone help me here?:confused:

(Please see last page on forum.)

(Please see last page on forum.)

forzainter

09-10-2007, 10:10 AM

try bbc bitesize, type into google 'bbc scotland bitesize' and then go into maths and if you are 14 - 15 - 16 click standard grade and if you are older click higher or advanced higher

ilovecarlos

09-10-2007, 10:44 AM

Is there a forum with homework help or can someone help me here?:confused:

Lotsa luck there...it took me 2 years to pass algebra (geometry is MUCH easier) and I'll tell you what....I've never ever used it...I do not care what

x =........it should be abolished...along with those, "two trains leave different stations at the same time" problems....Hey, I know..let's ban math period!!!;)

Lotsa luck there...it took me 2 years to pass algebra (geometry is MUCH easier) and I'll tell you what....I've never ever used it...I do not care what

x =........it should be abolished...along with those, "two trains leave different stations at the same time" problems....Hey, I know..let's ban math period!!!;)

Kobble

09-10-2007, 10:53 AM

Math for the sake of math had driven me crazy before college. Schools need to show people how to apply what they have learned. Check the web, though. You can find sites that break it down for you with good anaologies. My advice, learn the applications and it will stick better. Otherwise it is like learning a language without the intention of speaking it.

lethalfang

09-10-2007, 12:14 PM

Lotsa luck there...it took me 2 years to pass algebra (geometry is MUCH easier) and I'll tell you what....I've never ever used it...I do not care what

x =........it should be abolished...along with those, "two trains leave different stations at the same time" problems....Hey, I know..let's ban math period!!!;)

Are you the President by any chance?

x =........it should be abolished...along with those, "two trains leave different stations at the same time" problems....Hey, I know..let's ban math period!!!;)

Are you the President by any chance?

xtremerunnerars

09-10-2007, 01:30 PM

Post problems you have trouble with and someone (I'll help when it's something I know) will help you out.

Mastermind

09-10-2007, 01:30 PM

LMFAO @ lethal fang. :D

To the OP: if you can type it down I can probably help you, algebra isn't hard and I've taken Calculus.

To the OP: if you can type it down I can probably help you, algebra isn't hard and I've taken Calculus.

diegaa

09-10-2007, 01:43 PM

LMFAO @ lethal fang. :D

To the OP: if you can type it down I can probably help you, algebra isn't hard and I've taken Calculus.

ditto, it has happened before.

i had 2 courses of algebra and 3 calculus. I almost dont recall anything, but like mastermind said, it isnt that hard.

To the OP: if you can type it down I can probably help you, algebra isn't hard and I've taken Calculus.

ditto, it has happened before.

i had 2 courses of algebra and 3 calculus. I almost dont recall anything, but like mastermind said, it isnt that hard.

xtremerunnerars

09-10-2007, 01:50 PM

ditto, it has happened before.

i had 2 courses of algebra and 3 calculus. I almost dont recall anything, but like mastermind said, it isnt that hard.

Exactly: it's all relative. What's hard for you at age 11 is a joke when you're 18. I never really got how most tests could be hard because a lot of stuff you know it or you dont. Some things take a while and others demand a lot, but there's a pretty solid correlation between less intelligent people and how much they say stuff is hard.

Be glad if stuff comes naturally!

i had 2 courses of algebra and 3 calculus. I almost dont recall anything, but like mastermind said, it isnt that hard.

Exactly: it's all relative. What's hard for you at age 11 is a joke when you're 18. I never really got how most tests could be hard because a lot of stuff you know it or you dont. Some things take a while and others demand a lot, but there's a pretty solid correlation between less intelligent people and how much they say stuff is hard.

Be glad if stuff comes naturally!

dave333

09-10-2007, 04:17 PM

I'm takin AP Calc right now as a sophmore and lemme tell ya, if you have trouble with algebra (I breezed through it), you're in for a hell of a time.

Its a lot harder than pre-cal, alg II, alg but could be worse I guess.

Its a lot harder than pre-cal, alg II, alg but could be worse I guess.

shojun25

09-10-2007, 04:26 PM

I'm takin AP Calc right now as a sophmore and lemme tell ya, if you have trouble with algebra (I breezed through it), you're in for a hell of a time.

Its a lot harder than pre-cal, alg II, alg but could be worse I guess.

Wow, did you skip a grade or two? Because I'm a junior taking AP Calc, no sophomores at all.

By the way, did you start derivatives and limits? So far, Calc has been easy (can't say the same for the teacher).

And to the OP, algebra isn't really hard. Write down the problem and I'll try to help you with it (if I remember).

Its a lot harder than pre-cal, alg II, alg but could be worse I guess.

Wow, did you skip a grade or two? Because I'm a junior taking AP Calc, no sophomores at all.

By the way, did you start derivatives and limits? So far, Calc has been easy (can't say the same for the teacher).

And to the OP, algebra isn't really hard. Write down the problem and I'll try to help you with it (if I remember).

Noveson

09-10-2007, 04:30 PM

I'm takin AP Calc right now as a sophmore and lemme tell ya, if you have trouble with algebra (I breezed through it), you're in for a hell of a time.

Its a lot harder than pre-cal, alg II, alg but could be worse I guess.

Naw AP Calc is nothing, well now, last year it was very hard;) Though I had a much harder time with Geometry than anything I have had afterwards. Oh and AP US History is just insane. I hate it so much.....

Its a lot harder than pre-cal, alg II, alg but could be worse I guess.

Naw AP Calc is nothing, well now, last year it was very hard;) Though I had a much harder time with Geometry than anything I have had afterwards. Oh and AP US History is just insane. I hate it so much.....

Noveson

09-10-2007, 04:31 PM

Wow, did you skip a grade or two? Because I'm a junior taking AP Calc, no sophomores at all.

By the way, did you start derivatives and limits? So far, Calc has been easy (can't say the same for the teacher).

And to the OP, algebra isn't really hard. Write down the problem and I'll try to help you with it (if I remember).

Yeah, I took it as a sophomore too, and I'm not like a genius or anything lol. Must just be what school you're in, because here AP Calc as a sophomore is for the pretty damn smart kids, but it's not like super unusual.

By the way, did you start derivatives and limits? So far, Calc has been easy (can't say the same for the teacher).

And to the OP, algebra isn't really hard. Write down the problem and I'll try to help you with it (if I remember).

Yeah, I took it as a sophomore too, and I'm not like a genius or anything lol. Must just be what school you're in, because here AP Calc as a sophomore is for the pretty damn smart kids, but it's not like super unusual.

bluestreak711

09-10-2007, 07:27 PM

Post problems you have trouble with and someone (I'll help when it's something I know) will help you out.

Two missing factors

3x+4y=1

2x+2y=1

and for future reference how would i type brackets

Two missing factors

3x+4y=1

2x+2y=1

and for future reference how would i type brackets

bluestreak711

09-10-2007, 07:33 PM

Math for the sake of math had driven me crazy before college. Schools need to show people how to apply what they have learned. Check the web, though. You can find sites that break it down for you with good anaologies. My advice, learn the applications and it will stick better. Otherwise it is like learning a language without the intention of speaking it.

i searched for this specific kind of problem and it just returned two results

Two missing factors.

and i don't have a use for algebra unless i do take triginomitry or physics

beacsue when that is learned you can apply equations to tennis

and if all works well that is what i'll do

p.s. if i don't need but one between trig and physics i won't take but one

i searched for this specific kind of problem and it just returned two results

Two missing factors.

and i don't have a use for algebra unless i do take triginomitry or physics

beacsue when that is learned you can apply equations to tennis

and if all works well that is what i'll do

p.s. if i don't need but one between trig and physics i won't take but one

Court_Jester

09-10-2007, 07:34 PM

Two missing factors

3x+4y=1

2x+2y=1

Multiply the 2nd eqtn by -2 and you'll get the ff:

3x + 4y = 1

-4x -4y = -2

Add the two equations together to solve for x:

-x = -1

or

x = 1

Using this value for x, substitute this in either of the first two equations to solve for y (I'll use the 1st one):

3(1) + 4y = 1 => 4y = -2

or

y = -1/2

Solving systems of linear equations is so easy! :D

3x+4y=1

2x+2y=1

Multiply the 2nd eqtn by -2 and you'll get the ff:

3x + 4y = 1

-4x -4y = -2

Add the two equations together to solve for x:

-x = -1

or

x = 1

Using this value for x, substitute this in either of the first two equations to solve for y (I'll use the 1st one):

3(1) + 4y = 1 => 4y = -2

or

y = -1/2

Solving systems of linear equations is so easy! :D

bluestreak711

09-10-2007, 07:44 PM

Multiply the 2nd eqtn by -2 and you'll get the ff:

3x + 4y = 1

-4x -4y = -2

Add the two equations together to solve for x:

-x = -1

or

x = 1

Using this value for x, substitute this in either of the first two equations to solve for y (I'll use the 1st one):

3(1) + 4y = 1 => 4y = -2

or

y = -1/2

Solving systems of linear equations is so easy! :D

ouch idk if i got a bit of that but my work book went from a problem like

5x+7=8x+4

5x-5x+7-4=8x+4-4-5x

7-4=8x-5x

3=3x

3.....3x

- = .. -

3......3

1=x

to something like that last problem and i am lost

3x + 4y = 1

-4x -4y = -2

Add the two equations together to solve for x:

-x = -1

or

x = 1

Using this value for x, substitute this in either of the first two equations to solve for y (I'll use the 1st one):

3(1) + 4y = 1 => 4y = -2

or

y = -1/2

Solving systems of linear equations is so easy! :D

ouch idk if i got a bit of that but my work book went from a problem like

5x+7=8x+4

5x-5x+7-4=8x+4-4-5x

7-4=8x-5x

3=3x

3.....3x

- = .. -

3......3

1=x

to something like that last problem and i am lost

Court_Jester

09-10-2007, 07:53 PM

ouch idk if i got a bit of that but my work book went from a problem like

Just messin' with ya! This happens to be one of my favorite math topics when I was in high school.

5x+7=8x+4

5x-5x+7-4=8x+4-4-5x

7-4=8x-5x

3=3x

3.....3x

- = .. -

3......3

1=x

to something like that last problem and i am lost

You did this one correctly, for what it's worth.

Just messin' with ya! This happens to be one of my favorite math topics when I was in high school.

5x+7=8x+4

5x-5x+7-4=8x+4-4-5x

7-4=8x-5x

3=3x

3.....3x

- = .. -

3......3

1=x

to something like that last problem and i am lost

You did this one correctly, for what it's worth.

tennishead93

09-10-2007, 07:53 PM

gifted geometry sux

bluestreak711

09-10-2007, 07:54 PM

Just messin' with ya! This happens to be one of my favorite math topics when I was in high school.

You did this one correctly, for what it's worth.

well i didnt mean the problem i just did that i finally got that but the one with two equations i didnt understand a word

You did this one correctly, for what it's worth.

well i didnt mean the problem i just did that i finally got that but the one with two equations i didnt understand a word

Court_Jester

09-10-2007, 08:05 PM

well i didnt mean the problem i just did that i finally got that but the one with two equations i didnt understand a word

Oh,...

That problem you listed is what is called system of linear equation. There is a solution if the number of unknown variables (here, x and y) is equal to the number of equations given. The trick is to multiply one of the equations by a number so that when you add it to another one, one of the variables will be eliminated, thus simplifying the system to the point that you can solve for remaining variable.

Oh,...

That problem you listed is what is called system of linear equation. There is a solution if the number of unknown variables (here, x and y) is equal to the number of equations given. The trick is to multiply one of the equations by a number so that when you add it to another one, one of the variables will be eliminated, thus simplifying the system to the point that you can solve for remaining variable.

Tennis_Gnat

09-10-2007, 08:07 PM

Well......let's put it this way.

the 8x and the 5x are numbers that can be added and subtracted. Same with the 4 and the 7. you CANNOT add/subtract say 4 with 8x.

You want to solve for x. You want x to stand alone. So you need to combine the x terms.

It's also easier if you combine the non-x terms.

personally i think the way that's put is a bit confusing this is what i would write:

5x + 7 = 8x +4

5x goes across the river and changes signs (or gets subtracted off both sides....)

5x - 5x = 8x -5x + 4

OR (same thing, whichever makes more sense to you)

7 = 8x + 4 - 5x

(7 = 8x - 5x + 4)

7= 3x + 4

now the 4 goes across the river and changes signs

7-4 = 3x

3 = 3x

3/3 = x

1 = x

Now I feel intelligent again. thanks. ;)

the 8x and the 5x are numbers that can be added and subtracted. Same with the 4 and the 7. you CANNOT add/subtract say 4 with 8x.

You want to solve for x. You want x to stand alone. So you need to combine the x terms.

It's also easier if you combine the non-x terms.

personally i think the way that's put is a bit confusing this is what i would write:

5x + 7 = 8x +4

5x goes across the river and changes signs (or gets subtracted off both sides....)

5x - 5x = 8x -5x + 4

OR (same thing, whichever makes more sense to you)

7 = 8x + 4 - 5x

(7 = 8x - 5x + 4)

7= 3x + 4

now the 4 goes across the river and changes signs

7-4 = 3x

3 = 3x

3/3 = x

1 = x

Now I feel intelligent again. thanks. ;)

bluestreak711

09-10-2007, 08:07 PM

Oh,...

That problem you listed is what is called system of linear equation. There is a solution if the number of unknown variables (here, x and y) is equal to the number of equations given. The trick is to multiply one of the equations by a number so that when you add it to another one, one of the variables will be eliminated, thus simplifying the system to the point that you can solve for remaining variable.

Go on........

That problem you listed is what is called system of linear equation. There is a solution if the number of unknown variables (here, x and y) is equal to the number of equations given. The trick is to multiply one of the equations by a number so that when you add it to another one, one of the variables will be eliminated, thus simplifying the system to the point that you can solve for remaining variable.

Go on........

ffrpg

09-10-2007, 08:19 PM

well i didnt mean the problem i just did that i finally got that but the one with two equations i didnt understand a word

Alright, let's look at your two equations.

3x+4y=1

2x+2y=1

The idea is to find a common factor for your X and Y values. Y was solved for you earlier and the Y values have a common factor of 4. What is the common factor for X? 6 is the common factor for X because 3x and 2x are multiples of 6. Let's pretend you don't know that Y is equal to -1/2 and lets try to eliminate the X from both equations. 3x+4y=1 becomes 6x+8y=2. 2x+2y=1 becomes 6x+6Y=3. Remember that 6 is the common factor for X, so we multiply everything in the first equation by 2 and everything in the second equation by 3. Now that both equations have 6x, we can subtract the two equations and eliminate the X variable.

''' 6x+8y=2 (ignore the '''. It won't space properly unless I use them)

- 6x+6y=3

____________

2y=-1

divide the 2 and you get y=-1/2. Then you can plug that into your original first equation and you'll get 3x+4(-1/2)=1. That becomes 3x-2=1. Add two to the other side and you get 3x=3. Divide the 3 to isolate the x and you get x=3/3 which is x=1. It's pretty hard to show this online. Go ask someone to show you in person. Once you get the hang of it, it is really easy.

Alright, let's look at your two equations.

3x+4y=1

2x+2y=1

The idea is to find a common factor for your X and Y values. Y was solved for you earlier and the Y values have a common factor of 4. What is the common factor for X? 6 is the common factor for X because 3x and 2x are multiples of 6. Let's pretend you don't know that Y is equal to -1/2 and lets try to eliminate the X from both equations. 3x+4y=1 becomes 6x+8y=2. 2x+2y=1 becomes 6x+6Y=3. Remember that 6 is the common factor for X, so we multiply everything in the first equation by 2 and everything in the second equation by 3. Now that both equations have 6x, we can subtract the two equations and eliminate the X variable.

''' 6x+8y=2 (ignore the '''. It won't space properly unless I use them)

- 6x+6y=3

____________

2y=-1

divide the 2 and you get y=-1/2. Then you can plug that into your original first equation and you'll get 3x+4(-1/2)=1. That becomes 3x-2=1. Add two to the other side and you get 3x=3. Divide the 3 to isolate the x and you get x=3/3 which is x=1. It's pretty hard to show this online. Go ask someone to show you in person. Once you get the hang of it, it is really easy.

Court_Jester

09-10-2007, 08:24 PM

Let's take the problem you listed earlier:

3x + 4y = 1

2x + 2y = 1

You have a choice here, whether you want to eliminate the x first or the y. The question is how can we eliminate one of them. Let's say that we want to get rid of y first. Simply adding the two equations will not eliminate the y variable. However, if we multiply the 2nd equation by -2, the two equations will look like this:

3x + 4y = 1

-4x - 4y = -2

Note that any multiplier number other than -2 will not eliminate the y variable after adding the two equations. Now, by adding the two equations, note that the terms containing y will cancel out, leaving you with one equation with one unknown, the x variable.

3x - 4x + 4y - 4y = 1 - 2

or

-x = -1

You have now reduced the system of two linear equations to a single equation with one unknown, which can then be solved easily to give:

x = 1

Now that you have the value for x, substitute this value into either one of the original equations and you'll be able to solve for thte y variable. As a check, substitute the x and y values that you have obtained into any one of the equations. The value of the left hand side should equal the one on the right hand side.

Hope this clears it up.

3x + 4y = 1

2x + 2y = 1

You have a choice here, whether you want to eliminate the x first or the y. The question is how can we eliminate one of them. Let's say that we want to get rid of y first. Simply adding the two equations will not eliminate the y variable. However, if we multiply the 2nd equation by -2, the two equations will look like this:

3x + 4y = 1

-4x - 4y = -2

Note that any multiplier number other than -2 will not eliminate the y variable after adding the two equations. Now, by adding the two equations, note that the terms containing y will cancel out, leaving you with one equation with one unknown, the x variable.

3x - 4x + 4y - 4y = 1 - 2

or

-x = -1

You have now reduced the system of two linear equations to a single equation with one unknown, which can then be solved easily to give:

x = 1

Now that you have the value for x, substitute this value into either one of the original equations and you'll be able to solve for thte y variable. As a check, substitute the x and y values that you have obtained into any one of the equations. The value of the left hand side should equal the one on the right hand side.

Hope this clears it up.

Court_Jester

09-10-2007, 08:29 PM

bluestreak711,

ffrpg offers an alternative solution. In his case, he chose to eliminate the x variable first (I chose the y). As you can see, we both came up with the same solution.

ffrpg offers an alternative solution. In his case, he chose to eliminate the x variable first (I chose the y). As you can see, we both came up with the same solution.

bluestreak711

09-10-2007, 08:37 PM

cool i think i got it

now all i gotta do is practice similar equations for a couple weeks and go on to the next step

so i will be back that is for sure

and by the way

the only person i know that can show me in person showed me

how to solve the equation=equation

with one variable on each side

but after i showed him the next step he didnt have a clue(the one yall just showed me that is)

anyway wish me luck

too bad i didnt get this much help when i post a thread on my tennis game

now all i gotta do is practice similar equations for a couple weeks and go on to the next step

so i will be back that is for sure

and by the way

the only person i know that can show me in person showed me

how to solve the equation=equation

with one variable on each side

but after i showed him the next step he didnt have a clue(the one yall just showed me that is)

anyway wish me luck

too bad i didnt get this much help when i post a thread on my tennis game

Duzza

09-11-2007, 03:10 AM

Two missing factors

3x+4y=1

2x+2y=1

and for future reference how would i type brackets

Wow....I did that in my head mate, come on! Give us something harder!

3x+4y=1

2x+2y=1

and for future reference how would i type brackets

Wow....I did that in my head mate, come on! Give us something harder!

psp2

09-11-2007, 05:42 AM

Wow....I did that in my head mate, come on! Give us something harder!

Here's another easy one for you, Duzza.......

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

Here's another easy one for you, Duzza.......

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

The_1337

09-11-2007, 05:52 AM

^i think you could with pascals triangle.

anyways, dont you learn this stuff in school? i always found that listening to the teacher will allow you to remember how to do everything.

anyways, dont you learn this stuff in school? i always found that listening to the teacher will allow you to remember how to do everything.

lorenza

09-11-2007, 06:22 AM

Here's another easy one for you, Duzza.......

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

umm -64? i was bored and that's what i got, but i might have messed up somewhere with my negatives :)

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

umm -64? i was bored and that's what i got, but i might have messed up somewhere with my negatives :)

Trainer

09-11-2007, 07:23 AM

Damn, this stuff is nauseating. And I'm going to have to learn it again when my daughter starts it in 2 years....

jinsol

09-11-2007, 07:32 AM

lmao Trainer

idk geometry sucks since right now were doing pointless conditionals, conjectures, etc.

or

If right now were doing pointless conditionals, conjectures, etc., then geometry sucks. doesn't seem false to me...:)

its so boring we have to write it ALL out...

idk geometry sucks since right now were doing pointless conditionals, conjectures, etc.

or

If right now were doing pointless conditionals, conjectures, etc., then geometry sucks. doesn't seem false to me...:)

its so boring we have to write it ALL out...

beernutz

09-11-2007, 09:46 AM

www.algebra1.com

This is more helpful if you are actually using one of the textbooks related to this site but even if not there is some good stuff on them.

This is more helpful if you are actually using one of the textbooks related to this site but even if not there is some good stuff on them.

bluestreak711

09-11-2007, 12:24 PM

ok is this correct?

7x+10y=2

4x+8y=9

28x+40y=8

28x+56y=63

-16y=-55

-16y/-16=-55/-16

y=3.4375

7x+10(3.4375)=2

7x+34.375=2

7x+34.375-34.375=2-34.375

7x=-32.375

7x/7=-32.375/7

x=-4.625

don't forget to look for the negatives the symbol is sometimes hard to see it is this little dash -

oh by the way if ya think i did well give me a thumbs up

7x+10y=2

4x+8y=9

28x+40y=8

28x+56y=63

-16y=-55

-16y/-16=-55/-16

y=3.4375

7x+10(3.4375)=2

7x+34.375=2

7x+34.375-34.375=2-34.375

7x=-32.375

7x/7=-32.375/7

x=-4.625

don't forget to look for the negatives the symbol is sometimes hard to see it is this little dash -

oh by the way if ya think i did well give me a thumbs up

Court_Jester

09-11-2007, 12:29 PM

I was about to post that your x value should have been negative but it seems that you were able to edit it.

Good job! See how easy it is? :D

Two thumbs up!

Good job! See how easy it is? :D

Two thumbs up!

bluestreak711

09-11-2007, 12:37 PM

I was about to post that your x value should have been negative but it seems that you were able to edit it.

Good job! See how easy it is? :D

Two thumbs up!

thanks yea i did it right on paper but when i type it to the pc i always mistype something

i most of the time and get in a hurry and type oyu instead of you

but i'll be back for sure so stay subscribed to this thread please

Good job! See how easy it is? :D

Two thumbs up!

thanks yea i did it right on paper but when i type it to the pc i always mistype something

i most of the time and get in a hurry and type oyu instead of you

but i'll be back for sure so stay subscribed to this thread please

J-man

09-11-2007, 01:29 PM

Lotsa luck there...it took me 2 years to pass algebra (geometry is MUCH easier) and I'll tell you what....I've never ever used it...I do not care what

x =........it should be abolished...along with those, "two trains leave different stations at the same time" problems....Hey, I know..let's ban math period!!!;)LOL that would be nice.

x =........it should be abolished...along with those, "two trains leave different stations at the same time" problems....Hey, I know..let's ban math period!!!;)LOL that would be nice.

Duzza

09-11-2007, 09:34 PM

Here's another easy one for you, Duzza.......

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

Come on...I don't do complex numbers.....anymore.

Thanks for popping my expanding head though :D.

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

Come on...I don't do complex numbers.....anymore.

Thanks for popping my expanding head though :D.

dave333

09-12-2007, 12:54 PM

Wow, did you skip a grade or two? Because I'm a junior taking AP Calc, no sophomores at all.

By the way, did you start derivatives and limits? So far, Calc has been easy (can't say the same for the teacher).

And to the OP, algebra isn't really hard. Write down the problem and I'll try to help you with it (if I remember).

yes, we haved dived into the world of derivitives, limits, and other stuff...

as for geometry, yeah, i hated geometry. So I skipped it. So happy i did it, cause we had to do formal proofs.

I took algebra II and either got sleep or did my homework while my teacher blabbed on.

I skipped pre-cal cuz my parents told me to.

By the way, did you start derivatives and limits? So far, Calc has been easy (can't say the same for the teacher).

And to the OP, algebra isn't really hard. Write down the problem and I'll try to help you with it (if I remember).

yes, we haved dived into the world of derivitives, limits, and other stuff...

as for geometry, yeah, i hated geometry. So I skipped it. So happy i did it, cause we had to do formal proofs.

I took algebra II and either got sleep or did my homework while my teacher blabbed on.

I skipped pre-cal cuz my parents told me to.

HellBunni

09-13-2007, 09:19 AM

Two missing factors

3x+4y=1

2x+2y=1

and for future reference how would i type brackets

lol, I was confused by what you were trying to ask. Never really seen it mentioned as Two missing factors, plus the reference to brackets didn't help.

3x+4y=1

2x+2y=1

and for future reference how would i type brackets

lol, I was confused by what you were trying to ask. Never really seen it mentioned as Two missing factors, plus the reference to brackets didn't help.

jasoncho92

09-13-2007, 03:02 PM

yes, we haved dived into the world of derivitives, limits, and other stuff...

as for geometry, yeah, i hated geometry. So I skipped it. So happy i did it, cause we had to do formal proofs.

I took algebra II and either got sleep or did my homework while my teacher blabbed on.

I skipped pre-cal cuz my parents told me to.

Ive never heard of a school that just let you skip math courses just because you want to lol. And alg 1 was incredibly easy lol. I literally slept through most classes because i had it first period and i aced it easily lol. But what ap cal are you taking? AB or BC?

as for geometry, yeah, i hated geometry. So I skipped it. So happy i did it, cause we had to do formal proofs.

I took algebra II and either got sleep or did my homework while my teacher blabbed on.

I skipped pre-cal cuz my parents told me to.

Ive never heard of a school that just let you skip math courses just because you want to lol. And alg 1 was incredibly easy lol. I literally slept through most classes because i had it first period and i aced it easily lol. But what ap cal are you taking? AB or BC?

bluestreak711

09-14-2007, 09:50 AM

ok need help again

1/2x+1/3y=2

x-1/3y=1

:confused:

1/2x+1/3y=2

x-1/3y=1

:confused:

bluestreak711

09-14-2007, 09:52 AM

and another one

x-3y=7

2x-2y=18

x-3y=7

2x-2y=18

HellBunni

09-14-2007, 10:05 AM

ok need help again

1/2x+1/3y=2

x-1/3y=1

:confused:

add the 2 equations together

1/2x + 1/3y = 2

1x - 1/3y = 1

-----------------

(1/2+1)x + (1/3 - 1/3)y = 2+1

1.5 x = 3

x = 2

1/2x+1/3y=2

x-1/3y=1

:confused:

add the 2 equations together

1/2x + 1/3y = 2

1x - 1/3y = 1

-----------------

(1/2+1)x + (1/3 - 1/3)y = 2+1

1.5 x = 3

x = 2

HellBunni

09-14-2007, 10:08 AM

and another one

x-3y=7

2x-2y=18

either do substitution

x - 3y = 7 => x = 7 + 3y

so 2x - 2y = 18 => 2(7+3y) - 2y = 18

or multiple first equation by -2

-2*(x - 3y = 7) => -2x + 6y = - 14

then add the two equations together

-2x + 6y = -14

2x - 2y = 18

----------------

(-2 + 2)x + (6 - 2)y = -14 + 18

x-3y=7

2x-2y=18

either do substitution

x - 3y = 7 => x = 7 + 3y

so 2x - 2y = 18 => 2(7+3y) - 2y = 18

or multiple first equation by -2

-2*(x - 3y = 7) => -2x + 6y = - 14

then add the two equations together

-2x + 6y = -14

2x - 2y = 18

----------------

(-2 + 2)x + (6 - 2)y = -14 + 18

bluestreak711

09-14-2007, 10:10 AM

i didnt get a bit of that

HellBunni

09-14-2007, 10:13 AM

which didn't you get?

bluestreak711

09-14-2007, 10:21 AM

both of them

i was taught to find common factors in the equations then subtract

then solve for a variable then plug it in to the first equation the gon on and solve for the last variable

i was taught to find common factors in the equations then subtract

then solve for a variable then plug it in to the first equation the gon on and solve for the last variable

bluestreak711

09-14-2007, 10:22 AM

it might sound like i know what i am talking about but i half do and half dont

im lost

im lost

Flyingpanda

09-14-2007, 10:23 AM

That's fine. But if the numbers don't match you have to multiply as necessary to get eliminate one of the variables.

bluestreak711

09-14-2007, 10:32 AM

That's fine. But if the numbers don't match you have to multiply as necessary to get eliminate one of the variables.

yes which is the same as finding a common factor for one of the variables

so when you find a common factor for one variable then you can eliminate it

but basically i am having trouble with starting with a fractind and a variable

or a variable without a number beside it

such as the one starting out with a fraction

1/2x+1/3y=2

x-1/3y=1

yes which is the same as finding a common factor for one of the variables

so when you find a common factor for one variable then you can eliminate it

but basically i am having trouble with starting with a fractind and a variable

or a variable without a number beside it

such as the one starting out with a fraction

1/2x+1/3y=2

x-1/3y=1

HellBunni

09-14-2007, 10:35 AM

both of them

i was taught to find common factors in the equations then subtract

then solve for a variable then plug it in to the first equation the gon on and solve for the last variable

there are many ways to solve these.

given an equation like

x + y = z

you can look at that as

something = something

you can multiple/divide/add/subtract from somthing as long as you do it to both side of the "=" then it'll remind the same

3(something = something) => 3* something = 3* something.

now lets say you had

something = something

and

somethingelse = somethingelse

you are free to add/subtract the two equations from each other, just think of it as adding something to somethingelse.

something+somethingelse = something + somethingelse

so for like

x + y = z

2x - y = p

you can add "z" to the second equation

z + (2x - y) = z + p

since z also = x + y

(x+y) + (2x -y) = z + p

3x = z + p

x = (z+p)/3

when you add equations and/or do substitution you are trying to isolate "x" or "y" because then you can solve it. But you are free to minpulate the equations as much as you like as long as they remind =.

i was taught to find common factors in the equations then subtract

then solve for a variable then plug it in to the first equation the gon on and solve for the last variable

there are many ways to solve these.

given an equation like

x + y = z

you can look at that as

something = something

you can multiple/divide/add/subtract from somthing as long as you do it to both side of the "=" then it'll remind the same

3(something = something) => 3* something = 3* something.

now lets say you had

something = something

and

somethingelse = somethingelse

you are free to add/subtract the two equations from each other, just think of it as adding something to somethingelse.

something+somethingelse = something + somethingelse

so for like

x + y = z

2x - y = p

you can add "z" to the second equation

z + (2x - y) = z + p

since z also = x + y

(x+y) + (2x -y) = z + p

3x = z + p

x = (z+p)/3

when you add equations and/or do substitution you are trying to isolate "x" or "y" because then you can solve it. But you are free to minpulate the equations as much as you like as long as they remind =.

HellBunni

09-14-2007, 10:38 AM

yes which is the same as finding a common factor for one of the variables

so when you find a common factor for one variable then you can eliminate it

but basically i am having trouble with starting with a fractind and a variable

or a variable without a number beside it

such as the one starting out with a fraction

1/2x+1/3y=2

x-1/3y=1

it's not just common factors but also common multiple (most cases multiples are easlier).

a variable without a number beside it is the same as 1*variable.

so x - 1/3y = 1, how many "x"s are there? there's only 1.

edit:

as I said before you are free the manipulate the equations as much as you want.

so you don't like fractions?

what's the LSD of 1/2 and 1/3? multiple the first equation with that, no more fractions.

so when you find a common factor for one variable then you can eliminate it

but basically i am having trouble with starting with a fractind and a variable

or a variable without a number beside it

such as the one starting out with a fraction

1/2x+1/3y=2

x-1/3y=1

it's not just common factors but also common multiple (most cases multiples are easlier).

a variable without a number beside it is the same as 1*variable.

so x - 1/3y = 1, how many "x"s are there? there's only 1.

edit:

as I said before you are free the manipulate the equations as much as you want.

so you don't like fractions?

what's the LSD of 1/2 and 1/3? multiple the first equation with that, no more fractions.

bluestreak711

09-14-2007, 11:06 AM

well i'll do "I said" "Answer Book said"

I said

for the problem

1/2x+1/3y=2

x-1/3y=1

x+2/3y=4

x-1/3y=1

1/3y=3

1/3y over 1/3=3 over 1/3

y=9

plug that in the first equation

1/2x+1/3(9)=2

1/2x+3=2

1/2x+3-3=2-3

1/2x=2-3

1/2x= -1

1/2x over 1/2= -1 over 1/2

x= -2

The answer book said for the problem

1/2x+1/3y=2

x-1/3y=1

3/2x=3

2/3 times 3/2=3 times 2/3

x=2

1/2(2)+1/3y=2

1/3y=1

y=3

i don't uderstand

but i got this one right...

7x+10y=2

4x+8y=9

28x+40y=8

28x+56y=63

-16y=-55

-16y/-16=-55/-16

y=3.4375

7x+10(3.4375)=2

7x+34.375=2

7x+34.375-34.375=2-34.375

7x=-32.375

7x/7=-32.375/7

x=-4.625

help

I said

for the problem

1/2x+1/3y=2

x-1/3y=1

x+2/3y=4

x-1/3y=1

1/3y=3

1/3y over 1/3=3 over 1/3

y=9

plug that in the first equation

1/2x+1/3(9)=2

1/2x+3=2

1/2x+3-3=2-3

1/2x=2-3

1/2x= -1

1/2x over 1/2= -1 over 1/2

x= -2

The answer book said for the problem

1/2x+1/3y=2

x-1/3y=1

3/2x=3

2/3 times 3/2=3 times 2/3

x=2

1/2(2)+1/3y=2

1/3y=1

y=3

i don't uderstand

but i got this one right...

7x+10y=2

4x+8y=9

28x+40y=8

28x+56y=63

-16y=-55

-16y/-16=-55/-16

y=3.4375

7x+10(3.4375)=2

7x+34.375=2

7x+34.375-34.375=2-34.375

7x=-32.375

7x/7=-32.375/7

x=-4.625

help

HellBunni

09-14-2007, 11:20 AM

well i'll do "I said" "Answer Book said"

I said

for the problem

1/2x+1/3y=2

x-1/3y=1

x+2/3y=4

x-1/3y=1

1/3y=3

1/3y over 1/3=3 over 1/3

y=9

help

2/3y - (-1/3)y = 3/3 y = y

that's why you should stick with adding equations. So in this case if you took another step and multipy second equation by -1 then add, you can avoid those mistakes.

but do you see that you could have just added the 2 equations as they were to begin with to solve for "x" first? don't do that much extra work if you can help it.

I said

for the problem

1/2x+1/3y=2

x-1/3y=1

x+2/3y=4

x-1/3y=1

1/3y=3

1/3y over 1/3=3 over 1/3

y=9

help

2/3y - (-1/3)y = 3/3 y = y

that's why you should stick with adding equations. So in this case if you took another step and multipy second equation by -1 then add, you can avoid those mistakes.

but do you see that you could have just added the 2 equations as they were to begin with to solve for "x" first? don't do that much extra work if you can help it.

bluestreak711

09-14-2007, 11:27 AM

Court_Jester told me to subtract them but can you show me how to ad this equation that i am already familiar with

3x+4y=1

2x+2y=2

3x+4y=1

2x+2y=2

HellBunni

09-14-2007, 11:48 AM

Court_Jester told me to subtract them but can you show me how to ad this equation that i am already familiar with

3x+4y=1

2x+2y=2

I'll go through from the beginning then.

1) decide which variable you want to eliminate first. pick the easiest one to you.

--in this problem, I pick to eliminate "y" first because 4 is a multiple of 2, so I'll only have to change 1 equation.

2) now how do we get 2y to 4y? multiply second equation by 2

2* (2x + 2y = 2)

2(2x+2y) = 2*2

4x + 4y = 4

so now I have

3x + 4y = 1

4x + 4y = 4

so now I want to get rid of the "y"s. to use addition, multiple one of those equations by -1

-1 * (4x + 4y = 4)

-4x - 4y = -4

now

3x + 4y = 1

+(-4x - 4y = -4)

-----------------

-x = -3

x = 3

This is the same as subtraction

3x + 4y = 1

- (4x + 4y = 4)

-------------------

you just have to remember the ( ) around the equation being subtracted. so that equals

3x + 4y = 1

+ (-1)(4x + 4y = 4)

so by multiplying first then add, you are reminding yourself that signs will change.

3x+4y=1

2x+2y=2

I'll go through from the beginning then.

1) decide which variable you want to eliminate first. pick the easiest one to you.

--in this problem, I pick to eliminate "y" first because 4 is a multiple of 2, so I'll only have to change 1 equation.

2) now how do we get 2y to 4y? multiply second equation by 2

2* (2x + 2y = 2)

2(2x+2y) = 2*2

4x + 4y = 4

so now I have

3x + 4y = 1

4x + 4y = 4

so now I want to get rid of the "y"s. to use addition, multiple one of those equations by -1

-1 * (4x + 4y = 4)

-4x - 4y = -4

now

3x + 4y = 1

+(-4x - 4y = -4)

-----------------

-x = -3

x = 3

This is the same as subtraction

3x + 4y = 1

- (4x + 4y = 4)

-------------------

you just have to remember the ( ) around the equation being subtracted. so that equals

3x + 4y = 1

+ (-1)(4x + 4y = 4)

so by multiplying first then add, you are reminding yourself that signs will change.

D-man

09-14-2007, 01:39 PM

I said

for the problem

1/2x+1/3y=2

x-1/3y=1

x+2/3y=4

x-1/3y=1

1/3y=3

Don't know if this was addressed, but I think your problem here was you forget to reverse the [-1/3y] into a postive when you subtracted the second equation from the first, so that in fact you should add the [1/3y] to the [2/3y] which comes out to [1y or y].

x+2/3y=4

-(x-1/3y=1) which turns into (-x+1/3y=-1)

—————

y=3

Sorry if that was addressed, just randomly popped in.

for the problem

1/2x+1/3y=2

x-1/3y=1

x+2/3y=4

x-1/3y=1

1/3y=3

Don't know if this was addressed, but I think your problem here was you forget to reverse the [-1/3y] into a postive when you subtracted the second equation from the first, so that in fact you should add the [1/3y] to the [2/3y] which comes out to [1y or y].

x+2/3y=4

-(x-1/3y=1) which turns into (-x+1/3y=-1)

—————

y=3

Sorry if that was addressed, just randomly popped in.

bluestreak711

09-15-2007, 08:14 PM

Don't know if this was addressed, but I think your problem here was you forget to reverse the [-1/3y] into a postive when you subtracted the second equation from the first, so that in fact you should add the [1/3y] to the [2/3y] which comes out to [1y or y].

x+2/3y=4

-(x-1/3y=1) which turns into (-x+1/3y=-1)

—————

y=3

Sorry if that was addressed, just randomly popped in.

i don't belive it was thanks

by the way i don't care if i recieve algebra advice more than once

that makes it easier to understand

x+2/3y=4

-(x-1/3y=1) which turns into (-x+1/3y=-1)

—————

y=3

Sorry if that was addressed, just randomly popped in.

i don't belive it was thanks

by the way i don't care if i recieve algebra advice more than once

that makes it easier to understand

SoBad

09-15-2007, 08:19 PM

I don't think you're weird, maybe just a little tipsy at present. Chase your next shot with a nice juicy mango...;)

VadeRetro

09-16-2007, 03:31 PM

umm -64? i was bored and that's what i got, but i might have messed up somewhere with my negatives :)

Not the negatives, but i think you forgot the i

I think it is -64i

Not the negatives, but i think you forgot the i

I think it is -64i

VadeRetro

09-16-2007, 03:51 PM

Not the negatives, but i think you forgot the i

I think it is -64i

I was too quick

It is indeed in the negatives.

(1-i)^2 = -2i

(1-i)^3 = 2(1-i)

(1-i)^4 = -4i

(1-i)^5 = -4(1+i)

(1-i)^6 = -8

so (1-i)^12 must be 64

I think it is -64i

I was too quick

It is indeed in the negatives.

(1-i)^2 = -2i

(1-i)^3 = 2(1-i)

(1-i)^4 = -4i

(1-i)^5 = -4(1+i)

(1-i)^6 = -8

so (1-i)^12 must be 64

VadeRetro

09-16-2007, 04:35 PM

Mea culpa

'Errare humanem est' is no apology.

I am a fool,

I bow deeply to Lorenza for having it right in the first place.

(1-i)^2 = 1^2 -2i + i^2= 1-2i+(-1) = -2i

(1-i)^3 = -2i(1-i)=-2i+2i^2=-2i+(2(-1))=-2i-2=-2(1+i)

(1-i)^4 = -2(1+i)(1-i) = -2(1^2 -i^2) = -2(1-(-1))=-2(2) =-4

(1-i)^12 = -4 x -4 x -4 = -64

'Errare humanem est' is no apology.

I am a fool,

I bow deeply to Lorenza for having it right in the first place.

(1-i)^2 = 1^2 -2i + i^2= 1-2i+(-1) = -2i

(1-i)^3 = -2i(1-i)=-2i+2i^2=-2i+(2(-1))=-2i-2=-2(1+i)

(1-i)^4 = -2(1+i)(1-i) = -2(1^2 -i^2) = -2(1-(-1))=-2(2) =-4

(1-i)^12 = -4 x -4 x -4 = -64

iplaybetter

09-16-2007, 04:56 PM

i love algebra its so easy, i cant stand normal math anymore, but i am in algebra 2 honors in ninth grade

bluestreak711

09-17-2007, 04:22 AM

I'll go through from the beginning then.

1) decide which variable you want to eliminate first. pick the easiest one to you.

--in this problem, I pick to eliminate "y" first because 4 is a multiple of 2, so I'll only have to change 1 equation.

2) now how do we get 2y to 4y? multiply second equation by 2

2* (2x + 2y = 2)

2(2x+2y) = 2*2

4x + 4y = 4

so now I have

3x + 4y = 1

4x + 4y = 4

so now I want to get rid of the "y"s. to use addition, multiple one of those equations by -1

-1 * (4x + 4y = 4)

-4x - 4y = -4

now

3x + 4y = 1

+(-4x - 4y = -4)

-----------------

-x = -3

x = 3

This is the same as subtraction

3x + 4y = 1

- (4x + 4y = 4)

-------------------

you just have to remember the ( ) around the equation being subtracted. so that equals

3x + 4y = 1

+ (-1)(4x + 4y = 4)

so by multiplying first then add, you are reminding yourself that signs will change.

i have read this many times but still have limited understanding

1) decide which variable you want to eliminate first. pick the easiest one to you.

--in this problem, I pick to eliminate "y" first because 4 is a multiple of 2, so I'll only have to change 1 equation.

2) now how do we get 2y to 4y? multiply second equation by 2

2* (2x + 2y = 2)

2(2x+2y) = 2*2

4x + 4y = 4

so now I have

3x + 4y = 1

4x + 4y = 4

so now I want to get rid of the "y"s. to use addition, multiple one of those equations by -1

-1 * (4x + 4y = 4)

-4x - 4y = -4

now

3x + 4y = 1

+(-4x - 4y = -4)

-----------------

-x = -3

x = 3

This is the same as subtraction

3x + 4y = 1

- (4x + 4y = 4)

-------------------

you just have to remember the ( ) around the equation being subtracted. so that equals

3x + 4y = 1

+ (-1)(4x + 4y = 4)

so by multiplying first then add, you are reminding yourself that signs will change.

i have read this many times but still have limited understanding

bluestreak711

09-17-2007, 06:10 AM

Help:confused:

bluestreak711

09-17-2007, 06:53 AM

i still don't understand how to subtract an equation to solve for Two Missing Factors

bluestreak711

09-17-2007, 07:15 AM

my brain is getting fried idk if i did this simple one right

6x-4=8x+6

6x-4+4+8x=8x-8x+6+4

6x+8x=6+4

14x=10

14x/14=10/14

x=0.714285714

6x-4=8x+6

6x-4+4+8x=8x-8x+6+4

6x+8x=6+4

14x=10

14x/14=10/14

x=0.714285714

scineram

09-17-2007, 07:32 AM

Choked again. x=-5

bluestreak711

09-17-2007, 08:36 AM

Choked again. x=-5

well what steps did i mess up?

well what steps did i mess up?

bluestreak711

09-17-2007, 08:53 AM

Choked again. x=-5

i think i see

6x-4=8x+6

6x-4+4-8x=8x-8x+6+4

6x-8x=6+4

-2x=10

-2x/-2=10/-2

x=-5

i think i see

6x-4=8x+6

6x-4+4-8x=8x-8x+6+4

6x-8x=6+4

-2x=10

-2x/-2=10/-2

x=-5

bluestreak711

09-17-2007, 09:30 AM

Is there anyway to solve any further for

630x=9y

630x=9y

bluestreak711

09-17-2007, 09:31 AM

and i also need further explination on how to subtract two equations

surfryder

09-17-2007, 09:33 AM

Is there anyway to solve any further for

630x=9y

no.....ten chars

630x=9y

no.....ten chars

surfryder

09-17-2007, 09:34 AM

my brain is getting fried idk if i did this simple one right

6x-4=8x+6

6x-4+4+8x=8x-8x+6+4

6x+8x=6+4

14x=10

14x/14=10/14

x=0.714285714

6x-8x=6+4

-2x=10

x=-5

those are your steps

6x-4=8x+6

6x-4+4+8x=8x-8x+6+4

6x+8x=6+4

14x=10

14x/14=10/14

x=0.714285714

6x-8x=6+4

-2x=10

x=-5

those are your steps

bluestreak711

09-18-2007, 07:37 AM

what i need is further explination of how to subtract equations

1/2x+1/3y=2

x-1/3y=1

1/2x+1/3y=2

x-1/3y=1

HellBunni

09-19-2007, 09:35 AM

you line up the variables and subtract like terms.

SOY78

09-19-2007, 10:05 AM

what i need is further explination of how to subtract equations

1/2x+1/3y=2

x-1/3y=1

1 1/2x = 3

x=2

i think u can do the rest...

I love math/science/physics, that is why I am a structural engineer ;)

1/2x+1/3y=2

x-1/3y=1

1 1/2x = 3

x=2

i think u can do the rest...

I love math/science/physics, that is why I am a structural engineer ;)

bluestreak711

09-19-2007, 11:23 AM

1 1/2x = 3

x=2

i think u can do the rest...

I love math/science/physics, that is why I am a structural engineer ;)

don't quite understand

x=2

i think u can do the rest...

I love math/science/physics, that is why I am a structural engineer ;)

don't quite understand

D-man

09-19-2007, 11:35 AM

don't quite understand

He chose to add the equations.

....1/2x+1/3y=2

+(......x-1/3y=1)

——————————

The plus sign leaves all the signs the same. If you subtract the equation then you reverse the signs of the bottom equations. Add each part straight down.

1/2x + 1 x = 1 1/2x

+1/3y - 1/3y = 0

2 + 1 = 3

So we end up with,

1 1/2x + 0 = 3

or

3/2x = 3

We can simplify that by:

(2/3) * 3/2x = (2/3) * 3/1

6/6x = 6/3

x = 2

He chose to add the equations.

....1/2x+1/3y=2

+(......x-1/3y=1)

——————————

The plus sign leaves all the signs the same. If you subtract the equation then you reverse the signs of the bottom equations. Add each part straight down.

1/2x + 1 x = 1 1/2x

+1/3y - 1/3y = 0

2 + 1 = 3

So we end up with,

1 1/2x + 0 = 3

or

3/2x = 3

We can simplify that by:

(2/3) * 3/2x = (2/3) * 3/1

6/6x = 6/3

x = 2

bluestreak711

09-19-2007, 12:26 PM

and is this one close to being right?

4x=2y=9

7x+2y=8

x=1.continuous54

y= 1.40909091

4x=2y=9

7x+2y=8

x=1.continuous54

y= 1.40909091

HellBunni

09-19-2007, 01:11 PM

and is this one close to being right?

4x=2y=9

7x+2y=8

x=1.continuous54

y= 1.40909091

do you mean 4x - 2y = 9?

(4x - 2y) + (7x + 2y) = 9 + 8

11x = 17

x = 17/11 = 1.5454545...

4x - 2y = 9

(4x-9)/2 = y

2x - 9/2 = y

34/11 - 9/2 = y

y = 68/22 - 99/22

y = -31/22 = -1.4090909...

4x=2y=9

7x+2y=8

x=1.continuous54

y= 1.40909091

do you mean 4x - 2y = 9?

(4x - 2y) + (7x + 2y) = 9 + 8

11x = 17

x = 17/11 = 1.5454545...

4x - 2y = 9

(4x-9)/2 = y

2x - 9/2 = y

34/11 - 9/2 = y

y = 68/22 - 99/22

y = -31/22 = -1.4090909...

dave333

09-19-2007, 02:18 PM

Ive never heard of a school that just let you skip math courses just because you want to lol. And alg 1 was incredibly easy lol. I literally slept through most classes because i had it first period and i aced it easily lol. But what ap cal are you taking? AB or BC?

BC.

I didn't just say "i want to skip" and they moved me up, I had to study over the summer (i did it at home, but other people take courses at colleges) and answer selected questions from the midterm and final.

BC.

I didn't just say "i want to skip" and they moved me up, I had to study over the summer (i did it at home, but other people take courses at colleges) and answer selected questions from the midterm and final.

bluestreak711

09-19-2007, 05:30 PM

do you mean 4x - 2y = 9?

(4x - 2y) + (7x + 2y) = 9 + 8

11x = 17

x = 17/11 = 1.5454545...

4x - 2y = 9

(4x-9)/2 = y

2x - 9/2 = y

34/11 - 9/2 = y

y = 68/22 - 99/22

y = -31/22 = -1.4090909...

im sorry i mistyped the problem

i meant

4x+2y=9

7x+2y=8

(4x - 2y) + (7x + 2y) = 9 + 8

11x = 17

x = 17/11 = 1.5454545...

4x - 2y = 9

(4x-9)/2 = y

2x - 9/2 = y

34/11 - 9/2 = y

y = 68/22 - 99/22

y = -31/22 = -1.4090909...

im sorry i mistyped the problem

i meant

4x+2y=9

7x+2y=8

HellBunni

09-21-2007, 06:24 AM

im sorry i mistyped the problem

i meant

4x+2y=9

7x+2y=8

eq2 - eq1

(7x - 4x) + (2y - 2y) = 8 - 9

3x = -1

x = -1/3

4x + 2y = 9

y = (9-4x)/2

y = (9-4(-1/3))/2

y = (9 + 4/3))/2

y = 9/2 + 2/3 = 27/6+ 4/6 = 31/6

i meant

4x+2y=9

7x+2y=8

eq2 - eq1

(7x - 4x) + (2y - 2y) = 8 - 9

3x = -1

x = -1/3

4x + 2y = 9

y = (9-4x)/2

y = (9-4(-1/3))/2

y = (9 + 4/3))/2

y = 9/2 + 2/3 = 27/6+ 4/6 = 31/6

psp2

09-21-2007, 05:42 PM

Here's another easy one for you, Duzza.......

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

Here's the answer:

(1-i)^12 is equivalent to [(1-i)^2]^6

(1-i)^2 is "1-2i+i^2" (simple squaring of a binomial)

i^2 is -1.... -2i

now, raise "-2i" to the ^6 and you get 64i^6. i^6 is i^2 x i^2 x i^2, which is -1 x -1 x -1.

Answer: 64 x -1 = -64

What's the value of (1-i)^12, where i is the sqrt(-1)?

Let's see you expand that in your head :)

Here's the answer:

(1-i)^12 is equivalent to [(1-i)^2]^6

(1-i)^2 is "1-2i+i^2" (simple squaring of a binomial)

i^2 is -1.... -2i

now, raise "-2i" to the ^6 and you get 64i^6. i^6 is i^2 x i^2 x i^2, which is -1 x -1 x -1.

Answer: 64 x -1 = -64

psp2

09-21-2007, 05:58 PM

For any systems of linear equations, you guys will probably learn the "matrix" method later. This method is called the Cramer's Rule.

Cramer's Rule is here.

http://en.wikipedia.org/wiki/Cramer's_rule

Cramer's Rule is here.

http://en.wikipedia.org/wiki/Cramer's_rule

WhiteSox05CA

09-21-2007, 07:15 PM

I'm beginning to love algebra, F@(*$ Calculus!

krz

09-21-2007, 07:41 PM

im sorry i mistyped the problem

i meant

4x+2y=9

7x+2y=8

you could just solve for y and then plug it into x.

so

y= (8-7x)/2

So the equation for you to solve would be

4x+2(8-7x)/2=9

x should be -1/3

then just plug -1/3 in for x in any of the equations to get the y.

that seems like the easiest way to think about it, although probably not the fastest.

we do this kind of stuff in Econometrics although a slightly more complicated version and using different methods :P

i meant

4x+2y=9

7x+2y=8

you could just solve for y and then plug it into x.

so

y= (8-7x)/2

So the equation for you to solve would be

4x+2(8-7x)/2=9

x should be -1/3

then just plug -1/3 in for x in any of the equations to get the y.

that seems like the easiest way to think about it, although probably not the fastest.

we do this kind of stuff in Econometrics although a slightly more complicated version and using different methods :P

bluestreak711

09-25-2007, 10:04 AM

ok in my workbook i notice sometimes they add equations and sometimes subtract equations what is the basic rule to discover which i do

i am still talking about the equations with "Two Missing Factors"

i am still talking about the equations with "Two Missing Factors"

Court_Jester

09-25-2007, 12:26 PM

ok in my workbook i notice sometimes they add equations and sometimes subtract equations what is the basic rule to discover which i do

Whichever one is easier. Simple as that.

Whichever one is easier. Simple as that.

mucat

09-25-2007, 01:43 PM

For any systems of linear equations, you guys will probably learn the "matrix" method later. This method is called the Cramer's Rule.

Cramer's Rule is here.

http://en.wikipedia.org/wiki/Cramer's_rule

Oh yes, I remebemr my Linear Algebra 3rd yr math\computer class. Tons of students in the beginning then everyone drop out and the class shrink to around 5-7 students, me, my wife and the rest are a few honor students. Scary course. We skipped so many of the class it is not even funny. At the end, we passed :) What a relief...

Cramer's Rule is here.

http://en.wikipedia.org/wiki/Cramer's_rule

Oh yes, I remebemr my Linear Algebra 3rd yr math\computer class. Tons of students in the beginning then everyone drop out and the class shrink to around 5-7 students, me, my wife and the rest are a few honor students. Scary course. We skipped so many of the class it is not even funny. At the end, we passed :) What a relief...

bluestreak711

09-26-2007, 08:30 AM

:confused: Whichever one is easier. Simple as that.

i do it both ways and only one works but i dont always knoe which one but i think i have noticed it might be that you add the mathematical operations in both equations and like add or subtract or positive and negative and its the opposite of the answer

like positive and postive equals positive so then i would subtract

negative and negative equals positive so then i would subtract

positive and neagative equals neagative so then i would add

my work book seemed to follow that pattern

if that is not the case i am still having problems

first how do i know which is the easiest mathimatical operation

i would usually get to problem and wouldnt know to add or subtract and guess one and by chance do i get it right

help!:confused:

i do it both ways and only one works but i dont always knoe which one but i think i have noticed it might be that you add the mathematical operations in both equations and like add or subtract or positive and negative and its the opposite of the answer

like positive and postive equals positive so then i would subtract

negative and negative equals positive so then i would subtract

positive and neagative equals neagative so then i would add

my work book seemed to follow that pattern

if that is not the case i am still having problems

first how do i know which is the easiest mathimatical operation

i would usually get to problem and wouldnt know to add or subtract and guess one and by chance do i get it right

help!:confused:

bluestreak711

09-28-2007, 08:03 AM

help:confused: ^^^^^

HellBunni

09-28-2007, 10:59 AM

adding or subtracting doesn't matter.

they are the same more or less.

you do whichever is easier for you.

you mess up because you forget the signs when you subtract. You have to remember to carry out the signs.

examlpe

4x - 7y = 48

10x - 18y = 10

if you subtract, get into the habit of using ( ) whenever you substitute.

(4x - 7y) - (10x - 18y) = (48) - (10)

(4x - 7y) stays as 4x - 7y when you remove the ( )

- (10x - 18y) = -10x + 18y when you remove the ( )

4x - 7y - 10x + 18y = 48 - 10

and etc...

they are the same more or less.

you do whichever is easier for you.

you mess up because you forget the signs when you subtract. You have to remember to carry out the signs.

examlpe

4x - 7y = 48

10x - 18y = 10

if you subtract, get into the habit of using ( ) whenever you substitute.

(4x - 7y) - (10x - 18y) = (48) - (10)

(4x - 7y) stays as 4x - 7y when you remove the ( )

- (10x - 18y) = -10x + 18y when you remove the ( )

4x - 7y - 10x + 18y = 48 - 10

and etc...

bluestreak711

09-28-2007, 12:01 PM

adding or subtracting doesn't matter.

they are the same more or less.

you do whichever is easier for you.

you mess up because you forget the signs when you subtract. You have to remember to carry out the signs.

examlpe

4x - 7y = 48

10x - 18y = 10

if you subtract, get into the habit of using ( ) whenever you substitute.

(4x - 7y) - (10x - 18y) = (48) - (10)

(4x - 7y) stays as 4x - 7y when you remove the ( )

- (10x - 18y) = -10x + 18y when you remove the ( )

4x - 7y - 10x + 18y = 48 - 10

and etc...

ok why did you change 10x to a negative and then add it to 18y?

i dont understand yet but i am beginning to

they are the same more or less.

you do whichever is easier for you.

you mess up because you forget the signs when you subtract. You have to remember to carry out the signs.

examlpe

4x - 7y = 48

10x - 18y = 10

if you subtract, get into the habit of using ( ) whenever you substitute.

(4x - 7y) - (10x - 18y) = (48) - (10)

(4x - 7y) stays as 4x - 7y when you remove the ( )

- (10x - 18y) = -10x + 18y when you remove the ( )

4x - 7y - 10x + 18y = 48 - 10

and etc...

ok why did you change 10x to a negative and then add it to 18y?

i dont understand yet but i am beginning to

punch

09-28-2007, 01:13 PM

Man, I think you need to get a tutor or go ask your teacher for extra help. I don't think you should be relying on this forum, as it is very hard to explain things in full description.

D-man

09-28-2007, 01:42 PM

ok why did you change 10x to a negative and then add it to 18y?

i dont understand yet but i am beginning to

Very simple, when you put negative in front of stuff in parenthesis, for example:

- (9 - 3 + 4 - 2)

you are times-ing all the numbers inside by -1, so their polarity is reversed.

- (9 - 3 + 4 - 2) =

-9 +3 -4 +2

i dont understand yet but i am beginning to

Very simple, when you put negative in front of stuff in parenthesis, for example:

- (9 - 3 + 4 - 2)

you are times-ing all the numbers inside by -1, so their polarity is reversed.

- (9 - 3 + 4 - 2) =

-9 +3 -4 +2

bluestreak711

09-30-2007, 04:37 PM

Very simple, when you put negative in front of stuff in parenthesis, for example:

- (9 - 3 + 4 - 2)

you are times-ing all the numbers inside by -1, so their polarity is reversed.

- (9 - 3 + 4 - 2) =

-9 +3 -4 +2

ok but how do i know what goes in parenthesis

- (9 - 3 + 4 - 2)

you are times-ing all the numbers inside by -1, so their polarity is reversed.

- (9 - 3 + 4 - 2) =

-9 +3 -4 +2

ok but how do i know what goes in parenthesis

D-man

09-30-2007, 04:52 PM

ok but how do i know what goes in parenthesis

well like if you are subtracting equations the second bottom equation has a negative put around the whole thing, like the following. First you have the original problem:

3 + 2x = 5y

2 + 2x = 8y

so you subtract the second one like this:

3 + 2x = 5y

- (2 + 2x) = - (8y)

3 + 2x = 5y

-2 -2x = -8y

————————

1 + 0 = -3y

y = - 1/3

Like you decide whether to add or subtract the equations or whatever by whatever seems the easiest way to do

well like if you are subtracting equations the second bottom equation has a negative put around the whole thing, like the following. First you have the original problem:

3 + 2x = 5y

2 + 2x = 8y

so you subtract the second one like this:

3 + 2x = 5y

- (2 + 2x) = - (8y)

3 + 2x = 5y

-2 -2x = -8y

————————

1 + 0 = -3y

y = - 1/3

Like you decide whether to add or subtract the equations or whatever by whatever seems the easiest way to do

Gmedlo

09-30-2007, 05:06 PM

Reading through this, everything sounds overcomplicated and it will probably just confuse you. Just get help from a tutor from your local HS or ask for extra help from whoever is teaching you.

Flyingpanda

09-30-2007, 08:34 PM

I don't want to be mean, but the basics have already been explained pretty well in this thread. If you don't get it by now go ask your teacher.

bluestreak711

11-13-2007, 08:37 AM

Ok thanks for all your help I did eventually get someone to help me I am in home school right now because where I live the public schools are awful and my grandmother teaches me but this is one subject that goes over her head.

So I couldn't ask my teacher.

I do appreciate all your help.

I may eventually need to come back when I get something different and harder.:p

So I couldn't ask my teacher.

I do appreciate all your help.

I may eventually need to come back when I get something different and harder.:p

bluestreak711

11-26-2007, 01:20 PM

If anyone is still here i have one quick question.

Mononomials

i am a little confused about the degrees of a mononomial

7a-degree of 1

2x to the 3 power-degree of 3

4abc to the 3 power-degree of 5

Why is that last one to the degree of 5?

Mononomials

i am a little confused about the degrees of a mononomial

7a-degree of 1

2x to the 3 power-degree of 3

4abc to the 3 power-degree of 5

Why is that last one to the degree of 5?

tricky

11-26-2007, 01:44 PM

7a^1 = 7 * a = degree of 1

2x^3 = 2 * x * x * x = degree of 3

4ab(c^3) = 4 * a * b * c * c * c = degree of 5

2x^3 = 2 * x * x * x = degree of 3

4ab(c^3) = 4 * a * b * c * c * c = degree of 5

bluestreak711

11-27-2007, 11:19 AM

7a^1 = 7 * a = degree of 1

2x^3 = 2 * x * x * x = degree of 3

4ab(c^3) = 4 * a * b * c * c * c = degree of 5

Perfect thankyou very much.

I see

Basically you add up the exponents to get the degree and variables that doesnt have exponents are just to be assumed as its exponent to be 1

2x^3 = 2 * x * x * x = degree of 3

4ab(c^3) = 4 * a * b * c * c * c = degree of 5

Perfect thankyou very much.

I see

Basically you add up the exponents to get the degree and variables that doesnt have exponents are just to be assumed as its exponent to be 1

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