Solving for X (new to forums)

**wilder7bc** · January 9th 2010, 06:43 PM

I am reviewing math in the back of my physics book and I don’t have a college algebra book anymore. I tried goggling this for 2 hours or so and could not find anything similar. It seems like it’s a certain type of problem giving me problems.

Two problems I have the answer to both but cannot figure out how they came up with the answer.

Solve for x

ax – 5 = bx +2 Answer is: x = (7/(a-b))

I cannot figure this out I moved the 5 over that’s easy

ax = bx + 7

I then tried to move the “bx”

ax – bx = 7

I then divided 7 by -b

ax –x = (7/(-b))

I then divided by a

x – x = (7/(a-b))

However would not x – x = 0

I must not be getting some rule…. Even if when I moved the negative b over if I didn’t leave a neg it would be x +x which would = 2x
Right? I am confused on this and the other one I had problems with Is similar.

A = (1/(1+x)) Answer is x = ((1-a)/(a))

I think these are related because this one also I could not solve this one either… I solve all the normal x stuff but it has to be a rule I am missing…

I think I figured out the second problem as listed below I left it in just in case I made a mistake.. I have taken College Algebra, Trig, Calculus I but that was back in 1997. I am now enrolled in Engineering Physics, and Calculus II which start next week so trying to review and catch up and review stuff on my own before I am really hazy on this stuff so forgive any seeming ignorance on my part. I may be posting a lot of questions on here in the coming weeks. Next week I will have access to school tutors as well and I should be getting in my Calculus I, and II MathTutorDvD I ordered from a website... hopefully they are good if so I will order Calc III and Physics CD from them as well. I use to be really good at math but not feeling so good at it right now

a = (1/(1+x))

took (1+x) and since I was dividing I did the opposite and multiplied by both sides

a(1+x) = 1

I think that then breaks down to

1a + ax = 1

I then did opposite moving 1a over

ax = 1 – a (since “a” and “1a” are the same thing I just took labeled it as “a”)

I then divided by a

X = (1- a)/(a)

I don’t know why I got confused on that one now it seems easy

So its just that first one I am having a problem with ugggg I hate not being able to figure out something I will almost go till my nails are fingers are bleeding and I fall over in the chair from exhaustion lol. Learning to put things away to come back a different day is very hard for me lol

**I-Think** · January 9th 2010, 06:47 PM

$ax-5=bx+2$

$ax=bx+2+5$

$ax=bx+7$

$ax-bx=7$

$x(a-b)=7$

$x=\frac{7}{a-b}$

**S31J41** · January 9th 2010, 06:49 PM

Originally Posted by wilder7bc

I am reviewing math in the back of my physics book and I don’t have a college algebra book anymore. I tried goggling this for 2 hours or so and could not find anything similar. It seems like it’s a certain type of problem giving me problems.

Two problems I have the answer to both but cannot figure out how they came up with the answer.

Solve for x

ax – 5 = bx +2 Answer is: x = (7/(a-b))

I cannot figure this out I moved the 5 over that’s easy

ax = bx + 7

I then tried to move the “bx”

ax – bx = 7

I then divided 7 by -b

ax –x = (7/(-b))

I then divided by a

x – x = (7/(a-b))

However would not x – x = 0

I must not be getting some rule…. Even if when I moved the negative b over if I didn’t leave a neg it would be x +x which would = 2x
Right? I am confused on this and the other one I had problems with Is similar.

A = (1/(1+x)) Answer is x = ((1-a)/(a))

I think these are related because this one also I could not solve this one either… I solve all the normal x stuff but it has to be a rule I am missing…

I think I figured out the second problem as listed below I left it in just in case I made a mistake.. I have taken College Algebra, Trig, Calculus I but that was back in 1997. I am now enrolled in Engineering Physics, and Calculus II which start next week so trying to review and catch up and review stuff on my own before I am really hazy on this stuff so forgive any seeming ignorance on my part. I may be posting a lot of questions on here in the coming weeks. Next week I will have access to school tutors as well and I should be getting in my Calculus I, and II MathTutorDvD I ordered from a website... hopefully they are good if so I will order Calc III and Physics CD from them as well. I use to be really good at math but not feeling so good at it right now

a = (1/(1+x))

took (1+x) and since I was dividing I did the opposite and multiplied by both sides

a(1+x) = 1

I think that then breaks down to

1a + ax = 1

I then did opposite moving 1a over

ax = 1 – a (since “a” and “1a” are the same thing I just took labeled it as “a”)

I then divided by a

X = (1- a)/(a)

I don’t know why I got confused on that one now it seems easy

So its just that first one I am having a problem with ugggg I hate not being able to figure out something I will almost go till my nails are fingers are bleeding and I fall over in the chair from exhaustion lol. Learning to put things away to come back a different day is very hard for me lol

Alright with the first one, what you did initially was correct, so I am going to start with

ax – bx = 7

What you need to do next is realize that both (ax) and (bx) have an (x) term, so you can factor

ax - bx into x(a-b) I think that is the confusing part. once you do that
x(a-b) = 7, simply divide both sides by (a-b)

**wilder7bc** · January 9th 2010, 06:54 PM

Wow... that was fast...

I have a question though...

I am not sure I understand this:

Is there any documentation on now this works or what its called. I am trying to remember this but its not striking a bell.

I understand looking at it that x(a - b) = 7 is the same as xa - xb = 7

but I dont remember pulling an x out like that. (btw is that algebra I, II, or College Algebra? (my terminology is going to be yuck for awhile lol)

Hey thanks a ton for the prompt answer I am so glad I found this forum!

Sincerely,

Brian

**wilder7bc** · January 9th 2010, 06:58 PM

Ok I think S31J41 answered the second part of my question... I am so relieved I tried figuring that out for hours...

So thats called factoring out and that definitely sounds familiar and I am starting to recall a tad bit about that. Also can now maybe do searches on it if I need to.

Guys are awesome thanks!

Sincerely,

Brian

**jgv115** · January 11th 2010, 12:34 AM

I'll just contribute a tad more (cause i can :P)

For the second one you have

$a(1+x)=1$

You can simply divide the right hand side by a because $a(1+x)$ is $a * (1+x)$

$1+x = \frac{1}{a}$

**rowe** · January 11th 2010, 03:59 AM

Originally Posted by wilder7bc

Wow... that was fast...

I have a question though...

I am not sure I understand this:

Is there any documentation on now this works or what its called. I am trying to remember this but its not striking a bell.

I understand looking at it that x(a - b) = 7 is the same as xa - xb = 7

but I dont remember pulling an x out like that. (btw is that algebra I, II, or College Algebra? (my terminology is going to be yuck for awhile lol)

Hey thanks a ton for the prompt answer I am so glad I found this forum!

Sincerely,

Brian

This is called the Distributive Law, which, more generally is:

$x(a+b) = (xa+xb)$

Here are a couple of examples:

$2(x - 3) = (2x - 2\cdot 3) = (2x - 6)$

$x(a^2 + x) = (xa^2 + x^2)$

$(y+3a)(2 - 7a) = ((y+3a)2 - (y+3a)7a)$
$= ((2y+6a) - (7ay + 21a^2))$
$= 2y + 6a - 7ay - 21a^2$

Note that in the last example, $(y+3a)2 = 2(y+3a)$ due to the Commutativity of multiplication.

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nice one question

great!

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