Make x the subject of...


by White Rabbit
Tags: subject
White Rabbit
White Rabbit is offline
#1
Apr12-07, 09:17 PM
P: 11
Hi guys.

I have just recently gone back to school and I'm having some problems with "assumed knowledge" . They assume that I actually have some maths knowledge but it has been nearly 20 years since I have done any of this so the knowledge that I had has long since evaporated.

I'm having a problem with this equation.

I have to make x the subject. y=~5/1-x

~ means square root
There is an image of the equation here that I put up in case you can't decipher my translation up top.

http://i174.photobucket.com/albums/w...f?t=1176429869

I'd like an explanation on how to solve this myself rather than giving me the answer as I need to know this for myself.

Any help would be much appreciated.

Thanks
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danago
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#2
Apr12-07, 09:19 PM
PF Gold
P: 1,132
A good start would be to square both sides. See if you can go from there.
White Rabbit
White Rabbit is offline
#3
Apr12-07, 09:19 PM
P: 11
I'm not sure if this will work but....
Attached Thumbnails
y 5 1x.gif  

White Rabbit
White Rabbit is offline
#4
Apr12-07, 11:59 PM
P: 11

Make x the subject of...


How does this look?

y = ~5/(1-x)
=
y= 5 squared / (1-x) squared

y = 25 / (1-x) squared

y/(1-x) squared = 25

(1-x) squared = 25y

(1-x) = ~25y

x = ~25y+1

Thanks
danago
danago is offline
#5
Apr13-07, 12:13 AM
PF Gold
P: 1,132
Whatever you do to one side of the equal sign, you MUST do to the other, if you want to maintain that equality.
Data
Data is offline
#6
Apr13-07, 01:10 AM
P: 998
Furthermore, if you square

[tex]\sqrt{\frac{5}{1-x}},[/tex]

do you really get

[tex]\frac{5^2}{(1-x)^2}[/tex]

?


Here's something to think about:

You have

[tex]y = \sqrt{\frac{5}{1-x}}.[/tex]

So, [itex]y[/itex] is something remotely like [itex]\sqrt{1/x}[/itex]. If [itex]y[/itex] is remotely like the square root of something, then that other thing must be remotely like the square of [itex]y[/itex]. In other words, you should expect [itex]1/x[/itex] to be something remotely like [itex]y^2[/itex], and so [itex]x[/itex] should be remotely like [itex]1/y^2[/itex].

Once you learn how to manipulate these things, arguments like the one I just made won't be needed for simple problems. However, they are sometimes still useful to get a rough idea of how an answer should look in much more complex problems.
Data
Data is offline
#7
Apr13-07, 01:46 AM
P: 998
Here's a little bit of a tutorial on manipulating equations, since you seem to be a little bit rusty.

Alright, let's say we have an equation

[tex]a=b.[/tex]

That means that a is the same as b.

It's worth noting a special property of equalities right away. Suppose [itex]a=b[/itex] and [itex]b=c[/itex] are equalities. Then [itex]a=c[/itex] is also a valid equality. This is called the "transitive property," and can be very important.

Now, let's suppose [itex]a=b[/itex] is an equality. I can add something else, call it c, to both sides, to get

[tex]a+c = b+c.[/tex]

Since a is the same as b, it is easy to see that a+c should be the same as b+c. Thus, you can manipulate an equality by adding something to both sides, and the result will be another equality.

Similarly, you could multiply both sides by c to get

[tex]ac = bc.[/tex]

Again, this is a valid equality. So far, then, given an equation, you can: 1) Add the same thing to both sides ; or, 2) Multiply both sides by the same thing; and you will keep equality.

There are other things that you can do, though. Suppose you have two equations,

[tex]a=b, \ \mbox{and } \ x = y.[/tex]

Here a is the same as b, and x is the same as y. So, it certainly makes sense that ax is the same as by!

Thus we have another rule: 3) Given equalities a=b and x=y, ax=by is also a valid equality. While we're at it, we had might as well note that a+x = b+y is also clearly a valid equality!

There are some general patterns you can notice here: Suppose you have an equation [itex]a=b[/itex]. As long as you do exactly the same thing to both sides of an equation, the equality remains valid. For an example: if [itex]a=b[/itex] is an equality, then [itex]a^2 = b^2[/itex] is also an equality because I've done exactly the same thing to both sides.

This is subject to a many conditions, though, depending on the context. For example, suppose [itex]a=b[/itex] is an equality. Is [itex]1/a = 1/b[/itex] a valid equality also? The answer is almost always yes. When is the answer no? Well, what if [itex]a=0[/itex]? Then [itex]b=0[/itex] also, since [itex]a=b[/itex]. But 1/0 is undefined. So, if a=0, then [itex]1/a = 1/b[/itex] doesn't mean anything at all (it says that two undefined quantities are equal, which is fairly meaningless).

There are other such conditions depending on what you want to do to both sides. Explaining why, and how to deal with them, would take too much time for me to do here; if you have questions, I suggest asking the instructor of your course for some help.

There's also the second pattern. If a=b and x=y are equations, then you can use [itex]x[/itex] the same way as [itex]y[/itex] when you're manipulating [itex]a=b[/itex]. Thus, for example, x/a = y/b is also an equality, as long as [itex]a \neq 0[/itex] (read: a is not equal to 0). Similarly [itex]a^x = b^y[/itex] is also a valid equality, etc.


For now though, you should remember the fundamental rule: In general, when manipulating an equation, you must do the same thing to both sides at each step.

You have great freedom in what you can do to the equation. But you have to do the same thing to each side, or else the result just isn't an equality anymore!
Feldoh
Feldoh is offline
#8
Apr13-07, 02:44 PM
P: 1,345
Quote Quote by Data View Post
Furthermore, if you square

[tex]\sqrt{\frac{5}{1-x}},[/tex]

do you really get

[tex]\frac{5^2}{(1-x)^2}[/tex]
To build off what he said or if you still don't get it, consider this:

[tex]\sqrt{\frac{5}{1-x}},[/tex] is the same as: [tex]({\frac{5}{1-x}})^{1/2}[/tex]

So if you were to square that what would you get?
OrbitalPower
OrbitalPower is offline
#9
Apr13-07, 11:55 PM
P: 88
[tex]y = \sqrt{\frac{5}{1-x}}[/tex]

[tex] y^2 = \frac{5}{1-x}[/tex] (square both sides)

[tex](1-x)(y^2) = 5[/tex] (times by 1-x)

[tex]-x = \[ (\frac{5}{y^2}) - 1\][/tex] (divide by y^2 and subtract 1 to get x by itself)

[tex]x = 1 - \frac{5}{y^2}[/tex] where y [tex]\ge[/tex] 0
White Rabbit
White Rabbit is offline
#10
Apr14-07, 01:23 AM
P: 11
Wow, thanks guys I think I actually get it, how about that. I see where I was going wrong with the squaring. Like I said it's been a loooong time since I was in school, infact I dont even remeber any of this from school.

Thanks
Again.
White Rabbit
White Rabbit is offline
#11
Apr14-07, 02:28 AM
P: 11
Would there be any chance of putting up some more of these types of equations for me so that I can see if I do get it.
Or can you point me in the directions of a site that would have some.

Thanks.
OrbitalPower
OrbitalPower is offline
#12
Apr14-07, 02:43 AM
P: 88
When you square both sides it undoes the RHS in the original equation:

[tex](\sqrt{x})^2[/tex] = x

Here's what Feldoh was talking about.

[tex]x^\frac{m}{n} = \[ (x^\frac{1}{n})^m\] =[/tex]

[tex](x^m)^\frac{1}{n} =[/tex]

[tex] = \[(\sqrt[n]{x})^m = \sqrt[n]{x^m}[/tex]

What you need is an algebra cheat sheet.
Data
Data is offline
#13
Apr14-07, 02:44 AM
P: 998
White Rabbit: Consider [itex]\sqrt{2}[/itex]. What's [itex](\sqrt{2})^2[/itex]?
White Rabbit
White Rabbit is offline
#14
Apr14-07, 03:21 AM
P: 11
=2.

Thanks, now I understand, when you multiply a square root by itself you eliminate the square root.
White Rabbit
White Rabbit is offline
#15
Apr14-07, 03:23 AM
P: 11
so [tex](\sqrt{\frac{5}{1-x}})^2,[/tex] is 5/1-x and not 5/1-x times 5/1-x because that would give you 25/1-x. Or 25/1-x^2?

Thanks
OrbitalPower
OrbitalPower is offline
#16
Apr14-07, 03:47 AM
P: 88
You're not squaring the 5/(1-x), you're squaring the sqrt(5/1-x). In the process of involution (repeated multiplication), you're multiplying a quantity by itself the number of times indicated by the exponent.

x^3
3 is the exponent. so it means x * x * x
A law of exponents state that you add the exponents when you multiply so x ^1 * x^ 1 * x^1 = x ^3

Perhaps you might think of it as you square the 5/(1-x), and you get your second answer -- assuming you mean 5^2/(1-x) -- and then if you take the square root of that, you get the 5/(1-x).

what math are you in if you don't mind me asking?

edit -- "assuming you mean 5^2/(1-x)^2"
OrbitalPower
OrbitalPower is offline
#17
Apr14-07, 04:07 AM
P: 88
Quote Quote by White Rabbit View Post
Would there be any chance of putting up some more of these types of equations for me so that I can see if I do get it.
Or can you point me in the directions of a site that would have some.

Thanks.
All elementary mathematics (Relevant chapter found here)

Introduction to Mathematics and Algebra

But really, nothing beats a good book (with answers to some of the problems to check yourself on).
White Rabbit
White Rabbit is offline
#18
Apr14-07, 04:15 AM
P: 11
The maths course I'm doing is a uni prep class one night a week. I'm not the only one that is stuggling with the course though. The problem I have is that there is some assumed knowledge, as you can see I have no maths background. Having said that, I'm not struggling with all the course but just square roots as they are assumed knowlege. It's something i need to get my head around because the keep cropping up.
The subjects that I have covered so far are expanding, percentages, factoring, geometry, equations (last week) and a few others. We basically cover one or two of these subjects a week so it's pretty fast, and not well taught in my opinion and the opinion of just about everyone I've spoken to. I dont actully get much out of the class because they fly through it. There is only time to write it down but not to understand, but I guess thats UNI so I should get use to it.
This is revision of the stuff we will need before starting calculus. Like I said most of it I'm ok with but square roots have not be taught at all.

Thanks


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