Basic Mathematic Questions

[LJP07]

I can't figure out these questions, so I need help as to how to figure them out:
1.

SQUARE ROOT of X + Y = 1/2
X - Y

The square root is over both the x + y and x - y, and the question is to express y as the subject of the formula, by the way the X-Y is a fraction under X+Y, it won't submit that in the thread though , the answer at the back of the book is -3x
5

2.
Another one I'm stuck on is :

T = 2Pi Square Root of L
g
To express g in terms of the other variables, the answer at the back of the book is g = 4 pi squared l
T squared.

3.
The surface area S, of a cylinder is given by S = 2 PI r squared + 2 Pi r h
It's volume V is given by V = Pi R squared h
Express V in terms of S and r only?

[moo]

Just a posting tip: You can use the "code" tags to line things up visually. Sometimes it helps to use an editor with a fixed width font to view it first though, or you can just keep previewing the post until it's correct.

SQUARE ROOT of X + Y  = 1/2
               X - Y

The tags I used are visible by hitting the "quote" button.

moo

[LJP07]

Ok, thanks, pity you couldn't be as helpful for solving the problems lol.

[moo]

Ok, thanks, pity you couldn't be as helpful for solving the problems lol.

You're welcome. Pity you didn't use the info I offered to clarify the problems lol.

moo

[CraigD]

Taking your first question first,

Originally Posted by Prolu2007

1.
SQUARE ROOT of X + Y = 1/2
X - Y

First, a note on writing math:
“Function of …” isn’t a very good way to do it.

You could better write the problem
Sqr((X+Y)/(X-Y)) = 1/2,
or
((X+Y)/(X-Y))^.5 = 1/2,
Or, if you feel like learning a little LaTeX (click the “quote” button to see how this simple one is done, or see the LaTeX thread for an introduction),


Now, a hint on how to solve it
((X+Y)/(X-Y))^.5 = 1/2
is hard to manipulate, because of the ^.5 (square root) exponent, so begin by raising each side to the 2nd power (squaring) it
(((X+Y)/(X-Y))^.5)^2 = (1/2 )^2
(X+Y)/(X-Y) = 1/4

You should be able to take it from there, and solve for X in terms of Y, or Y in terms of X.
A hint: the solution will look like Y = (?/?)X - no exponents, just a simple equation of a line.

Post your progress, or if you get stuck again.

[cwes99_03]

Well done, Craig. Hope we see an answer soon from Prolu

[LJP07]

That's as far as I got CraigD, and got:

x + y = 1/4(x-y)
x + y = 1/4x - 1/4y
y + 1/4y = -3/4x
y( 1 + 1/4 ) = -3/4x
y = -3/4x
1 + 1/4

y = -3/4x
5/4

y = -3/5x wrong as the answer is -3x

[CraigD]

Originally Posted by Prolu2007

y = -3/5x wrong as the answer is -3x

is correct.

Reread your first post. You appear to have missed the underlining and the 5 due to it not aligning properly on the page.

[sebbysteiny]

Hi prolu,

Craig D removed my reply from this post. I think it was quite reasonable to do so because my post went into it at a higher level than what you need to get full marks.

But I would like to say, even if you don't need it for that question, you should always be very careful when you squre or take the square root of both sides of an equation.

This is because sqr (x^2) = + or - x. When you square both sides you should gain a solution. eg, y = x. Square both sides. Y^2 = x^2. But one solution here is y=1, x = -1. This is not a solution of the original equation. It is a solution of the corresponding equation y = -x.

Similarly, solutions are lost when you take the (positive) square root of both sides.


I wouldn't worry about this in too much detail. Just be aware that taking the squares and square roots of both sides is a bit dodgy and await with glea the time when you will be shown how to do such opperations correctly.

If you want to get a taster into how this works, we are discussing exactly this in the physics and Mathematics forum.

[sebbysteiny]

Originally Posted by prolu2007

Glad my question posed some debate, but as regards my other thread, can ye solve problem two?

Originally Posted by prolu2007

2.
Another one I'm stuck on is :

T = 2Pi Square Root of L
g
To express g in terms of the other variables, the answer at the back of the book is g = 4 pi squared l
T squared.

3.
The surface area S, of a cylinder is given by S = 2 PI r squared + 2 Pi r h
It's volume V is given by V = Pi R squared h
Express V in terms of S and r only?

Yes I can solve problem 2. The question is, can you?

It's not difficult. It's simply a matter of rearranging the equation to make g the subject of the equation. Short of actually doing the question for you, there is not much more I can say.

If you need any futher help, ask, but I'm sure you can do it on your own.

There are not even any complexities about -ve roots here.

3. This is simply substitution. Make R the subject of the Surface Area equation; sub R for S into the volume equation.

You would be adviced to know 3 backwards and in your sleep. Solving simultaneous equations by substitution is one of the most useful parts of mathematics you learn.

If you want to get top grades, you could try using elimination here as well. To do very well, you need to be fluent in both methods.