# Algebra (easy for you, difficult for me)

1. Sep 29, 2008

### synthetic.

I havent done Maths for some time, and my Algebra needs some work.

1. The problem statement, all variables and given/known data

Two questions in particular;
1. The problem statement, all variables and given/known data

Two questions in particular;

a) Rearrange for x in terms of y and t, to its simplest form.

t(t-x) = 3y(3y-x) [3y-t =/= 0]

b) Solve for x

(x+4)/x - 14/(2x-1) = 0

2. Relevant equations

3. The attempt at a solution

a) I have taken a couple of different approaches, but here is one

t^2 - tx = 9y^2 - 3yx
3yx = 9y^2 - t^2 + tx
x = (9y^2 - t^2 + tx) / 3y
x = 3y - t^2 + tx
tx = t^2 + x - 3y
x = t + x/t - 3y/t
x - x/t = t - 3y/t

And when i try to take it any farther i end up going round in circles, or finding fanciful methods of ending up where i started.

b) Multiplying both Numerators by (x) and (2x-1), and cancelling, leaves

(x+4)(2x-1)=14x
2x^2 - x +8x - 4 = 14x
2x^2 - 7x = 4
x (2x-7) = 4
x = 4/(2x-7)
4/x - 2x = -7

Similarly, i cant see how to go any farther without going round in circles. Can i go to 2x^-1 - x = 7/2? Even then i wouldnt know where to go.

Thanks to anyone who offers help.

2. Sep 29, 2008

### St. Aegis

t(t-x) = 3y(3y-x) [3y-t =/= 0]
ok so let me try this
t^2-xt=9y^2-3yx
move x to one side
(t^2-9y^2)=xt-3yx)
factor out x
xt-3yx=x(t-3y)
so distribute it and solve for x
(t^2-9y^2)/(t-3y)

3. Sep 29, 2008

### St. Aegis

b) Solve for x

(x+4)/x - 14/(2x-1) = 0
ok.....
move and seperate the two
(x+4)/x=14/(2x-1)
multiply them out
(2x-1)(x+4)=14x
2x^2+8x-x-4=14x
2x^2-7x-4=0
using FOIL
(2x+1)(x-4)=0
and so if you take each seperate factor
x=-1/2 and 4, make sure however that they dont contradict the statment,(ie. in original solution, denominator =0) which it doesnt

4. Sep 29, 2008

### HallsofIvy

Staff Emeritus
You are almost there. Your objective is to get x by itself. Now you have x- xt. Well, that's equal to x(1- 1/t). x(1- 1/t)= t- 3y/t. You can get x by itself by dividing both sides by 1- 1/t. That will, however, give you a messy fraction on the right. Since I don't like fractions, I would suggest multiplying both sides of the equation by t first.

b) Multiplying both Numerators by (x) and (2x-1), and cancelling, leaves

(x+4)(2x-1)=14x
2x^2 - x +8x - 4 = 14x
2x^2 - 7x = 4
x (2x-7) = 4[/quote]
No point in factoring there. The only reason you might want to factor polynomials is because "if ab= 0 the either a= 0 or b= 0". You need to get "= 0" first. 2x^2- 7x- 4= 0. Now either factor if you can do so easily, or use the quadratic formula.

5. Sep 29, 2008

### Mentallic

a)
expanding out - $$t^{2}-tx=9y^{2}-3yx$$

moving all x terms to one side - $$3yx-tx=9y^{2}-t^{2}$$

factoring the x terms out - $$x(3y-t)=9y^{2}-t^{2}$$

dividing both sides to make x the subject - $$x=\frac{9y^{2}-t^{2}}{3y-t}$$
note how the question stated that $$3y-t\neq0$$, since in maths one cannot divide by 0, this is to help you. If the question never stated that, you would have to add it in yourself in your solution.

If you understand how to factorise with "the difference of 2 squares", then it should be noticeable that the answer can be simplified.

b)
Multiply by $$x(2x-1)$$ - $$(x+4)(2x-1)-14x=0$$

expanding out and collecting like terms - $$2x^{2}-7x-4=0$$

The solution can be factorised, but if you are unsure how to do this, you can use the quadratic formula to find the solutions:

For a quadratic polynomial $$ax^{2}+bx+c=0$$

The solutions to x can be found using $$x=\frac{-b\pm\sqrt{b^{2}-4ac}{2a}$$

6. Sep 30, 2008

### synthetic.

Thanks to you all, much.

Where would one be without PF.