# Thread: Solving equations

1. ## Re: Solving equations

Originally Posted by David Green
Just having another go at latex?

$/frac{2x}{x^22+1}$

$\frac{2x}{x^2}{2+1}$
You can only have two terms in each "frac" (I still use dfrac myself but it appears it's much the same thing now)

$$\frac{2x}{x^2(2x+1)}$$ = $\frac{2x}{x^2(2x+1)}$

$$\lim_{x \to \infty} \left(1+ \dfrac{1}{x}\right)^x = e$$ is $\lim_{x \to \infty} \left(1+ \dfrac{1}{x}\right)^x = e$

If you want to test your latex you can create a thread in http://www.mathhelpforum.com/math-help/f47/ if you title it "test"

2. ## Re: Solving equations

Originally Posted by Quacky
In the future, would you mind posting some working out stages rather than jumping immediately several stages ahead? It would help us to follow your logic and check your solution.

Firstly, you have no reason to do anything with the denominator. Once you've found the common denominator, just leave it as it is. Expanding it is just wasting your energy (and potentially exam time!).

When we combine everything, we get:

$\frac{2x(x+1)+3(x+1)(2x-3)-2x(2x-3)}{(2x-3)(x+1)}$

This is certainly a monster, so don't try to rush it. Take it term by term. I'd suggest starting by rewriting the $3(x+1)$ towards the middle as $(3x+3)$

Then, start chronologically from the beginning and expand, using FOIL if necessary.
At this moment in time it's not working for me as this is a lot to take in. At this moment I seem to be multiplying out and everything is getting bigger rather than smaller?

3. ## Re: Solving equations

It will get bigger at first, and then you can collect like terms.

4. ## Re: Solving equations

Originally Posted by Quacky
It will get bigger at first, and then you can collect like terms.
I'll put down what I have done but can't get it right?

2x(x+1) + 3(x+1)(2x - 3) - 2x(2x - 3) = in stages;

2x(x+1) = 2x^2 + 2x

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

3(2x^2 - 5x - 3) = 6x^2 -15x + 9

- 2x(2x - 3) = - 4x^2 + 6

combine terms

2x^2 + 2x + 6x^2 -15x + 9 + - 4x^2 + 6 = 8x^2 - 13x + 9

then the denominator;

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

The result I get;

8x^2 - 13x + 9 / 2x^2 -x - 3

4x^2 + x + 1

But I am not right?

5. ## Re: Solving equations

Why are you still expanding the denominator? You should leave the denominator factored, as we advised previously.

Originally Posted by David
$3(x+1)(2x-3) = 2x^2 - 3x + 2x - 3$
is where you start to get confused.

I suggested writing this as $(3x+3)(2x-3)$ in a previous reply also. Then, when you foil it, you get $6x^2+6x-9x-9$

This is the only mistake I can see, though - the rest was very good.

6. ## Re: Solving equations

Originally Posted by David Green
I'll put down what I have done but can't get it right?

2x(x+1) + 3(x+1)(2x - 3) - 2x(2x - 3) = in stages;

2x(x+1) = 2x^2 + 2x

3(x+1)(2x-3) = 2x^2 - 3x + 2x - 3 [1]

3(2x^2 - 5x - 3) = 6x^2 -15x + 9

- 2x(2x - 3) = - 4x^2 + 6[2]

combine terms

2x^2 + 2x + 6x^2 -15x + 9 + - 4x^2 + 6 = 8x^2 - 13x + 9 [3]

then the denominator;

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

The result I get;

8x^2 - 13x + 9 / 2x^2 -x - 3

But I am not right?
[1]
Note your signs: $-3x+2x = -x \neq -5x$. Consequently the next line has a carried error and should read $6x^2 - 3x - 9$

[2]
You need to distribute the x on the second term: $-2x \cdot -3 = 6x$ instead of just 6

[3]
Look at your coefficients of x^2. $2 + 6 -4 = 4 \neq 8$

The numerator, once expanded is below. I've put in brackets to show each term for clarity's sake. They are not mandatory.
$(2x^2+2x) + (6x^2-3x-9) + (-4x^2+6x)$

Collecting like terms: $2x^2 + 6x^2 - 4x^2 = 4x^2$
$2x-3x+6x = 5x$
$-9 = -9$

Thus your numerator is $4x^2+5x-9$

7. ## Re: Solving equations

Originally Posted by Quacky
Why are you still expanding the denominator? You should leave the denominator factored, as we advised previously.

is where you start to get confused.

I suggested writing this as $(3x+3)(2x-3)$ in a previous reply also. Then, when you foil it, you get $6x^2+6x-9x-9$

This is the only mistake I can see, though - the rest was very good.
Please advise so I understand why is 3(x+1)(2x-3) worked out as;

2x^2 - 3x + 2x - 3 = 2x^2 -x - 3

3(2x^2 -x -3) = 6x^2 -3x +9 wrong

I don't understand how I would know that I should put the 3(x+1) inside the bracket as (3x +1) because working with it the results are different.

also in your example; (3x + 3)(2x - 3) = 6x^2 + 6x - 9x - 9

then I get (3x + 3)(2x - 3) = 6x^2 - 9 + 6x - 9

I assume then that you tidy the result up as 6x^2 + 6x -9x -9

Thanks

David

8. ## Re: Solving equations

Originally Posted by David Green
Please advise so I understand why is 3(x+1)(2x-3) worked out as;

2x^2 - 3x + 2x - 3 = 2x^2 -x - 3

3(2x^2 -x -3) = 6x^2 -3x +9 wrong
Only the sign on 9: it should be $6x^2-3x-9$

I don't understand how I would know that I should put the 3(x+1) inside the bracket as (3x +1) because working with it the results are different.
You put in (3x+3) in the bracket. Personally I prefer to leave multiplying by 3 until last but both methods are good.

also in your example; (3x + 3)(2x - 3) = 6x^2 + 6x - 9x - 9

then I get (3x + 3)(2x - 3) = 6x^2 - 9 + 6x - 9

I assume then that you tidy the result up as 6x^2 + 6x -9x -9

Thanks

David
You can further simplify 6x-9x to -3x

9. ## Re: Solving equations

Originally Posted by David Green
Please advise so I understand why is 3(x+1)(2x-3) worked out as;

2x^2 - 3x + 2x - 3 = 2x^2 -x - 3

3(2x^2 -x -3) = 6x^2 -3x -9 wrong? Q:I made a correction here to a sign error.

I don't understand how I would know that I should put the 3(x+1) inside the bracket as (3x +3) because working with it the results are different. Again, corrected a small error.

also in your example; (3x + 3)(2x - 3) = 6x^2 + 6x - 9x - 9

then I get (3x + 3)(2x - 3) = 6x^2 - 9x + 6x - 9 The terms in red, when multiplied, give -9x.

I assume then that you tidy the result up as 6x^2 + 6x -9x -9 =6x^2-3x-9

Thanks

David
Your method was fine, I didn't understand it, that's all, and there was a minor sign error. It was my mistake, rather than yours, for the most part. I'd suggest looking over $e^{i\pi}$'s post in detail.

Edit: I was beaten to it.

10. ## Re: Solving equations

Originally Posted by Quacky
Your method was fine, I didn't understand it, that's all, and there was a minor sign error. It was my mistake, rather than yours, for the most part. I'd suggest looking over $e^{i\pi}$'s post in detail.

Edit: I was beaten to it.
So in conclusion;

2 / x + 1 = 2x / 2x - 3 + 3 = 4x^2 + 5x - 9 / (2x - 3)(x+1)

Now I will have a go at practicing latex as a previous poster created me a link to work at it.

Thank you all very much.

If there is anything wrong with my final answer please let me know.

David

11. ## Re: Solving equations

Originally Posted by David Green
So in conclusion;

2 / x + 1 = 2x / 2x - 3 + 3 = 4x^2 + 5x - 9 / (2x - 3)(x+1)

Now I will have a go at practicing latex as a previous poster created me a link to work at it.

Thank you all very much.

If there is anything wrong with my final answer please let me know.

David
Hang on, you've not solved it until you have values for x. You've merely simplified the equation

We have:
$\dfrac{4x^2+5x-9}{(2x-3)(x+1)} = 0$

Multiply through by $(2x-3)(x+1)$. I've also factored the numerator to make solving it easier

$\dfrac{(4x+9)(x-1)}{(2x-3)(x+1)} \times (2x-3)(x+1) = 0 \times (2x-3)(x+1) = 0$

Thus $(4x+9)(x-1) = 0$. Solving for x should be trivial at this point

12. ## Re: Solving equations

Originally Posted by e^(i*pi)
Hang on, you've not solved it until you have values for x. You've merely simplified the equation

We have:
$\dfrac{4x^2+5x-9}{(2x-3)(x+1)} = 0$

Multiply through by $(2x-3)(x+1)$. I've also factored the numerator to make solving it easier

$\dfrac{(4x+9)(x-1)}{(2x-3)(x+1)} \times (2x-3)(x+1) = 0 \times (2x-3)(x+1) = 0$

Thus $(4x+9)(x-1) = 0$. Solving for x should be trivial at this point
Little bit confused at the moment because I thought I had already factored out those brackets as;

4x^2 + 5x - 9

and I thought that previously one of the helpers told me to leave the demoninators alone?

David

13. ## Re: Solving equations

denominators, yes. numerators, no...factoring the numerator is helpful.

14. ## Re: Solving equations

Originally Posted by David Green
Little bit confused at the moment because I thought I had already factored out those brackets as;

4x^2 + 5x - 9

and I thought that previously one of the helpers told me to leave the demoninators alone?

David
You don't have to factor. If you didn't you'd be left with $4x^2+5x-9 = 0$ which is a quadratic you can solve using your favourite method.

15. ## Re: Solving equations

Originally Posted by e^(i*pi)
You don't have to factor. If you didn't you'd be left with $4x^2+5x-9 = 0$ which is a quadratic you can solve using your favourite method.
Now I understand

All I have to do is get my + and - mixed up again

Thanks

David

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