# Solving equations

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• Nov 17th 2011, 08:05 AM
David Green
Solving equations
...(2) = ...(2x) + 3
(x + 1) (2x - 3)

The dots don't mean anything I am trying to make it look OK?

I am asked to solve the above equation!

I think my first step is to determine the denominators, so I think;

common denominators are; 3(x+1)(2x-3)

when I expand these denominators I get;

3(x+1)(2x-3) = 2x^2 - 3x + 2x = 3(2x^2 - 3x +2x) = 6x^2 - 9x + 6x = 6x^2 - 3x

So my common denominators are 3 and 6

am I right so far?

Thanks

David(Happy)
• Nov 17th 2011, 08:07 AM
Quacky
Re: Solving equations
Is the problem statement:

$\displaystyle \frac{2}{x+1}=\frac{2x+3}{2x-3}$?
• Nov 17th 2011, 08:12 AM
David Green
Re: Solving equations
Quote:

Originally Posted by Quacky
Is the problem statement:

$\displaystyle \frac{2}{x+1}=\frac{2x+3}{2x-3}$?

nearly, I wished I could present the work as you do then I would not make things difficult for people?

........2x..... + 3
......2x - 3
Sorry I hope this is clearer

David
• Nov 17th 2011, 08:32 AM
Deveno
Re: Solving equations
this forum implements a math-formatting system called "latex". to write a fraction in latex, you type:

$$\frac{expression one}{expression two}$$

so $$\frac{2x}{x^2+1}$$

produces $\displaystyle \frac{2x}{x^2 + 1}$.

read this thread, it has more (note: we don't use the "math" tags anymore, we use "tex" tags).
• Nov 17th 2011, 08:43 AM
David Green
Re: Solving equations
Quote:

Originally Posted by Deveno
this forum implements a math-formatting system called "latex". to write a fraction in latex, you type:

$\displaystyle \frac{expression one}{expression two}$

so $\displaystyle \frac{2x}{x^2+1}$

produces $\displaystyle \frac{2x}{x^2 + 1}$.

read this thread, it has more (note: we don't use the "math" tags anymore, we use "tex" tags).

I am just going to try it copying your example.

{tex}\frac{2x}{x^22+1}{tex}

• Nov 17th 2011, 08:51 AM
Deveno
Re: Solving equations
use square brackets [], not curly ones {} around the tex tags. remember you start with [ tex] and end with [/tex]. the curly brackets seperate different things for the \frac function inside the tex tags.
• Nov 17th 2011, 09:02 AM
David Green
Re: Solving equations
Quote:

Originally Posted by Deveno
use square brackets [], not curly ones {} around the tex tags. remember you start with [ tex] and end with [/tex]. the curly brackets seperate different things for the \frac function inside the tex tags.

OK here I go?

$$\frac[2x][x^2][2+1][tex] There must be something else involved? I still can't get it(Thinking) • Nov 17th 2011, 09:26 AM Quacky Re: Solving equations [Tex]\frac{2x}{x^2+1}[\tex] But use / instead of \ Anyway, so we have: \displaystyle \frac{2x}{x+1}=\frac{2x}{2x-3}+3 I think. There are a few ways that you could go about this. I'd start by collecting everything on one side of the equation: \displaystyle 0=\frac{2x}{2x-3}+3-\frac{2x}{x+1} Now, you could, if you wanted to, find a common denominator. The lowest common denominator isn't actually affected by the \displaystyle 3 though because we can manipulate this to match anything we want. Just look at the algebraic fractions.\displaystyle (x+1)(2x-3) is the lowest common denominator for both of them. So I multiply both the numerator and denominator by the missing components, as it were, one at a time. \displaystyle 0=\frac{2x{\color{red}(x+1)}}{(2x-3){\color{red}(x+1)}}+\frac{3{\color{red}(x+1)(2x-3)}}{{\color{red}(x+1)(2x-3)}}-\frac{2x{\color{red}(2x-3)}}{(x+1){\color{red}(2x-3)}} Hopefully you should be able to see that the parts in red cancel to give us exactly what we had in the stage above. I've multiplied through both the numerator and denominator by the same value, so I haven't changed anything - the denominators are now exactly the same. Can you combine and expand them? You need to rewrite everything as one fraction. You don't need to change the denominators at all. • Nov 17th 2011, 09:35 AM David Green Re: Solving equations Quote: Originally Posted by Quacky [Tex]\frac{2x}{x^2+1}[\tex] But use / instead of \ Anyway, so we have: \displaystyle \frac{2x}{x+1}=\frac{2x}{2x-3}+3 I think. There are a few ways that you could go about this. I'd start by collecting everything on one side of the equation: \displaystyle 0=\frac{2x}{2x-3}+3-\frac{2x}{x+1} Now, you could, if you wanted to, find a common denominator. The lowest common denominator isn't actually affected by the \displaystyle 3 though because we can manipulate this to match anything we want. Just look at the algebraic fractions.\displaystyle (x+1)(2x-3) is the lowest common denominator for both of them. So I multiply both the numerator and denominator by the missing components, as it were, one at a time. \displaystyle 0=\frac{2x{\color{red}(x+1)}}{(2x-3){\color{red}(x+1)}}+\frac{3{\color{red}(x+1)(2x-3)}}{{\color{red}(x+1)(2x-3)}}-\frac{2x{\color{red}(2x-3)}}{(x+1){\color{red}(2x-3)}} Hopefully you should be able to see that the parts in red cancel to give us exactly what we had in the stage above. I've multiplied through both the numerator and denominator by the same value, so I haven't changed anything - the denominators are now exactly the same. Can you combine and expand them? You need to rewrite everything as one fraction. You don't need to change the denominators at all. I have copied the info to do as you ask, but just one question regarding the RED bits which you advise cancel, should these now be cancelled or combined and expanded? Thanks David • Nov 17th 2011, 09:39 AM Deveno Re: Solving equations combined and expanded (and by this i mean: add the numerators). your main concern is finding out what the numerator over a common denominator will be, because THAT numerator has to equal 0. • Nov 17th 2011, 09:51 AM David Green Re: Solving equations Quote: Originally Posted by Quacky [Tex]\frac{2x}{x^2+1}[\tex] But use / instead of \ Anyway, so we have: \displaystyle \frac{2x}{x+1}=\frac{2x}{2x-3}+3 I think. There are a few ways that you could go about this. I'd start by collecting everything on one side of the equation: \displaystyle 0=\frac{2x}{2x-3}+3-\frac{2x}{x+1} Now, you could, if you wanted to, find a common denominator. The lowest common denominator isn't actually affected by the \displaystyle 3 though because we can manipulate this to match anything we want. Just look at the algebraic fractions.\displaystyle (x+1)(2x-3) is the lowest common denominator for both of them. So I multiply both the numerator and denominator by the missing components, as it were, one at a time. \displaystyle 0=\frac{2x{\color{red}(x+1)}}{(2x-3){\color{red}(x+1)}}+\frac{3{\color{red}(x+1)(2x-3)}}{{\color{red}(x+1)(2x-3)}}-\frac{2x{\color{red}(2x-3)}}{(x+1){\color{red}(2x-3)}} Hopefully you should be able to see that the parts in red cancel to give us exactly what we had in the stage above. I've multiplied through both the numerator and denominator by the same value, so I haven't changed anything - the denominators are now exactly the same. Can you combine and expand them? You need to rewrite everything as one fraction. You don't need to change the denominators at all. I don't think I will be anywhere near right to be honest, and if I put my full working out in now I am sure it would be very confusing? However I have ended up with; 8^2 - 7 - 2x / 2x^2 - x - 7 ???????? • Nov 17th 2011, 10:03 AM David Green Re: Solving equations Just having another go at latex? \displaystyle /frac{2x}{x^22+1} [tex\frac{2x}{x^2}{2+1}$$
• Nov 17th 2011, 10:03 AM
Quacky
Re: Solving equations
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:

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

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

Then, start chronologically from the beginning and expand, using FOIL if necessary.
• Nov 17th 2011, 10:04 AM
e^(i*pi)
Re: Solving equations
Quote:

Originally Posted by David Green
I don't think I will be anywhere near right to be honest, and if I put my full working out in now I am sure it would be very confusing?

However I have ended up with;

8^2 - 7 - 2x / 2x^2 - x - 7

????????

The numerator is $\displaystyle 2x(x+1) + 3(2x-3)(x+1) - 2x(2x-3)$. Expand and collect like terms (and be aware of the rule regarding subtracting negatives in the last term)

The denominator is best left as $\displaystyle (2x-3)(x+1)$ since we'll multiply both sides by this to clear the fraction and since anything multiplied by 0 is 0 we need only look at the numerator
• Nov 17th 2011, 10:05 AM
Quacky
Re: Solving equations
Quote:

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

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

This is fine except that your slash is the wrong direction in the "/frac" - it should be "\frac" and I think it should be x^2 instead of x^22
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