Solving equation with fraction & bracket help required please

[tony66]

Hi, I am having awful trouble working this out and have been going round in circles. Can you help me please?
Its:
x/x+1 - 4/x - 2 = 2

 

[Prove It]

Seeing as your question says that there should be brackets in the original equation, please put there where they should be so that the equation can actually be read...

 

[sarannie]

Seeing as your question says that there should be brackets in the original equation, please put there where they should be so that the equation can actually be read...

Whoops, no brackets, sorry.
I have got this far but whether it's right or not I'm not sure?
I have canceled out 'x' from the first fraction so I'm left with 1
1 - 4/x - 2 = 2
Then, add 2 next?
1 - 2 = - 1 and 4/x - 2 + 2 = 2
So I have -1 - 4/x = 2
Am I anywhere near or hopelessly wrong?
Thank you

 

[Jameson]

Hi sarannie, (Wave)

Welcome to MHB!

First question, what is your starting equation? Is it this?

$$\frac{x}{x+1}-\frac{4}{x-2}=2$$

 

[sarannie]

Yes that's the one. Sorry for the delay in replying, I had to go out.
Thank you

 

[Jameson]

Yes that's the one. Sorry for the delay in replying, I had to go out.
Thank you

No worries. I post and come back later very often as well.

Ok, what you need to do here is find a common denominator to combine terms on the left-hand side. If you've never done this before then it will look weird but here's what to do.

LHS: $$\frac{x}{x+1}-\frac{4}{x-2}=\frac{x}{x+1} \left( \frac{x-2}{x-2} \right)-\frac{4}{x-2} \left( \frac{x+1}{x+1} \right)=\frac{x(x-2)-4(x+1)}{(x+1)(x-2)}$$

Then you set that equal to the right-hand side and go from there. If that seems too complicated then you could also try just multiplying every term (there are three) by $(x+1)(x-2)$ and go from there.

$$\frac{x}{x+1}-\frac{4}{x-2}=2$$

$$\frac{x}{x+1} \big[ (x+1)(x-2) \big]-\frac{4}{x-2} \big[ (x+1)(x-2) \big]=2 \big[ (x+1)(x-2) \big]$$

 

[sarannie]

Thank very much, I'll give all that a go and show you where I get to. Maths is a mystery to me but I have to do it to get to where I want to be so I just keep trying. Thank you again

 

[sarannie]

HI, can I cancel out (x - 2) here?

x(x - 2) - 4(x + 1)
(x + 1) (x - 2)

Leaving -4x + 1
x^2 + 1Thank you

 

[SuperSonic4]

HI, can I cancel out (x - 2) here?

x(x - 2) - 4(x + 1)
(x + 1) (x - 2)

Leaving -4x + 1
x^2 + 1Thank you

No you wouldn't because you'd need (x-2) in both numerator terms to do that. Your first expression is in it's simplest form and cannot be canceled further

 

[sarannie]

What should I do next?
Thank you

 

[HallsofIvy]

I "thanked" Jameson without reading his post completely and just duplicated his last paragraph!

 

[SuperSonic4]

What should I do next?
Thank you

Use your first expression and set it equal to 2

$$\dfrac{x(x-2) - 4(x+1)}{(x+1)(x-2)} = 2$$

I would suggest multiplying both sides by $$(x+1)(x-2)$$ to get

$$x(x-2) - 4(x+1) = 2(x+1)(x-2)$$

Next expand the brackets and collect like terms (you may find it best to put it in the standard quadratic form [math]ax^2+bx+c=0[/math])

 

[Jameson]

Hi sarannie, (Wave)

The above post has very good advice. What do you get after expanding the following and trying to solve for $x$?

$\displaystyle x(x-2) - 4(x+1) = 2(x+1)(x-2)$

 

[sarannie]

Not sure about this:
x^2 - 2x - 4x + -3 = 2x + 3 Then I seem to have 2x left on its own.
I was wondering whether I can now collect like terms:
x^2 - 6x + -3 = 4x + 3 ?

Thanks again guys for your help, I know it's like drawing teeth with me :(

 

[Jameson]

Part of that is right. :)

$$x(x-2) - 4(x+1) = 2(x+1)(x-2)$$

$$x^2-2x-4x-4=2(x^2-2x+x-2)$$

Can you make some progress from here?

 

[sarannie]

I deal with the brackets first?
x^2 - 2x - 4x - 4 = 2(x^- 2x + x -2 )
x^2 - 2x - 4x - 4 = x^2 + x -2

Then
x^2 -6x - 4 = x^2 + x -2

(hopeful face)

 

[Jameson]

You have a "2" on the right-hand side that disappeared. Let's focus just on the right hand side.

$$2(x^2-2x+x-2)=2x^2-4x+2x-4$$

You can combine two terms here, which both contain an "$x$". What does it look like after doing so?

 

[sarannie]

x^2 - 2x - 4x - 4 = 2(x^- 2x + x -2 )I multiply x^ by 2 and then 2 x -2x + 2x - 4
So I have 2x^ + 2x -4 ?
Thank you

 

[Jameson]

I was helping you simplify just one side of the equation. Let's go back to what we need to actually solve.

$$x^2-2x-4x-4=2(x^2-2x+x-2)$$

$$x^2-6x-4=2x^2-2x-4$$

$$-x^2-4x=0$$

$$x^2+4x=0$$

What should we do now?

(Is this for a class you're taking? These equations can be tricky if no one has gone through examples with your before. I'm more than happy to help but hope you also have a good teacher than you can talk to. It's great to have a handful of resources. :) )

 

[sarannie]

Thank you for your patience. I am a mature student and I am working my way through a distance learning course. There is little help and I can't get to tutorials. At present my tutor is ill so I am struggling.

I am thinking that the common factor is x so do I divide by x?
x^2 divided by x gives x and 4x divided by x is 4.

 

[MarkFL]

I see Jameson is offline so I'll jump in...

Rather than dividing through by $x$, I would suggest factoring and then applying the zero-factor property. Basically this means that if you have one or more factors equal to zero, then the equation is true when any of the factors is equal to zero.

So factor the left side, and then equate each factor in turn to zero to find all of the solutions.

 

[sarannie]

So that x = 0 or x = 4?

Thank you

 

[MarkFL]

So that x = 0 or x = 4?

Thank you

You have:

$$x^2+4x=0$$

So, on factoring, we obtain:

$$x(x+4)=0$$

Now, equating the first factor to zero, we immediately get:

$$x=0$$

That's one solution. Equating the second factor to zero, we have:

$$x+4=0$$

Solve that for $x$, and you will see that one of the solutions you gave was incorrect. :D

 

[sarannie]

Thank you so much, you have taught me something useful today :)