More adding & subtracting fractions

Ok, one more. I can only get so far with this one too.

(x-2)/(x²+2x)

plus

3/(x²+3x)

minus

(x+4)/(x²+5x+6)

Solution = 0.

I have a LCM of x(x+2)(x+3)

The closest I can get is = 4/[(x+2)(x+3)] which obviously is not zero! But I'm obviously going wrong.

If the numerator is zero but the denominator is an expression does that mean the solution is zero. Seems like a crude process for such a subject.
:s-smilie:

Reply 1

Indeterminate

Original post by DeccyJ

\displaystyle \frac{x-2}{x(x+2)} + \frac{3}{x(x+3)} - \frac{x+4}{(x+3)(x+2)}

= \displaystyle \frac{x(x-2)(x+3) + 3x(x+2) - x^2(x+4)}{x^2(x+3)(x+2)}

Reply 2

Aquar

\displaystyle \frac{x-2}{x^2+2x}+\frac{3}{x^2+3x}-\frac{x+4}{x^2+5x+6}

\displaystyle =\frac{x-2}{x^2+2x}+\frac{3}{x^2+3x}-\frac{x+4}{x^2+5x+6}

I'd start with the first two; it's easier to get the common denominator for two than for one.

\displaystyle \frac{(x-2)(x^2+3x)}{(x^2+2x)(x^2+3x)}+ \frac{3(x^2+2x)}{(x^2+3x)(x^2+2x)}-\frac{x+4}{x^2+5x+6}

Simplify:

\displaystyle \frac{(x-2)(x^2+3x)+3(x^2+2x)}{(x^2+2x)(x^2+3x)}-\frac{x+4}{x^2+5x+6}

Expanding brackets, simplifying and getting a common denominator between the remaining two fractions:

\displaystyle \frac{x^3+4x^2}{x^4+5x^3+6x^2}-\frac{x^2(x+4)}{x^2(x^2+5x+6)}

You should be able to see where this is going. :smile:

Reply 3

DeccyJ

Sorry, I was just having some lunch. I'll take a look at these now.

Reply 4

DeccyJ

Not sure I see where it's going at all, I'm afraid.

I thought the method dictated you reduce all terms to a single term and simplify?

Am I missing something? I can follow both posts but (and I'm going out on a limb here) I think they're wrong!

Reply 5

davros

Original post by Indeterminate

\displaystyle \frac{x-2}{x(x+2)} + \frac{3}{x(x+3)} - \frac{x+4}{(x+3)(x+2)}

= \displaystyle \frac{x(x-2)(x+3) + 3x(x+2) - x^2(x+4)}{x^2(x+3)(x+2)}

That looks good although you've unnecessarily introduced a factor x^2 into the denominator of the first 2 terms - a simple 'x' factor is enough!

Original post by DeccyJ

See above update - both answers look OK to me and reduce to 0 as required.

Reply 6

Indeterminate

Original post by davros

That looks good although you've unnecessarily introduced a factor x^2 into the denominator of the first 2 terms - a simple 'x' factor is enough!

See above update - both answers look OK to me and reduce to 0 as required.

That's easily dealt with! :smile:

\displaystyle \frac{x(x-2)(x+3) + 3x(x+2) - x^2(x+4)}{x^2(x+3)(x+2)}

\displaystyle =\frac{x}{x} \times \frac{(x-2)(x+3) + 3(x+2) - x(x+4)}{x(x+3)(x+2)}

\displaystyle =\frac{(x-2)(x+3) + 3(x+2) - x(x+4)}{x(x+3)(x+2)}

(edited 11 years ago)

Reply 7

uncapitalised

No need to do all that just add the first two.

Reply 8

ghostwalker

Original post by Indeterminate

EDIT:

Spoiler

Oh, dear!

Reply 9

Indeterminate

Original post by ghostwalker

Oh, dear!

Yes, I realised that it's false but then got busy and forgot to remove it from the post :smile:

Reply 10

DeccyJ

Ok, I have it and it's working out. I was going wrong by failing to multiply out the brackets of the first (x-2)(x+3) in the factorised numerator. Long story.

I'll surely be back on here soon. I'm trying to teach myself A-level standard maths and it's tricky sometimes when all you have is a book, some online resources, but no tutor to ask questions of!

Thanks for all your help everyone.

But one final question on this particular equation: is it considered bad practice to find your equation reduced to 0/[x(x+3)(x+2)] for example. Or should you always have terms which cancel each other out. I can't see how that would work, but I've got no one to ask.

Thanks again!