[MATH]\frac{1}{m-1}+\frac{9}{2m+3}-\frac{8}{m+4}[/MATH]

=

[MATH]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/MATH]=

[MATH]\dfrac{2m^2+8m+3m+12+9m^2-9m+36m-36-16m^2+m-24m+24}{(m-1)(2m+3)(m+4)}[/MATH]=

[MATH]\frac{5m(3-m)}{(m-1)(2m+3)(m+4)}[/MATH]

What next should be done?

Edited July 13, 201411 yr by Chikis

Check your numerator again to be sure, I think it could be wrong. I also don't think (mod the possible mistake) that you can simplify this any further.

Edited July 13, 201411 yr by ajb

Will echo AJB's comments. Both of them.

 

My maths is old-fashioned enough to be very longhand and methodical. I still think it is the best way -- my version of your simplification is six lines long; perhaps you have put too many steps together and missed your arithmetic.

 

You have FOIL multiplied out the brackets and multiplied by the outside constant in one go; the third expansion seems wrong.

!

Moderator Note

 

Acme - we don't give direct answers even if they are readily available on the net. Per the rules at the top of this forum I have hidden your post. Thanks

 

!

Moderator Note

 

Acme - we don't give direct answers even if they are readily available on the net. Per the rules at the top of this forum I have hidden your post. Thanks

 

Well, you guys gave direct answers so I don't really see the problem. But whatever. You could at least leave my reference to Wolfram Alpha as it can be a valuable resource to student.

Check your numerator again to be sure, I think it could be wrong. I also don't think (mod the possible mistake) that you can simplify this any further.

I go again:

[MATH]\dfrac{(2m+3)(m+4)+(9m-9)(m+4)-(16m+24)(m-1)}{(m-1)(2m+3)(m+4)}[/MATH]

=

[MATH]\dfrac{46m+11m^2-32}{(m-1)(2m+3)(m+4)}[/MATH]

Is the numerator okay now?

Edited July 14, 201411 yr by Chikis

I go again:

[math]\dfrac{(2m+3)(m+4)+(9m-9)(m+4)-(16m+24)(m-1)}{(m-1)(2m+3)(m+4)}

=

[math]\dfrac{46m+11m^2-32}{(m-1)(2m+3)(m+4)}

Is the numerator okay now?

No; the numerator is still incorrect. It should have two terms in the variable m and no constant. The denominator is correct.

Edited July 14, 201411 yr by Acme

Please help me. What mistake am making that am not getting the right thing?

Just go through it slowly. You will have to show your working out here for us to spot a mistake. Typically mistakes are due to minus signs and silly mistakes with multiplication. You are on the right lines with this, so don't give up.

Please help me. What mistake am making that am not getting the right thing?

[math]\dfrac{(2m+3)(m+4)+(9m-9)(m+4)-(16m+24)(m-1)}{(m-1)(2m+3)(m+4)}[/math]

 

Be sure to watch the signs as ajb pointed out.

In the above, (2m+3)(m+4) is correct.

For (9)(m-1)(m+4) I suggest doing the (m-1)(m+4) first and then multiply through by 9.

Same for the (-8)(2m+3)(m-1). Do the (2m+3)(m-1) first and then multiply through by the -8.

 

You laid it out right in the first post with this:

[math]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/math]

Edited July 15, 201411 yr by Acme

[MATH]\frac{1}{m-1}+\frac{9}{2m+3}-\frac{8}{m+4}[/MATH]

=

[MATH]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/MATH]

=

[MATH]\dfrac{2m^2+8m+3m+12+9m^2+36m-9m-36-16m^2+16m-24m+24}{(m-1)(2m+3)(m+4)}[/MATH]

 

=

[MATH]\frac{30m-5m^2}{(m-1)(2m+3)(m+4)}[/MATH]

=

[MATH]\frac{5m(6-m)}{(m-1)(2m+3)(m+4)}[/MATH]

Am confident that this one is very correct.

Just go through it slowly. You will have to show your working out here for us to spot a mistake. Typically mistakes are due to minus signs and silly mistakes with multiplication. You are on the right lines with this, so don't give up.

Thanks for the caoching.

[math]\dfrac{(2m+3)(m+4)+(9m-9)(m+4)-(16m+24)(m-1)}{(m-1)(2m+3)(m+4)}[/math]Be sure to watch the signs as ajb pointed out.In the above, (2m+3)(m+4) is correct.For (9)(m-1)(m+4) I suggest doing the (m-1)(m+4) first and then multiply through by 9.Same for the (-8)(2m+3)(m-1). Do the (2m+3)(m-1) first and then multiply through by the -8.You laid it out right in the first post with this:[math]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/math]

Thanks for the tip.

Edited July 15, 201411 yr by Chikis

That looks correct to me. Well done, you managed to do it with a few hints and tips only.

Nice one.