Simplify 1/(m-1)+9/(2m+3) - 8/(m+4)

[Chikis]

[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, 2014 by Chikis

[ajb]

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, 2014 by ajb

[imatfaal]

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.

[imatfaal]

!

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

 

[Acme]

!

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.

[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.

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, 2014 by Chikis

[Acme]

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, 2014 by Acme

[Chikis]

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

[ajb]

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.

[Acme]

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, 2014 by Acme

[Chikis]

[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, 2014 by Chikis

[ajb]

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

[imatfaal]

Nice one.