Question 24

[Leodis80]

Hi, Please could somebody walk me through how to solve the following question (Exercise 14.3 Question 24) in the 'Algebra: A complete introduction' book by Hugh Neil.

Given (P - 1/2Q) = b(Q - 1/2P), rearrange the terms so as to express P in terms of the other quantities.

I've had a few goes, however I can't get near the answer given in the back of the book which is P=2b+1 over 2+b then multiplied by Q (apologies, I haven't worked out how to display fractions)

Thanks in advance,

M

 

[lex]

If you post a photo of your work we can have a look at it.
(I would multiply out the bracket on the right for starters, and then multiply the whole equation by 2, so it's easier to work with).

 

[Harry_the_cat]

Is it
\(\displaystyle (P - \frac{1}{2}Q) = b(Q - \frac{1}{2}P)\) ?

 

[Otis]

P=2b+1 over 2+b then multiplied by Q … haven't worked out how to display fractions

P = (Q)(2b + 1)/(2 + b)

[imath]\;[/imath]

 

[Leodis80]

Hi, thanks for your replies. I've tried Lex's suggestion, however I am still stuck. I've attached a pic of what I have done so far and hopefully somebody could point out where I am going wrong please? Many thanks. M

 

[Deleted member 4993]

Hi, thanks for your replies. I've tried Lex's suggestion, however I am still stuck. I've attached a pic of what I have done so far and hopefully somebody could point out where I am going wrong please? Many thanks. M

Please check the third line of your work.

Please tell me - exactly what are you trying to do there.

In other words, in my opinion, the third line does not follow from the 2 nd line.

 

[BigBeachBanana]

Hi, thanks for your replies. I've tried Lex's suggestion, however I am still stuck. I've attached a pic of what I have done so far and hopefully somebody could point out where I am going wrong please? Many thanks. M

[math]\left(P - \frac{1}{2}Q\right) = b\left(Q - \frac{1}{2}P\right)\\ \text{Following Lex's suggestion, multiply the equation by 2:}\\ \blue{2}\left(P - \frac{1}{2}Q\right) = \blue{2}b\left(Q - \frac{1}{2}P\right)\\[/math]Now, continue by distribute the 2 using the distributive property.

 

[Leodis80]

Thanks for your reply. I've had another go and still can't get it. I've attached another note with what I did step by step.

 

[Deleted member 4993]

Thanks for your reply. I've had another go and still can't get it. I've attached another note with what I did step by step.

Check the 3 rd line again of your new response. Remember

2 * (1/2) b * P = b * P

 

[BigBeachBanana]

Thanks for your reply. I've had another go and still can't get it. I've attached another note with what I did step by step.

Edited:
You didn't distribute correctly on the right-hand side.

 

[Steven G]

bQ+b is not b(2Q)

Is 3*4 +3 = 3(2*4)?? Note that I am using 3 for b and 4 for Q.

 

[Leodis80]

Hi, thanks for all your replies. Unfortunately I'm still not getting anywhere with this. I've been doing really well with the book up until now and whenever I've become stuck I have managed to get there in the end. I know this is a big ask but is there any chance somebody could post a pdf or word doc that shows me line by line how to get the solution please? Thanks for all your help. M

 

[lex]

You have made a mistake multiplying out the bracket on the right hand side in the line marked above.

[imath]\hspace20ex 2b(Q-\tfrac{1}{2}P) \text{ means }2b(Q)-2b(\tfrac{1}{2}P)[/imath]
[imath]\hspace40ex = 2bQ-bP[/imath]
So the complete line should read:
[imath]2P - Q = 2bQ - bP[/imath]

Then you don't seem to know the procedure for getting P=...
You must have all terms with P, on the left and all terms without P, on the right.
Therefore there are 2 terms on the wrong side:
[imath]2P \boxed{-Q} = 2bQ \boxed{-bP}[/imath]
You have dealt with -Q, but not with -bP

Try to fix that first.
Do you then know how to get P=...?

 

[Leodis80]

Hi everybody. I think I have solved it. Please see attached. Thank you all so much for you patience as I was getting close to throwing in the towel!!!

 

[Deleted member 4993]

Hi everybody. I think I have solved it. Please see attached. Thank you all so much for you patience as I was getting close to throwing in the towel!!!

You have done it - it took a bit of time but you got it !!!

I have to tell you that your presentation is super-excellent. The letters and groups are nicely spaced. The consecutive lines are parallel and nicely spaced - ad you patience to take advice, implement and finish it.....

Very nice work...

 

[lex]

Hi everybody. I think I have solved it. Please see attached. Thank you all so much for you patience as I was getting close to throwing in the towel!!!

You have done it. Well done!
(It should be noted that for this expression, you need [imath]b\not =-2[/imath]).