The Student Room Group

Help Needed with Algebra

Really struggling with these questions would really appreciate some help,
answers need to be given with positive indices.

maths.jpg
Reply 1
Original post by John Robbo
Really struggling with these questions would really appreciate some help,
answers need to be given with positive indices.

maths.jpg


Use similar application from these examples, for
q1, consider this xn×xm=xm+nx^n \times x^m=x^{m+n} note that the only different part about this on the first part is that you need to also multiply the coefficients of each part together 9 and 2, and add a slightly more complicated power together.

q2, a very similar idea to the first here, note that x=x1x=x^{1} use the above rule to solve this problem by multiplying each part of the bracket by the outside.

q3, another rule here, (xm)n=xmn (x^{m})^n=x^{mn}, this is a very fast solve tbh
(edited 11 years ago)
Reply 2
Original post by Robbie242
Use similar application from these examples, for
q1, consider this xn×xm=xm+nx^n \times x^m=x^{m+n} note that the only different part about this on the first part is that you need to also multiply the coefficients of each part together 9 and 2, and add a slightly more complicated power together.

q2, a very similar idea to the first here, note that x=x1x=x^{1} use the above rule to solve this problem by multiplying each part of the bracket by the outside.

q3, another rule here, (xm)n=xmn (x^{m})^n=x^{mn}, this is a very fast solve tbh


Thanks for your help is there any chance you could give me answers too each question too help me revise them.
Reply 3
Original post by John Robbo
Thanks for your help is there any chance you could give me answers too each question too help me revise them.
Okay, although it isn't the standard policy I've put the answers in the Spoiler

Spoiler

(edited 11 years ago)
Original post by John Robbo
Thanks for your help is there any chance you could give me answers too each question too help me revise them.


If you tell us what answers you get, we can tell you if they are correct, and if necessary help you get to the correct answer.
Original post by Robbie242
Okay, although it isn't the standard policy I've put the answers in the Spoiler

Spoiler



On the third one, you forgot to apply the power to the 9 as well :tongue:. It should have been:

Spoiler

(edited 11 years ago)
Reply 6
Original post by brittanna
If you tell us what answers you get, we can tell you if they are correct, and if necessary help you get to the correct answer.
Dumb question but look at my spoiler please, is my 3rd answer wrong? I was wondering if I do 9^-1/2 as well? Aha nevermind, that was a silly mistake!
Reply 7
Original post by Robbie242
Okay, although it isn't the standard policy I've put the answers in the Spoiler

Spoiler



so with positive indices would the answers be :

1. 18/a^4/7

2. 3/x^2 - 12/x^1

3. 9/x^4
Reply 8
Original post by John Robbo
so with positive indices would the answers be :

1. 18/a^4/7

2. 3/x^2 - 12/x^1

3. 9/x^4


3rd one I did wrong, it was 13x4\frac{1}{3}x^{-4} since 9^-1/2=1/3 And yup other two are right
(edited 11 years ago)
Reply 9
how do work out the 1/3x bit?
Original post by John Robbo
how do work out the 1/3x bit?


(9x8)12=912(x8)12(9x^8)^{-\frac{1}{2}}=9^{-\frac{1}{2}}*(x^8)^{-\frac{1}{2}}

=191x8=\dfrac{1}{\sqrt{9}} * \dfrac{1}{\sqrt{x^8}}

=131x4=\dfrac{1}{3} * \dfrac{1}{x^4}

=13x4=13x4=\dfrac{1}{3x^4} = \boxed{\dfrac{1}{3}x^{-4}}
(edited 11 years ago)
Reply 11
Original post by John Robbo
how do work out the 1/3x bit?

Given you have (9x8)12(9x^8)^{\frac{-1}{2}} you need to use your algebra rules on the power of 8 as well as 9, as both are intricate parts of the bracket affected by this negative power
notice that 912=199^\frac{-1}{2}=\frac{1}{\sqrt 9} which is 1/3

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