anonymous
  • anonymous
8m-16 ÷ 3m-6 -20 40
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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asnaseer
  • asnaseer
I am not sure I understand you question - can you please clarify?
asnaseer
  • asnaseer
is this:\[(8m-16)\div(3m-6)\]with two choices for the answer: -20 or 40?
anonymous
  • anonymous
no the 20 should have a line above it but under the 8m-16 and the same with the 40 and 3m-16

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asnaseer
  • asnaseer
ok, so question is:\[\frac{8m-16}{20}\div\frac{3m-6}{40}\]correct?
anonymous
  • anonymous
yes it wouldnt let me type it like that but thats correct
asnaseer
  • asnaseer
ok, do you the rule for dividing fractions? i.e. this rule:\[\frac{a}{b}\div\frac{p}{q}=\frac{a}{b}\times\frac{q}{p}\]
asnaseer
  • asnaseer
*do you know the rule...
anonymous
  • anonymous
yes
asnaseer
  • asnaseer
so we can make use of this to get:\[\frac{8m-16}{20}\div\frac{3m-6}{40}=\frac{8m-16}{20}\times\frac{40}{3m-6}\]
asnaseer
  • asnaseer
you can also factorise a bit to get:\[\frac{8m-16}{20}\times\frac{40}{3m-6}=\frac{8(m-2)}{20}\times\frac{40}{3(m-2)}\]
asnaseer
  • asnaseer
you now be able to cancel some things out. are you able to work it out yourself from here?
anonymous
  • anonymous
ok once i cancel out the m-2 i get lost
anonymous
  • anonymous
bec ause the 20 is negative
asnaseer
  • asnaseer
ok, let me show you the steps...
asnaseer
  • asnaseer
oh - so the equation I wrote was not correct?
anonymous
  • anonymous
im sorry i thought u put a -20
asnaseer
  • asnaseer
are you saying it should be:\[\frac{8(m-2)}{-20}\times\frac{40}{3(m-2)}\]
anonymous
  • anonymous
yes
asnaseer
  • asnaseer
ok , I'll try and illustrate the steps to you...
asnaseer
  • asnaseer
|dw:1327017741147:dw|
asnaseer
  • asnaseer
|dw:1327017793288:dw|
asnaseer
  • asnaseer
|dw:1327017851696:dw|
anonymous
  • anonymous
omg thanks so much..... i just have one more problem that im stuck on
asnaseer
  • asnaseer
so you end up with:\[\frac{8}{-1}\times\frac{2}{3}=\frac{16}{-3}=-\frac{16}{3}=-5\frac{1}{3}\]
asnaseer
  • asnaseer
just post your new question to the left and I'm sure there will be lots of people willing to help. unfortunately I need to leave now otherwise I would have assisted you.
anonymous
  • anonymous
thanks though
asnaseer
  • asnaseer
yw

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