anonymous
  • anonymous
Hi All, is there anyone who might be able to show me how to check for accuracy on this problem I worked? 2y/3-3/4=5/12. The answer is y=7/4 so I do know I need to replace y with 7/4 and work it but I can't seem to figure out how to lay that out? Anyone? Really appreciate any help:) Denise
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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UnkleRhaukus
  • UnkleRhaukus
\[\frac{2y}{3}-\frac{3}{4}=\frac{5}{12}\]\[ \qquad\downarrow \]\[\frac{2\times(7/4)}{3}-\frac{3}{4}=\frac{5}{12}\]\[\frac{14}{12}-\frac{3}{4}=\frac{5}{12}\]\[\frac{14}{12}-\frac{9}{12}=\frac{5}{12}\]\[14-9=5\] which is true. therefore y = 7/4 is a solution
anonymous
  • anonymous
oh thanks, sorry, got stuck on the phone, thank you soooooooooooo much, Denise:)
UnkleRhaukus
  • UnkleRhaukus
do you follow the method?

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anonymous
  • anonymous
no, how did you get 14/12 UR:)?
anonymous
  • anonymous
I understand the LCD;) but don't see how you got the 14/12:)
anonymous
  • anonymous
and we just cancel out the 12's as denominators, I get that part too:)
UnkleRhaukus
  • UnkleRhaukus
\[\frac{2\times 7/4}{3}\]\[=\frac{14/3}{4}\]\[=\frac{14}{3}\frac{1}{4}\]\[=\frac{14}{12}\]
UnkleRhaukus
  • UnkleRhaukus
generalizing \[\frac{a/b}{c/d}=\frac{ad}{bc}\]
anonymous
  • anonymous
Oh, that is something I totally forgot how to do, I see what you did though. Can't say I could do another one but I'll try another and see;) practice practice right;)
anonymous
  • anonymous
Thank you again, I think that was the last one I had to check and my homework is done:) Denise:)
UnkleRhaukus
  • UnkleRhaukus
simplify \[\frac{1}{3/b}\]
anonymous
  • anonymous
sec, let me think a minute:)
anonymous
  • anonymous
1/3 first right?
UnkleRhaukus
  • UnkleRhaukus
the simplified form is a single fraction
UnkleRhaukus
  • UnkleRhaukus
where has the b gone?
UnkleRhaukus
  • UnkleRhaukus
you can use the generalized formula above
anonymous
  • anonymous
1/3 over b, nope, Im not getting it.
anonymous
  • anonymous
right, I have it written here on my paper, sec
anonymous
  • anonymous
arrrrrrrrrrrrrg:(
UnkleRhaukus
  • UnkleRhaukus
ok so the step is to multiply both the numerator and denominator by the same thing,
anonymous
  • anonymous
you have to eliminate the b right?
UnkleRhaukus
  • UnkleRhaukus
\[\frac{1}{3/b}\] times both top and bottom by b , what do you get?
anonymous
  • anonymous
ohhhhhhhhhhhhhhhhhhhhhhh
anonymous
  • anonymous
1b/3b and cancel out b's
UnkleRhaukus
  • UnkleRhaukus
\[ \frac{1}{3/b}\]\[ =1\times\frac{1}{3/b}\]\[ =\frac{b}{b}\times\frac{1}{3/b}\]\[=...\]
UnkleRhaukus
  • UnkleRhaukus
the simplified form will have a b in it
anonymous
  • anonymous
oh:( Thought I had it, sec
anonymous
  • anonymous
B1/3?
UnkleRhaukus
  • UnkleRhaukus
oh that is looking much better
UnkleRhaukus
  • UnkleRhaukus
remember\[1\times b=b\],so you can simplify one bit more
UnkleRhaukus
  • UnkleRhaukus
\[\frac{1}{3/b}=\frac{b\times1}{3}=...\]
anonymous
  • anonymous
so it is just B/3!!
UnkleRhaukus
  • UnkleRhaukus
!!!!yes
anonymous
  • anonymous
LOL:) Thank you, you are so kind to teach me that:) That will come in handy I can tell you, thank you so much:)
UnkleRhaukus
  • UnkleRhaukus
we had a fraction in a fraction, and we pulled the denominator in the denominator, into the numerator
anonymous
  • anonymous
Ok, well, I better get finished up here. I will be back tomorrow night I suspect. You are not in Texas I hope?? Storms/Tornados?
UnkleRhaukus
  • UnkleRhaukus
hmm, no. i live in Australia
anonymous
  • anonymous
Oh that is something, I can tell folks I got help all the way from Australia on my Algebra!! Nite UR, hope to see you again;) Denise
UnkleRhaukus
  • UnkleRhaukus
Until next time, all the best
anonymous
  • anonymous
same to you:)

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