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How do I rearrange this for T?

Screenshot_9.png
I divide both sides by t first, which gives me A/t = π +5

However, dividing by A then gives me: 1/t = (π + 5)/A

How do I then get the answer (in red)? And why can you do that?
Original post by vector12
Screenshot_9.png
I divide both sides by t first, which gives me A/t = π +5

However, dividing by A then gives me: 1/t = (π + 5)/A

How do I then get the answer (in red)? And why can you do that?


Start by factorising.
Original post by vector12
Screenshot_9.png
I divide both sides by t first, which gives me A/t = π +5

However, dividing by A then gives me: 1/t = (π + 5)/A

How do I then get the answer (in red)? And why can you do that?


Your method is fine (if a bit unusual). Just take reciprocals of each side.
Reply 3
Avatar for meenu89
meenu89
19
Original post by vector12
Screenshot_9.png
I divide both sides by t first, which gives me A/t = π +5

However, dividing by A then gives me: 1/t = (π + 5)/A

How do I then get the answer (in red)? And why can you do that?


What do you notice about the RHS? Can you factorise it?

Edit: As others have answered I will not post again!
(edited 6 years ago)
You need to factorise to get: A = t(π + 5)
Then divide A by (π + 5) to get A/(π + 5) = t
Original post by vector12
Screenshot_9.png
I divide both sides by t first, which gives me A/t = π +5

However, dividing by A then gives me: 1/t = (π + 5)/A

How do I then get the answer (in red)? And why can you do that?


so basically factorise a t out on the right hand and the remainder you have in brackets you can then divide to get the red writing. SEE BELOW .
sketch 4.jpg224.8KB
As the others say - can you see a way to get just one t on the RHS?
Hope that helps mate :smile:
Original post by vector12
Screenshot_9.png
I divide both sides by t first, which gives me A/t = π +5

However, dividing by A then gives me: 1/t = (π + 5)/A

How do I then get the answer (in red)? And why can you do that?


May be wrong but:
1) factorise to A = t(π+5)
2) divide by the whole bracket on both sides so left with A/(π+5) = t
Reply 9
Avatar for vector12
vector12
OP
10
Original post by Mathematical_A
Your method is fine (if a bit unusual). Just take reciprocals of each side.


Okay, so I have 1/t = (π + 5)/A

How do I take reciprocals of both sides? And why are you able to do that?
Original post by vector12
Okay, so I have 1/t = (π + 5)/A

How do I take reciprocals of both sides? And why are you able to do that?


Taking reciprocals means flipping the numerators and denominators of fractions on both sides of an equation.

e.g. 23=4060\displaystyle \frac{2}{3}=\frac{40}{60}

and 32=6040\displaystyle \frac{3}{2}=\frac{60}{40}
Reply 11
Avatar for vector12
vector12
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Original post by Throwaway740
You need to factorise to get: A = t(π + 5)
Then divide A by (π + 5) to get A/(π + 5) = t


Thanks. How did you know to factorise it?
Reply 12
Avatar for vector12
vector12
OP
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Original post by Mathematical_A
Taking reciprocals means flipping the numerators and denominators of fractions on both sides of an equation.

e.g. 23=4060\displaystyle \frac{2}{3}=\frac{40}{60}

and 32=6040\displaystyle \frac{3}{2}=\frac{60}{40}


Oh okay, thank you! I get it now, then you have t/1 on the LHS (which is obviously t).

So when you flip the numerators and deniminators of both sides, does that only work if you have fractions on both sides? And if you had something without a fraction on one side, could you just put that side with 1 as a denominator, and then flip the fractions?

eg. If I made up this situation, is this how it would be (if I worked backwards sort of thing)? Am I able to do this?
IMG_1420.jpg

And if so, could I then say that 1 = hb/(a+c)? (if I wanted it in that form)

NB: To clarify, thats a b and not a 6.
(edited 6 years ago)
Original post by vector12
Oh okay, thank you! I get it now, then you have t/1 on the LHS (which is obviously t).

So when you flip the numerators and deniminators of both sides, does that only work if you have fractions on both sides? And if you had something without a fraction on one side, could you just put that side with 1 as a denominator, and then flip the fractions?

eg. If I made up this situation, is this how it would be (if I worked backwards sort of thing)? Am I able to do this?
IMG_1420.jpg

And if so, could I then say that 1 = hb/(a+c)? (if I wanted it in that form)


Yes to all of those questions.
Reply 14
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vector12
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Original post by Mathematical_A
Yes to all of those questions.


Brilliant, thank you! I suppose taking a reciprocol just means making both sides the numerators and putting 1 as the denominator, right? And you can do that in any situation, right?
Original post by vector12
Brilliant, thank you! I suppose taking a reciprocol just means making both sides the numerators and putting 1 as the denominator, right? And you can do that in any situation, right?


That's not really the definition of a reciprocal.

If x and y are reciprocals, then their product (the number you get when you multiply them) is 1.
Reply 16
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vector12
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Original post by Mathematical_A
That's not really the definition of a reciprocal.

If x and y are reciprocals, then their product (the number you get when you multiply them) is 1.


Ok, but if I were to take a reciprocol, you are essentially putting 1/each side on each side aren't you?
Original post by vector12
Ok, but if I were to take a reciprocol, you are essentially putting 1/each side on each side aren't you?


Yes, you essentially put both sides as denominators and the numerator as 1.

Then obviously you can simplify, but we skip the former step.
Reply 18
Avatar for vector12
vector12
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Original post by RDKGames
Yes, you essentially put both sides as denominators and the numerator as 1.

Then obviously you can simplify, but we skip the former step.


Got it. Thanks a lot! Have a good evening!

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