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Rearranging formula question help

I need help solving this equation
P = 5xy + y
I have tried to approach this by dividing each side by y so we end up with
P/y = 5x
Its obvious right now that I have divide by 5
Is the answer x = P/5y? Or is it x= (P/y)/5?

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Reply 1
Original post by socialgain
I need help solving this equation
P = 5xy + y
I have tried to approach this by dividing each side by y so we end up with
P/y = 5x
Its obvious right now that I have divide by 5
Is the answer x = P/5y? Or is it x= (P/y)/5?

The bold line is wrong. What happened to the + y ?

Which variable are you trying to make the subject?
(edited 10 years ago)
Original post by socialgain
I need help solving this equation
P = 5xy + y
I have tried to approach this by dividing each side by y so we end up with
P/y = 5x
Its obvious right now that I have divide by 5
Is the answer x = P/5y? Or is it x= (P/y)/5?


Factorising 5xy+y gives y(5x+1).
Reply 3
Original post by notnek
The bold line is wrong. What happened to the + y ?

Which variable are you trying to make the subject?

Whoops its meant to be
P = 5xy + y
Make x the subject
P - y = 5xy
(P - y)/y = 5x
Where should I go from here?
(edited 10 years ago)
Reply 4
Original post by majmuh24
Factorising 5xy+y gives y(5x+1).

Sorry I should of specified that I am trying to make X the subject
Original post by socialgain
Sorry I should of specified that I am trying to make X the subject


I know, and you can do that by factorising. Instead of 5xy+y, rearrange P=y(5x+1) to get x (it should be a lot easier)
Reply 6
P = 5xy + y
= y(5x + 1)
P/y = 5x + 1
5x = P/y - 1
x = (
P/y - 1)/5
Reply 7
Original post by socialgain
Whoops its meant to be
P = 5xy + y
Make x the subject
P - y = 5xy
(P - y)/y = 5x
Where should I go from here?


When you divided by y you should have divided by 5y
Reply 8
Original post by TenOfThem
When you divided by y you should have divided by 5y


So would the final answer be
P = 5xy + y
P-y = 5xy
(P-y)/5y = X
X = (p-y)/5y
Thanks for the help also could I pm you for help on one more question?
Reply 9
Original post by socialgain
So would the final answer be
P = 5xy + y
P-y = 5xy
(P-y)/5y = X
X = (p-y)/5y



Yes


Thanks for the help also could I pm you for help on one more question?


Can you ask on here
Reply 10
Original post by TenOfThem
Yes



Can you ask on here

Could you check if this is right?
W = 2y/3 (1+P)
3W=2y(1+p)
3w=2y+2yp
3w/2y=2+p
3w/2y - 2 = p
Reply 11
W = 2y/3 (1+P)
3W=2y(1+p)
3W/2y=(1+P)
P = 3W/2y - 1
Original post by socialgain
Could you check if this is right?
W = 2y/3 (1+P)
3W=2y(1+p)
3w=2y+2yp
3w/2y=2+p
3w/2y - 2 = p


Assuming this is 2y3(1+p)\dfrac{2y}{3}(1+p)

Then your error is here

2y/2y is 1 not 2


Though to be honest you should just have divided by 2y at this point
Reply 13
Original post by TenOfThem
Assuming this is 2y3(1+p)\dfrac{2y}{3}(1+p)

Then your error is here

2y/2y is 1 not 2


Though to be honest you should just have divided by 2y at this point

3W = 2y(1+p)
3W/2y = (1+p)
3W/2y = 1+p
3w/2y - 1 = p

Is that right?
Original post by socialgain
3W = 2y(1+p)
3W/2y = (1+p)
3W/2y = 1+p
3w/2y - 1 = p

Is that right?


yes
Reply 15
Original post by TenOfThem
yes

Thanks for all the help I appreciate it!
Reply 16
Original post by mc1996
W = 2y/3 (1+P)
3W=2y(1+p)
3W/2y=(1+P)
P = 3W/2y - 1

Thanks for all the help I appreciate it!
Reply 17
Original post by socialgain
Thanks for all the help I appreciate it!


No Problem
Reply 18
Original post by mc1996
No Problem


Sorry about this but could you help me with one more?
T = Square root 2s/g
Make g the subject
Could you should me the method in solving this? thanks
Reply 19
Original post by socialgain
Sorry about this but could you help me with one more?
T = Square root 2s/g
Make g the subject
Could you should me the method in solving this? thanks


T = Square root 2s/g
T2 = 2s/g
g T2=2s
2s/T2 = g

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