The Student Room Group
Reply 1
Thing to do is get all the terms you want the subject to be on one side, so in this case B.

Multiply both sides by the denominator b-5 and then move all B's to one side and factorise
Reply 2
jumblebumble
I'm normally good at these, but I'm tired and my brain is just not working:

Make b the subject of the formula.
a = 2-7b over b-5

No brackets.

Thank you. :smile:


a=27bb5\displaystyle a = \frac{2-7b}{b-5}

First multiply by the underside of the fraction

a(b5)=27b\displaystyle a (b-5 ) = 2 - 7b

Then expand the brackets, and collect all b terms on one side of the equation

ab5a+7b=2\displaystyle ab - 5a + 7b = 2

Collect all non-b terms on the right hand side, and factorise the left hand side

b(a+7)=2+5a\displaystyle b ( a + 7) = 2 + 5a

And divide through by the factor of b

b=5a+2a+7\displaystyle b = \frac{5a + 2}{a + 7}

Done.
Reply 3
Thank you very much - it was the last question of my paper, which is now done. :wink:
Reply 4
Get all the b's to one side to begin with so....a(b-5)=2-7b
>> ab+7b=2+5a.....try from there :smile:.....(putting in brackets might help :smile:
Reply 5
timotiis
a=27bb5\displaystyle a = \frac{2-7b}{b-5}


Then expand the brackets, and collect all b terms on one side of the equation

ab5b+7b=2\displaystyle ab - 5b + 7b = 2

Done.



Sorry to be a pain, but where has the 5a gone?

x
Reply 6
jumblebumble
Sorry to be a pain, but where has the 5a gone?

x


Thank you, my mistake. I'll just go edit that...
Reply 7
Thank you. :smile:
Reply 8
jumblebumble
Thank you. :smile:


No problem. Slightly shameful really, I should know better than to make that sort of error.
Reply 9
Haha. Well you did better than me - i managed all the histograms, circle theorems, sine graph translations...just not this! :biggrin:
Where did you get the add 7b from?