Rearrange the formula?

Question

rearrange the formula to make g the subject
B= t^2 - g
divide line
t+g

Answer 1

B = (t^2-g)/(t+g)
B(t+g)=t^2-g
Bt+Bg=t^2-g
Bg+g=t^2-Bt
g(B+1)=t^2-Bt

g= t^2-Bt
divide line
b+1

Answer 2

So you have, apparently. B = [t^2 - g]/[t + g]
Note: whenever's written in the following some operation eq.(=equation) it means the operation must be done in both sides of the equation):

== Multiply eq. by denominator: B(t + g) = t^2 - g

==Add +g to eq. and open parentheses: Bg + g + Bt = t^2

== Factor out g in left side and add -Bt to eq: (B + 1)g = t^2 - Bt

== Divide eq. by B + 1, assuming B is NOT -1:
g = [t^2 - Bt]/(B + 1)

Regards
Tonio

Answer 3

So this is the problem: B=(t^2-g)/(t+g) then
Bt+Bg=t^2-g => move all g on the same side
=> Bg+g=t^2-Bt =>
=>g(B+1)=t^2-Bt =>
=> g= (t^2-Bt)/(B+1)

therefore g= (t^2-Bt)/(B+1)

Answer 4

i assumed the problem to read

B = (t^2 - g) / (t + g)

multiply both sides by (t + g) to cancel the denominator..

youre left with ::

B(t+g) = (t^2 - g)

right side expands by the difference of two squares factoring pattern

B(t + g) = (t + g)(t - g)

(t + g) cancels on both sides

left with B = t - g

- g = B - t

multiply both sides by -1

g = -B + t

i hope making g the subject means getting g by itselff, cause thats what i did. :)

Answer 5

Did you mean B = (t^2 - g) / (t + g)?

If yes, B = (t + g) (t - g) / (t + g) = t - g

B = t - g can be rearranged as

g = t - B

Answer 6

B= t^2 - g/(t+g)
multiply (t+g) to other side
(t+g)B=t^2-g
add g to other side
(t+g)B+g=t^2
multiply out
tB+gB+g=t^2
subtract tB to other side
gB+g=t^2-tB
factor out g
g(B+1)=t^2-tB
divide out (B+1)
g=t^2-tB/(B+1)

Answer 7

Assume expression is:-
B = (t² - g) / (t + g)
Bt + Bg = t² - g
Bg + g = t² - Bt
g.(B + 1) = t.(t - B)
g = t.(t - B) / (B + 1)

Answer 8

g (t+g)=t^2-B
gt + g^2= t^2-B
where t=1
g + g^2= -B
g=-B/ g^2