how do you solve 's' for the equation r=(fs)/(s+a)?

Question

step-by-step please...

Answer 1

becareful the positive and negative, and ( ) as well!

some of the answers above are collect, but for better and clear step, use ( ).

r = (fs) / (s + a)
rs + ar = fs
rs - fs = -ar
s (r - f) = -ar
s = (-ar) / (r-f)

or,

s = (ar) / (f-r)

good luck!

Answer 2

:/


I would cry :'(


Erm,


Try reversing the equation and muddling it around, so "s" is on one side of the equation... i think

search the topic on google.


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Answer 3

r = (fs)/(s+a)

multiple both sides by (s+a) to get:

r(s+a) = (fs)

multiply r(s+a) to (rs) + (ra) to get:

rs + ra = fs

subtract fs from both sides to get:

rs - fs + ra = 0

subtract ra from both sides to get:

rs - fs = -ra

factor s from rs - fs to s(r-f) to get:

s(r-f) = -ra

divide both sides by (r-f) to get:

s = (-ra)/(r-f)

Answer 4

James has got it right, Mellissas forgot a negative sign somewhere, easily done.

Answer 5

r=(fs)/(s+a)
(s+a)r=fs
sr+ar=fs
ar=fs-sr
ar=s(f-r)
(ar)/(f-r)=s

Basically, you need to get all like terms together, then get s by itself.

Answer 6

rs + ra = fs
(r - f) s = - ra
s = - ra / (r - f)

Answer 7

r=(fs)/(s+a)
r(s+a)=fs
rs+ra=fs
rs-fs+ra=0
s(r-f)+ra=0
s(r-f)= -ra
s= -ra/(r-f)

Answer 8

that sum is impossible

Answer 9

r=(fs)/(s+a)

rs+ra=fs

(r-f)s+ra=0

rs-fs= -ra

(r-f)s= -ra

s= -(ra)/(r-f)=(ra)/(f-r)