Algebra Help Please, solve for a when s=H/b-a <===== Fraction?

Question

Help please

Sorry about the way I wrote this equation, it should be,

s=H/(b-a)

Answer 1

the expression of your equation is confusing. It has 2 interpretation but I will solve the two for sure.

your equation is s=H/b-a

1. consider that variable a is not included in the fraction.
first, transpose the term H/b to the left that will be

s - H/b = -a

then, divide both sides by -1 to make the variable a
positive that will give you :

a = H/b - s

2. consider that a is included in the fraction
your equation will be:

s = H/(b - a)

first, by cross multiplication you will get

sb - sa = H

next, transpose the term sb to the right

- sa = H - sb

then divide by - s to get the value of a

you will get a = (sb - H)/s

Im sure any of the two interpretation I get from it
it is in you to decide what you are going to choose.

Answer 2

First We need a on Left hand side

SO multiply throughout by b-a

s(b-a)=H (b-a) in denominator cancels with b-a in numerator after multiplication

expand bracket

sb-sa=H

bring H on other side and sa on right side to avoid -ve

sb-H=sa

divide by s to get a

a=(sb-H)/s

further solve right hand side

a= b-(H/s)

Answer 3

Multiply both sides of the equation by (b-a)
So s (b-a) = H
Divide both sides by s
So b-a = H/s
Subtract b from both sides
So -a = H/s - b
Multiply both sides by (-1)
So a = b - H/s

Answer 4

s = H/b - a

sb = b(H/b) - ab

sb = H - ab

sb - H = H - ab - H

sb - H = - ab

sb/b - H/b = - ab/b

s - H/b = a

The answer is a = s - H/b

- - - - -s-

Answer 5

s=H/(b-a)
=>s*(b-a) = H
=>s*b - s*a = H
=> - s*a = H - s*b
=> s*a = s*b - H
=> a = (s*b - H)/s = b - (H/s)