How do you solve #S= n/s (A+t)# for A?

1 Answer
Sep 19, 2016

#color(green)(A=S * s/n -t)#

Explanation:

If
#color(white)("XXX")S=n/s(A+t)#
then
#color(white)("XXX")S * s/n = cancel(s)/cancel(n) * cancel(n)/cancel(s) (A+t)#
and
#color(white)("XXX")S * s/n - t =A#

Note: if #S# and #s# are suppose to be the same variable then:
#color(white)("XXX")A=(s^2)/n-t#