How do you solve for s in #a=s-sr #?

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Answer 1 · 2016-03-19T15:30:47

#s=a/(1-r)#

#color(blue)("Shortcut method")#

By inspection, write as:#" "a=s(1-r)#

Take the #(1-r)# to the other side

#a/(1-r)=s#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("From first principles")#

Note that:
#" "sxx1 = s#
#" "sxx (-r)=-sr#

So for the right hand side; factor out the #s#

#" "s(1-r)#

This means everything inside the bracket is multiplied by #s#

So now we have

#" "a=s(1-r)#

Divide both sides by #(1-r)# giving

# " "a/(1-r)" "=" "sxx(1-r)/(1-r)#

But #(1-r)/(1-r) = 1# so we now have

#" "a/(1-r)" "=" "sxx1#

But #sxx1 = s# so we have

#" "a/(1-r)=s#

Swap sides

#" "s=a/(1-r)#