How do you solve #A = P + Prt# for #t#?

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Answer 1 · 2017-03-02T17:11:18

See the entire solution process below:

Step 1) Subtract #color(red)(P)# from each side of the equation to isolate the #r# term while keeping the equation balanced:

#A - color(red)(P) = -color(red)(P) + P + Prt#

#A - P = 0 + Prt#

#A - P = Prt#

Now, divide each side of the equation by #color(red)(P)color(blue)(t)# to solve for #r# while keeping the equation balanced:

#(A - P)/(color(red)(P)color(blue)(t)) = (Prt)/(color(red)(P)color(blue)(t))#

#(A - P)/(Pt) = (color(red)(cancel(color(black)(P)))rcolor(blue)(cancel(color(black)(t))))/(cancel(color(red)(P))cancel(color(blue)(t)))#

#(A - P)/(Pt) = r#

#r = (A - P)/(Pt)#

Or

#r = A/(Pt) - P/(Pt)#

#r = A/(Pt) - color(red)(cancel(color(black)(P)))/(color(red)(cancel(color(black)(P)))t)#

#r = A/(Pt) - 1/t#

Answer 2 · 2017-03-02T17:19:33

#t=(A-P)/(Pr)#

Given:#" "A=P+Prt#

Factor out the #P#

#A=P(1+rt)#

Divide both sides by P

#A/P=1+rt#

Subtract 1 from both sides

#A/P-1=rt#

Divide both sides by r

#A/(Pr)-1/r=t" "->" "t=A/(Pr)-1/r#

Or could write this as:

#t=(A-P)/(Pr)#