How do you solve #P= \frac { b } { d + t }# for #t#?

2 Answers
Jan 28, 2017

#t = \frac {b} {p}# - #d#

Explanation:

  1. Cross multiply #P# and #d+t# to get:
    #d+t# = # \frac {b} {P}#

  2. To make #t# the subject, substract #d# (move #d# to the other
    side!) :
    #t# = #\frac {b} {P}# - #d#

Jan 28, 2017

#t=b/P-d#

Explanation:

We can #color(blue)"cross multiply"# to 'eliminate' the fraction.

We treat equations with letters only #color(blue)"literal equations"# in exactly the same way as normal equations.

#rArrP/1=b/(d+t)#

#rArrP(d+t)=b#

divide both sides by P

#(cancel(P) (d+t))/cancel(P)=b/P#

#rArrd+t=b/P#

To solve for t, subtract d from both sides.

#cancel(d)cancel(-d)+t=b/P-d#

#rArrt=b/P-d" is the solution"#