How do you solve #5/2t-t=3+3/2t#?

1 Answer
Jan 14, 2017

Answer:

See entire solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(2)# to eliminate the fraction and keep the equation balanced:

#color(red)(2)(5/2t - t) = color(red)(2)(3 + 3/2t)#

#(color(red)(2) xx 5/2t) - (color(red)(2) xx t) = (color(red)(2) xx 3) + (color(red)(2) xx 3/2t)#

#(cancel(color(red)(2)) xx 5/color(red)(cancel(color(black)(2)))t) - 2t = 6 + (cancel(color(red)(2)) xx 3/color(red)(cancel(color(black)(2)))t)#

#5t - 2t = 6 + 3t#

#3t = 6 + 3t#

We can now subtract #color(red)(3t)# from each side of the equation:

#3t - color(red)(3t) = 6 + 3t - color(red)(3t)#

#0 = 6 + 0#

#0 != 6#

Because #0# does not equal #6# we know there is no solution for #t# for this problem other than the null set or #t = {O/}#