How do you solve #3( 2p + 19) + 2p = 8p#?

2 Answers
Apr 22, 2017

p does not exist.


This is the step-by-step procedure of solving for p:
(ignore this if you are familiar with algebra)

Expand the the bracket:
32p = 6p
19 = 57

Then put it altogether
6p + 57 + 2p = 8p.

Put all values with p to one side of the equation
6p - 8p+ 2p = -57
and as 57 is positive, changing it to the other side with make it negative, the same applies to 8p.

Then simplify:
0 = -57.
As this is false, p does not exist.

It would not be within the field of real numbers.

Apr 22, 2017

See explanation


Begin by distributing #3# to #2p# and #19#

#color(green)3(color(blue)(2p+19)) +2p=8p -> 6p+57+2p=8p#

Combine like terms:

#color(blue)(6p)+38+color(blue)(2p)=8p -> 8p+57=8p#

We can subtract #8p# from both sides to keep the variable on one side but if we do, our variable no longer has any value and thus we cannot complete the problem.

#cancel(8p-8p)+57=8p-8p -> 57=0p -> 57=0#

There is nothing that would make #57=0# true. There is perhaps a typo in the question or there is no solution.