SOLUTION: Divide 56 Into Two Parts Such That Three Times The First Parts Exceeds One Third Of The Second By 48. The Parts Are

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Divide 56 Into Two Parts Such That Three Times The First Parts Exceeds One Third Of The Second By 48. The Parts Are      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 732140: Divide 56 Into Two Parts Such That Three Times The First Parts Exceeds One Third Of The Second By 48. The Parts Are
Found 2 solutions by lynnlo, ikleyn:
Answer by lynnlo(4176) About Me  (Show Source):
Answer by ikleyn(53472) About Me  (Show Source):
You can put this solution on YOUR website!
.
Divide 56 Into Two Parts Such That Three Times The First Parts Exceeds One Third Of The Second By 48.
The Parts Are
~~~~~~~~~~~~~~~~~~~~~~~~ 

Let x be first part; then the second part is (56-x).


The imposed condition is

    3x - %2856-x%29%2F3 = 48.


To solve this equation, multiply all the terms by 3

    9x - (56-x) = 48*3,

    9x + x = 144 + 56,

      10x  =    200,

        x  =    200/10 = 20.


ANSWER.  The parts are 20 and 56-20 = 36.

Solved.

The answer in the post by @lynnlo is incorrect.

Simply ignore his post.