Lesson Solving word problems in three unknowns by the backward method

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Solving word problems in three unknowns by the backward method


Problem 1

Alice,  Ben and  Carl collect stamps.  They exchange stamps among themselves according to the following scheme:
    (1)   Alice gives  Ben as many stamps as  Ben has and  Carl as many stamps as  Carl has.
    (2)   After that,  Ben gives  Alice and  Carl as many stamps as each of them has,
    (3)   and then  Carl gives  Alice and  Ben as many stamps as each has.
If each finally has  80  stamps,  with how many stamps does  Ben start?

Solution

This problem can be solved by reducing to system of three equations in three unknown.
But this reduction is very labor-intensive and may lead to many errors.
The solution of the system of 3 equations in 3 unknown is also labor-intensive.
I will solve the problem using the backward method
It allows to get the solution with much lesser efforts. 

                FIRST BACKWARD STEP, inverse to the base step 3


Let "a", "b" and "c" be the numbers of stamps that Alice, Ben and Carl had after step 2 immediately before step 3.


Step 3 is this transformation  (a, b, c) ---> (2a, 2b, c-a-b).

So, the description of step 3 gives us these equations

    2a = 80,

    2b = 80,

    c-a-b = 80.


Their solution is  a = 80/2 = 40,  b = 80/2 = 40,  c = 80 + a + b = 80 + 40 + 40 = 160.


          Thus, immediately before step 3 and after step 2 
     Alice has 40 stamps, Ben has 40 stamps, Carl has 160 stamps.



                SECOND BACKWARD STEP, inverse to the base step 2


Let "a", "b" and "c" be the numbers of stamps that Alice, Ben and Carl had after step 1 immediately before step 2.


Step 2 is this transformation  (a, b, c) ---> (2a, b-a-c, 2c).

So, the description of step 2 gives us these equations

    2a = 40,

    b-a-c = 40,

    2c = 160.


Their solution is a = 40/2 = 20, c = 160/2 = 80, b = 40 + a + c = 40 + 20 + 80 = 140.


          Thus, immediately before step 2 and after step 1 
    Alice has 20 stamps, Ben has 140 stamps, Carl has 80 stamps.



                THIRD BACKWARD STEP, inverse to the base step 1


Let "a", "b" and "c" be the numbers of stamps that Alice, Ben and Carl had before step 1 (i.e. initially). 


Step 1 is this transformation  (a, b, c) ---> (a-b-c, 2b, 2c).

So, the description of step 1 gives us these equations

    a-b-c = 20,

    2b = 140,

    2c = 80.


Their solution is b = 140/2 = 70,  c = 80/2 = 40,  a = 20 + b + c = 20 + 70 + 40 = 130.


          Thus, immediately before step 1 (initially) 
    Alice has 130 stamps, Ben has 70 stamps, Carl has 40 stamps.


ANSWER.  Ben starts with 70 stamps.


CHECK.  Step 1:  (130, 70,  40) ---> (130-70-40,      140,         80)  = (20, 140,  80).

        Step 2:  (20, 140,  80) ---> (       40, 140-20-80,       160)  = (40,  40, 160).

        Step 3:  (40,  40, 160) ---> (       80,        80, 160-40-40)  = (80,  80,  80).    ! correct !


My other lessons in this site on determinants of  3x3-matrices and the Cramer's rule for solving systems of linear equations in three unknowns are
    - Determinant of a 3x3 matrix
    - Co-factoring the determinant of a 3x3 matrix
    - HOW TO solve system of linear equations in three unknowns using determinant (Cramer's rule)
    - Solving systems of linear equations in three unknowns using determinant (Cramer's rule)
    - Solving word problems by reducing to systems of linear equations in three unknowns
    - The tricks to solve some word problems with three and more unknowns using mental math
    - Solving systems of non-linear equations in three unknowns using Cramer's rule
    - Sometime two equations are enough to find three unknowns by an UNIQUE way
    - Two very different approaches to one word problem
    - Solving system of linear equation in 17 unknowns
    - Solving system of linear equation in 19 unknowns
    - OVERVIEW of LESSONS on determinants of 3x3-matrices and Cramer's rule for systems in 3 unknowns
under the current topic  Matrices, determinant, Cramer rule  of the section  Algebra-II.

My other lessons in this site on solving systems of linear equations in three unknowns are
    - Solving systems of linear equations in 3 unknowns by the Substitution method,
    - BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method,
    - Solving systems of linear equations in 3 unknowns by the Elimination method  and
    - BRIEFLY on solving systems of linear equations in 3 unknowns by the Elimination method

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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