Lesson Using hidden symmetry to solve systems of linear equations in three unknowns

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Using hidden symmetry to solve systems of linear equations in three unknowns

In this lesson, I collected problems on solving linear systems of three equations in three unknowns,
that admit tricks making their solutions short, easy and attractive. The method uses
hidden symmetry in equations, which is sometimes very effective trick.

Problem 1

Al, Bill and Carl went shopping. When they compared the totals that each had spent they found that:
Al and Bill together spent a total of Php 1 200;
Bill and Carl together spent a total of Php 1 800.00; and
Al and Carl together spent a total of Php 1 000.
How much did each person spend?

Solution

Let  'a'  be the amount that Al spent (in Php);

     'b'  be the amount that Bill spent (in Php), and 

     'c'  be the amount that Carl spent (in Php). 


Then you have 3 equations in 3 unknowns:

    a + b     = 1200, (1)

        b + c = 1800, (2)

    a     + c = 1000. (3)


Add all three equations (which means "add their left sides and add their right sides"). 
You will get 

    2*(a + b + c) = 1200 + 1800 + 1000 = 4000, or

    a + b + c = 2000. (4)



Now, subtract equation (1) from (4). You will get

    c = 2000 - 1200 = 800. Hence, Carl spent 800 Php.


Next, subtract equation (2) from (4). You will get

    a = 2000 - 1800 = 200. Hence, Al spent 200 Php.


Finally, subtract equation (3) from (4). You will get

    b = 2000 - 1000 = 1000. Hence, Bill spent 1000 Php.


Answer. Al spent 200 Php, Bill spent 1000 Php, and Carl spent 800 Php.

To see other similar problems, look at the lesson The tricks to solve some word problems with three and more unknowns using mental math in this site.

Problem 2

A store has a sale on almonds, pecans and pistachios. One pound of almonds, one pound of pecans
and one pound of pistachios cost $12. Two pounds of almonds and three pounds of pecans cost $16.
Three pounds of pecans and two pounds pistachios cost $24. Find the price of each kind of nut.

Solution

Meanwhile, the problem has a much simpler solution, which teaches you to think and to approach the problem
in a new way.

Let  'a'  be  the pounds of almonds,
     'b'  be  the pounds of pecans, and
     'c'  be  the pounds of pistachios.


You have 3 equations that need to be solved simultaneously

     a +  b + c  = 12,     (1)
    2a + 3b      = 16,     (2)
         3b + 2c = 24.     (3)


This system of equations has a hidden symmetry, which will help us to solve.


Add equations (2) and (3)  (both sides separately).  You will get

    2a + 6b + 2c = 40.     (4)


From equation (4), subtract equation (1), multiplied by 2.  
The terms with  'a'  and  'c'  will cancel, and you will get

    6b - 2b = 40 - 2*12,

or

       4b = 16.


Hence,  b = 16/4 = 4.



Now, substitute b=4 into equation (2) and find  'a'

    2a + 3*4 = 16  --->  2a = 16 - 12 = 4  --->  a = 4/2 = 2.



Next, substitute b=4 into equation (3) and find  'c'

    3*4 + 2c = 24  --->  2c = 24 - 12 = 12  --->  c = 12/2 = 6.


At this point, the problem is solved completely.


ANSWER.  a = $2 per pound (almonds);  b = $4 per pound (pecans),  and  c = $6 per pound (pistachios).

The lesson to learn

If you have to solve a system of three equations with three unknowns,
look if there is a hidden symmetry in it that could help.

Problem 3

Hector, Kenny and Elliot went shopping with $2190.
Kenny spent $80 and Hector spent 2/5 of his money.
Hector spent half as much as Elliot and had $160 less than what Elliot had left.
In the end, Kenny and Hector had the same amount of money left.
How much money had Elliot at first?

Solution

This problem has unexpectedly twisted formulation (which is a rare case in such problems).
So, the major step forward is to algebraize it.

Let H, K and E be the amounts the persons had originally.


First equation is obvious

   H + K + E = 2190.         (1)



Hector spent 2/5 of his money; so, he left with 3/5 of his money.

Kenny spent $80;  so Kenny left with (K-80) dollars.

At the end, Kenny and Hector had the same amount, which gives us the second equation

     = K - 80

or

    3H = 5K - 400.           (2)



Hector spent half as much as Elliot, which means that Elliot spent twice as much as Hector.

Hence, Elliot spent .

It means that Elliot left with E -   dollars.


From the problem, at the end Hector had left $160 less than what Elliot had left.

It gives us the last, third equation

     + 160 = E - ,

or

    7H - 5E = -800.          (3)



Thus we have the system of 3 equations in 3 unknowns


    H +  K +  E = 2190.      (1)

   3H - 5K      = -400.      (2)

   7H      - 5E = -800.      (3)


So the problem is algebraized.


To solve this system of equations, add equations (2) and (3).
Keep equation (1) as is

    H +  K +  E =  2190      (1')

  10H - 5K - 5E = -1200      (4)


Now multiply equation (1') by 5 and add with equation (4).  The terms with 'K' and 'E' will cancel
and you will get

    15H = 2190*5 - 1200 = 9750,

      H                 = 9750/15 = 650.


Now from equation (2) we find

    3*650 - 5K = -400  --->  3*650 + 400 = 5K  --->  2350 = 5K  --->  K = 2350/5 = 470;


from equation (3) we get

    7*650 - 5E = -800  --->  7*650 + 800 = 5E  --->  5150 = 5E  --->  E = 5350/5 = 1070.


ANSWER : H= 650;  K = 470;  E = 1070.

Problem 4

Solve a system

    3x + 2y - 8z =  29,
    9x -  y + 2z =  23,
    -x - 2y + 8z = -11.

Solution

        When an assignment is to solve 3x3-system of equations, there are two different ways.
        One way is to use standard substitution or elimination, step by step.
        Another way is to use/(to find) some trick which simplifies the task.
        It is possible to do if you see some useful pattern.

        In this system, the first and the third equations have similar parts "2y - 8z" and "-2y + 8z",
        so if we add these equations, we eliminate 'y' and 'z' simultaneously and easy will find 'z'.

        It is the key to start.

So, your original system of equation is

    3x + 2y - 8z =  29    (1)
    9x -  y + 2z =  23    (2)
    -x - 2y + 8z = -11    (3)


Add equations (1) and (3).  The terms with 'y' and 'z' will cancel each other, and you will get then

    3x - x = 29 - 11,  --->  2x = 18,  --->  x = 18/2 = 9.


Now substitute this value x = 9 into the first and second equations

    3*9 + 2y - 8z = 29     (1')
    9*9 -  y + 2z = 23     (2')


Simplify

          2y - 8z =   2    (1'')
          -y + 2z = -58    (2'')


Now we are on the straight finish line: we only need to solve one 2x2-syatem of equations (1''), (2'').

Solve it by the Elimination method. For it, multiply equation (2'') by 2 (both sides) and add to equation (1'').
You will get

             - 8z + 4z = 2 + 2*(-58),  --->  -4z = -114,  z = (-114)/(-4) = 28.5.


Now from equation (2'')

           y = 58 + 2*28.5 = 115.


ANSWER.  The solution to the given system is x= 9,  y = 115,  z = 28.5.

Problem 5

Solve this system of equations

    3x + 2y + z =   8  
    2x - 3y +2z = -16
     x + 4y - z =  20

Solution

It can be solved using s standard elimination procedure by eliminating one unknown after another.
This way is a standard, but not very effective.
There is more effective and more impressive way.
See my solution below.

So, your starting equations are

    3x + 2y +  z =   8,    (1)
    2x - 3y + 2z = -16,    (2)
     x + 4y -  z =  20.    (3)


Add equation (2) and (3)

    3x +  y +  z =   4.    (4)


Subtract equation (4) from equation (1).  The terms with both unknowns 'x' and 'z' will be eliminated simultaneously.
You will get then

          y      =   4.


Now substitute this value, y = 4, in equations (1) and (3).  You will get then

    3x + z = 8 - 2y = 8 - 2*4 = 0,

     x - z = 20 - 4y = 20 - 4*4 = 4.


Thus we reduced the original system of three equations in three unknowns to the system of two equations in two unknowns

    3x + z =  0,    (5)
     x - z =  4.    (6)


Add equations (5) and (6) to get

    4x = 4,  --->  x = 4/4 = 1.


Find z from equation (6)

    z = x - 4 = 1 - 4 = -3.


ANSWER.  x = 1,  y = 4,  z = -3.

My other lessons in this site on solving systems of linear equations in three unknowns are
    - 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 word problems in three unknowns by the backward method
    - 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

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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