Linear cyclic systems

Aim: This topic aims at extending the knowledge students have about systems of equations and forming skills for solving them in extracurricularly in the eighth grade.

Systems of the following form

tema2_EN_files\tema2_EN_MathML_0.jpg (1) 


tema2_EN_files\tema2_EN_MathML_1.jpg (2) 

or ones reduced to these two types will be called linear cyclic systems. Here tema2_EN_files\tema2_EN_MathML_2.jpg are unknowns and tema2_EN_files\tema2_EN_MathML_3.jpg are given constants.

Analyzing system (1) we find the unknown quantities occur as summands in the left sides of the equations constituting it the same number of times, i.e.. they reoccur twice. Even if we change the places of the unknowns in each of the equations the system does not change its type and also contains the unknown quantities twice. This gives rise to the idea that adding together both sides of these equations we get for both cases

tema2_EN_files\tema2_EN_MathML_4.jpg (3) 

Using consecutively the first, second and third equation of the system and equation (3) we realize that




Think whether the solution of the system would change if we changed tema2_EN_files\tema2_EN_MathML_7.jpg with tema2_EN_files\tema2_EN_MathML_8.jpg, tema2_EN_files\tema2_EN_MathML_9.jpg with tema2_EN_files\tema2_EN_MathML_10.jpg and tema2_EN_files\tema2_EN_MathML_11.jpg with tema2_EN_files\tema2_EN_MathML_12.jpg.

Analyzing system (2) we discover that here tema2_EN_files\tema2_EN_MathML_13.jpg are multipliers which are also repeated twice and the system does not change its type if we change the places of the unknowns in each its equations. This suggests the idea of multiplying together both sides of the equation. We get tema2_EN_files\tema2_EN_MathML_14.jpg, after which tema2_EN_files\tema2_EN_MathML_15.jpg is found. Using the result for tema2_EN_files\tema2_EN_MathML_16.jpg and each of the equations of (2) tema2_EN_files\tema2_EN_MathML_17.jpg are easily found.

Some of the systems in this group contain a cubic equation or one with a higher exponent, but as the way to solve them requires only knowledge about how to solve linear equation systems, linear cyclic systems would be easy to cope with.

Example 1. Solve the system:

tema2_EN_files\tema2_EN_MathML_18.jpg (4) 

This system is of the type (1), that is why we add together the sides of the equations in system (4). We get tema2_EN_files\tema2_EN_MathML_19.jpg.

From the conditions we know that tema2_EN_files\tema2_EN_MathML_20.jpg. Then tema2_EN_files\tema2_EN_MathML_21.jpgi.e. tema2_EN_files\tema2_EN_MathML_22.jpg. From tema2_EN_files\tema2_EN_MathML_23.jpg we get that tema2_EN_files\tema2_EN_MathML_24.jpg and from tema2_EN_files\tema2_EN_MathML_25.jpg we find tema2_EN_files\tema2_EN_MathML_26.jpg. The ordered triple found (-2;6;-3) is the solution to (4).

Knowing the solution of system (4) find the solutions to the following two systems:

tema2_EN_files\tema2_EN_MathML_27.jpg (5) 
tema2_EN_files\tema2_EN_MathML_28.jpg (6) 

without solving them.

Example 2. Solve the system

tema2_EN_files\tema2_EN_MathML_29.jpg (7) 

Hint : The system is of the type (2) After multiplying together the equations in (7) we get tema2_EN_files\tema2_EN_MathML_30.jpg

After which the solutions to (7) are found: (2;3;1), (-2;-3;-1).

Example 3. Without solving the system

tema2_EN_files\tema2_EN_MathML_31.jpg (8) 

find its solutions.

These ideas for solving system (1) and (2) can be used for solving the following problems:

Example 4. Solve the system

tema2_EN_files\tema2_EN_MathML_32.jpg (9) 

In system (9) tema2_EN_files\tema2_EN_MathML_33.jpg are not equal to zeros. We write (9) down in the form

tema2_EN_files\tema2_EN_MathML_34.jpg (10) 

System (10) is similar to (1). Then we use already known way of solving - adding together each member of the equations. We get

tema2_EN_files\tema2_EN_MathML_35.jpg (11) 

By means of equations (10) and (11) we have

tema2_EN_files\tema2_EN_MathML_36.jpg (12) 

This system (12) is of the type (2). It is solved using multiplication of the equations, i.e. tema2_EN_files\tema2_EN_MathML_37.jpg and in combination with each equation of (12) we find that


from where it is easy to find the ordered triples of numbers which are solutions to (9)


Problem 1. tema2_EN_files\tema2_EN_MathML_40.jpg .

Hint: The system is of type (1). After suitable transformations it is reduced to the following system


i.e. to a system of type (2).

Problem 2. tema2_EN_files\tema2_EN_MathML_42.jpg .

Hint: After determining what values tema2_EN_files\tema2_EN_MathML_43.jpg and tema2_EN_files\tema2_EN_MathML_44.jpg can assume present the system in the following form

tema2_EN_files\tema2_EN_MathML_45.jpg (13) 

System (13) is of type (1).

Problem 3. tema2_EN_files\tema2_EN_MathML_46.jpg .

Hint: After dividing the first equation by 2, the second by 3, and the third by 6 the system is reduced to type (1) (see Example 3 above).

Problem 4. tema2_EN_files\tema2_EN_MathML_47.jpg .

Hint: Represent the system in this way tema2_EN_files\tema2_EN_MathML_48.jpg .

Use the hint for Problem 1.

Answers: (tema2_EN_files\tema2_EN_MathML_49.jpg

Problem 5. tema2_EN_files\tema2_EN_MathML_50.jpg

Hint: As tema2_EN_files\tema2_EN_MathML_51.jpgare different from 0, the system is of the following type tema2_EN_files\tema2_EN_MathML_52.jpg .

See the hint for Problem 4 above.

Problem 6. tema2_EN_files\tema2_EN_MathML_53.jpg.

Hint: We add together each member of the equations. We get tema2_EN_files\tema2_EN_MathML_54.jpgafter which we will get a system of type (1).

Problem 7. tema2_EN_files\tema2_EN_MathML_55.jpg

Hint: Open the brackets in each of the equations of the system. The system which we get is of type (1).

By Rumyana Mavrova, Plovdiv university,