Solving problems using substitutions

Essense of the method: The given problem is transformed into a new one by changing the region of solutions tema5_EN_files\tema5_EN_MathML_0.jpg into tema5_EN_files\tema5_EN_MathML_1.jpgso that every element tema5_EN_files\tema5_EN_MathML_2.jpg corresponds to an element tema5_EN_files\tema5_EN_MathML_3.jpgand follows a certain rule tema5_EN_files\tema5_EN_MathML_4.jpg. Hence the set of solutions to the problem tema5_EN_files\tema5_EN_MathML_5.jpg is transformed into a set of solutions tema5_EN_files\tema5_EN_MathML_6.jpg by means of tema5_EN_files\tema5_EN_MathML_7.jpg the sought set tema5_EN_files\tema5_EN_MathML_8.jpg for the given problem is found. This correspondence between tema5_EN_files\tema5_EN_MathML_9.jpg, defined by means of tema5_EN_files\tema5_EN_MathML_10.jpg, is called susbstitution.

1  Solving equations or inequalities by means of introducing a new unknown quantity which are then reduced again to equations or inequalities

Example 1.1. 

Solve the equation tema5_EN_files\tema5_EN_MathML_11.jpg

Solution The equation can be represented in the following way tema5_EN_files\tema5_EN_MathML_12.jpg( we use the identity tema5_EN_files\tema5_EN_MathML_13.jpg). What remains is for the reader to notice that tema5_EN_files\tema5_EN_MathML_14.jpg. Then the given equation assumes a more agreeable form tema5_EN_files\tema5_EN_MathML_15.jpg

Obviously we make the substitution tema5_EN_files\tema5_EN_MathML_16.jpg and after that we solve the equation tema5_EN_files\tema5_EN_MathML_17.jpg. Its roots are tema5_EN_files\tema5_EN_MathML_18.jpg and tema5_EN_files\tema5_EN_MathML_19.jpg. What remains is to solve the quadratic equations tema5_EN_files\tema5_EN_MathML_20.jpg and tema5_EN_files\tema5_EN_MathML_21.jpg and to unite their solutions.

Example 1.2. 

Solve the equation tema5_EN_files\tema5_EN_MathML_22.jpg


(variant 1) The arithmetic mean of 4,5,6 and 7 is 5.5. We make the substitution tema5_EN_files\tema5_EN_MathML_23.jpg and reduce the equation to the nicer tema5_EN_files\tema5_EN_MathML_24.jpg, the solution to which is easy to find.

(variant 2) Multiplying tema5_EN_files\tema5_EN_MathML_25.jpg and tema5_EN_files\tema5_EN_MathML_26.jpg the given equation is transformed into tema5_EN_files\tema5_EN_MathML_27.jpgWe have to good possibilities for substitution tema5_EN_files\tema5_EN_MathML_28.jpg or tema5_EN_files\tema5_EN_MathML_29.jpg

Example 1.3. 

Solve the equation tema5_EN_files\tema5_EN_MathML_30.jpg

If you notice that you can represent the equation in the form tema5_EN_files\tema5_EN_MathML_31.jpg then the substitution tema5_EN_files\tema5_EN_MathML_32.jpg will be usefultema5_EN_files\tema5_EN_MathML_33.jpg tema5_EN_files\tema5_EN_MathML_34.jpg

Example 1.4. 
Solve the inequality tema5_EN_files\tema5_EN_MathML_35.jpg.

With a bit more resourcefulness the given inequality will look like this tema5_EN_files\tema5_EN_MathML_36.jpgor tema5_EN_files\tema5_EN_MathML_37.jpg

A suitable subsititution is tema5_EN_files\tema5_EN_MathML_38.jpg. Now we will solve the quadratic inequality tema5_EN_files\tema5_EN_MathML_39.jpg

Going back to the unknown quantity tema5_EN_files\tema5_EN_MathML_40.jpg tema5_EN_files\tema5_EN_MathML_41.jpg it is necessary to solve a double inequality tema5_EN_files\tema5_EN_MathML_42.jpg which is the same as the following system tema5_EN_files\tema5_EN_MathML_43.jpg.

After finding the solutions to each of the inequalities and then take their intersection you get the answer tema5_EN_files\tema5_EN_MathML_44.jpg.

Example 1.5. 

Solve the equation tema5_EN_files\tema5_EN_MathML_45.jpg

Hint If we add to both sides of the expression tema5_EN_files\tema5_EN_MathML_46.jpg (doubled product) we get tema5_EN_files\tema5_EN_MathML_47.jpgor tema5_EN_files\tema5_EN_MathML_48.jpg

If you notice thattema5_EN_files\tema5_EN_MathML_49.jpg tema5_EN_files\tema5_EN_MathML_50.jpg, then the previous equation will look like this tema5_EN_files\tema5_EN_MathML_51.jpgOf course the substitution is tema5_EN_files\tema5_EN_MathML_52.jpg

Example 1.6. 

Solve the equation tema5_EN_files\tema5_EN_MathML_53.jpg

Hint We replace the equation with its equivalent tema5_EN_files\tema5_EN_MathML_54.jpgThe substitution is tema5_EN_files\tema5_EN_MathML_55.jpg.

Problems for individual work

Exercise 1.1. Solve the equation tema5_EN_files\tema5_EN_MathML_56.jpg.

Exercise 1.2. Solve the equation tema5_EN_files\tema5_EN_MathML_57.jpg.

Exercise 1.3. Solve the equation tema5_EN_files\tema5_EN_MathML_58.jpg.

Exercise 1.4. Solve the inequality tema5_EN_files\tema5_EN_MathML_59.jpg.

2  Solving equations by means of substitutions which are reduced to solving systems

Example 2.1. 

Solve the equation tema5_EN_files\tema5_EN_MathML_60.jpg.

(variant 1) A suitable substitution is tema5_EN_files\tema5_EN_MathML_61.jpg

We get the following system of equations tema5_EN_files\tema5_EN_MathML_62.jpg. Let tema5_EN_files\tema5_EN_MathML_63.jpg and tema5_EN_files\tema5_EN_MathML_64.jpg.

The system now looks like this tema5_EN_files\tema5_EN_MathML_65.jpg . This way it is pleasing to be looked at as well as to be solved.

(variant 2) We make the substitutions directly tema5_EN_files\tema5_EN_MathML_66.jpgи tema5_EN_files\tema5_EN_MathML_67.jpg and get the system tema5_EN_files\tema5_EN_MathML_68.jpg

After solving it the solutions to the given equation are found through the equations tema5_EN_files\tema5_EN_MathML_69.jpg and tema5_EN_files\tema5_EN_MathML_70.jpg

Example 2.2. 

Solve the equation tema5_EN_files\tema5_EN_MathML_71.jpgand tema5_EN_files\tema5_EN_MathML_72.jpg.

Solution Let tema5_EN_files\tema5_EN_MathML_73.jpg and tema5_EN_files\tema5_EN_MathML_74.jpg. Then the equation will look like this tema5_EN_files\tema5_EN_MathML_75.jpg The second equation for tema5_EN_files\tema5_EN_MathML_76.jpg and tema5_EN_files\tema5_EN_MathML_77.jpg we will get when we multiply tema5_EN_files\tema5_EN_MathML_78.jpgand tema5_EN_files\tema5_EN_MathML_79.jpg respectively with tema5_EN_files\tema5_EN_MathML_80.jpg and tema5_EN_files\tema5_EN_MathML_81.jpg and then add them together. The result is tema5_EN_files\tema5_EN_MathML_82.jpg, i.e. the system is tema5_EN_files\tema5_EN_MathML_83.jpg After determining tema5_EN_files\tema5_EN_MathML_84.jpg and tema5_EN_files\tema5_EN_MathML_85.jpg and going back to the substitutions it is easy to find that tema5_EN_files\tema5_EN_MathML_86.jpg.

Example 2.3. 

Solve the equation tema5_EN_files\tema5_EN_MathML_87.jpg.

Solution Let tema5_EN_files\tema5_EN_MathML_88.jpg and tema5_EN_files\tema5_EN_MathML_89.jpg Then the given equation will look like this tema5_EN_files\tema5_EN_MathML_90.jpg. The second equation for tema5_EN_files\tema5_EN_MathML_91.jpg and tema5_EN_files\tema5_EN_MathML_92.jpg we will get by adding together term by term tema5_EN_files\tema5_EN_MathML_93.jpg and tema5_EN_files\tema5_EN_MathML_94.jpg, i.e.tema5_EN_files\tema5_EN_MathML_95.jpg. The result is the following system tema5_EN_files\tema5_EN_MathML_96.jpg whose solutions are tema5_EN_files\tema5_EN_MathML_97.jpg. Going back to the substitutions we find that tema5_EN_files\tema5_EN_MathML_98.jpg.

Example 2.4. 

Solve the equation tema5_EN_files\tema5_EN_MathML_99.jpg.

Hint We make the substitutionstema5_EN_files\tema5_EN_MathML_100.jpg and tema5_EN_files\tema5_EN_MathML_101.jpg and solve the following system tema5_EN_files\tema5_EN_MathML_102.jpg.

Answer tema5_EN_files\tema5_EN_MathML_103.jpg.

Problems for individual work

Exercise 2.1. Solve the equation tema5_EN_files\tema5_EN_MathML_104.jpg.

Exercise 2.2. Solve the equation tema5_EN_files\tema5_EN_MathML_105.jpg.

Exercise 2.3. Solve the equation tema5_EN_files\tema5_EN_MathML_106.jpg.

3  Solving some systems of equations by means of partial substitution (in one of the equations of the system) or global substitution (in each of the equations of the system)

Example 3.1. 

Solve the system tema5_EN_files\tema5_EN_MathML_107.jpg.

Solution The system can be written down in the form tema5_EN_files\tema5_EN_MathML_108.jpg.

A suitable substitution is tema5_EN_files\tema5_EN_MathML_109.jpg. Then the second equation of the system will look like this tema5_EN_files\tema5_EN_MathML_110.jpg. Later on the set of solutions of the system becomes a union of the sets of solutions of the systems tema5_EN_files\tema5_EN_MathML_111.jpg:

Example 3.2. 
Solve the system tema5_EN_files\tema5_EN_MathML_112.jpg

Solution Notice that the left sides of both equations are symmetric expressions in regard to tema5_EN_files\tema5_EN_MathML_113.jpg and tema5_EN_files\tema5_EN_MathML_114.jpg. To solve such systems substitutionstema5_EN_files\tema5_EN_MathML_115.jpg and tema5_EN_files\tema5_EN_MathML_116.jpg are used. Before that we will replace the system with its equivalent system tema5_EN_files\tema5_EN_MathML_117.jpg(We bring to notice thattema5_EN_files\tema5_EN_MathML_118.jpg and tema5_EN_files\tema5_EN_MathML_119.jpg).

We make the substitutions tema5_EN_files\tema5_EN_MathML_120.jpg and tema5_EN_files\tema5_EN_MathML_121.jpg and solve the system tema5_EN_files\tema5_EN_MathML_122.jpg

In the end we have to solve the systems tema5_EN_files\tema5_EN_MathML_123.jpg Notice that if tema5_EN_files\tema5_EN_MathML_124.jpg and tema5_EN_files\tema5_EN_MathML_125.jpg are the respective sets of solutions to the last two systems, then the solution of the given system is the set tema5_EN_files\tema5_EN_MathML_126.jpg.

Problems for individual work

Exercise 3.1. Solve the system tema5_EN_files\tema5_EN_MathML_127.jpg

Hint We replace tema5_EN_files\tema5_EN_MathML_128.jpg and tema5_EN_files\tema5_EN_MathML_129.jpg and the system is reduced to the following system tema5_EN_files\tema5_EN_MathML_130.jpg. If you add together the equations of the new system you will get a quadratic equationtema5_EN_files\tema5_EN_MathML_131.jpg.

Exercise 3.2. Solve the system tema5_EN_files\tema5_EN_MathML_132.jpg .

Hint Again replace tema5_EN_files\tema5_EN_MathML_133.jpg and tema5_EN_files\tema5_EN_MathML_134.jpg and keep in mind that tema5_EN_files\tema5_EN_MathML_135.jpg.

By Rumiana Mavrova and Ilia Makrelov, Plovdiv university, ,