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Lecture 1

Fractional Equations

Denition 1.1. An equation is called a fractional equation if it contains frac-tional expressions where the unknown variables are appeared in their denomina-tors.

The basic approach for solving a fractional equation is to remove the denomi-nator if possible. However, in many cases this cannot be done by simply multiply-ing the equation by the least common multiple of the denominators, since it willlead to an equation of a high degree. Therefore the following techniques are oftenneeded.

1. Use partial fraction techniques .

2. Use techniques of telescopic sum .

3. Use substitutions of variables or expressions .Sometimes, it is advisable to manipulate the expressions before the substitu-tion is discovered and applied.

Examples

Example 1. Solve equationx C7x

C8

x C8x

C9

x C5x

C6 C

x C6x

C7 D0.

Solution The given equation can be simplied to

1 1x C8

1 1x C9

1 1x C6 C

1 1x C7 D

0;

1

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2 Lecture 1 Fractional Equations

so1

x C8 1

x C9 D

1

x C6 1

x C7;

1x 2 C17x C72 D

1x 2 C13x C42

;

x 2 C17x C72 Dx2

C13x C42; ) x D152

:

Example 2. Solve the following equation

1x 2

C2x C

1x 2

C6x

C8 C

1x 2

C10x

C24 D

15

1x 2

C14x

C48

:

Solution By moving the term 1

x 2 C14x C48to left hand side, it follows

that

1x 2 C2x C

1x 2 C6x C8 C

1x 2 C10x C24 C

1x 2 C14x C48 D

15

;

1x.x C2/ C

1.x C2/.x C4/ C

1.x C4/.x C6/ C

1.x C6/.x C8/ D

15

;

12

1x

1x C2 C

1x C2

1x C4 C C

1x C6

1x C8 D

15

;

1x

1x C8 D

25

; so x 2 C8x 20 D0;then .x 2/.x C10/ D0, namely x 1 D2 and x2 D 10 .Example 3. (CHINA/2005) Solve equation jx 3j jx C1j

2jx C1j D1.

Solution Splitting the left hand side to two terms, then

jx 3j2jx C1j

12 D1;

x 3x C1

D3;x 3x C1 D

3 orx 3x C1 D

3;

x 3 D3x C3 ) x D 3 and x 3 D 3x 3 ) x D0:Thus, x D 3 or x D0.

Substitutions of variables or expressions play important role in solving frac-tional equations, as shown in the following examples.

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Lecture Notes on Mathematical Olympiad 3

Example 4. Solve equation2x 2 C1

x C2 C 2x C42x 2 C1 D

3.

Solution Let y D 2x 2 C1

x C2, then the given equation becomes y C

2y D3.

y C 2y D3 ) y

2 3y C2 D0 ) .y 2/.y 1/ D0:

When y D1, then 2x2 x 1 D0, so x 1 D1; x 2 D

12

.

When y D2, then 2x2

2x 3 D0, so x3

D 1

2 .1p

7/; x4

D 1

2 .1 Cp

7/ .Example 5. (SSSMO(J)/2006) Suppose that the two roots of the equation

1x 2 10x 29 C

1x 2 10x 45

2x 2 10x 69 D0

are and . Find the value of C .Solution Let y Dx

2 10x 45 , the given equation then becomes

1y C16 C

1y D

2y 24

;

y C8y 2 C16y D

1y 24

;

y 2 16y 192 Dy2

C16y ) y D 6;x 2 10x 45 D 6 ) x

2 10x 39 D0:

Thus, by Vietes Theorem, C D10 .Sometimes the manipulations on the given equations are needed for nding

the desired substitution, as shown in the following examples.

Example 6. (CHINA/2000) Solve the system for .x; y/ :

xy3x C2y D

18

; xy

2x C3y D 17

:

Solution By taking reciprocals to two sides of each equation, it follows that

2x C

3y D8;

3x C

2y D7:

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Letting u D 1x

; v D 1y

, the system for .u; v/ is

2u C3v D8;3u C2v D7:

By solving them, it is obtained that u D1; v D2. Then, returning to .x; y/ ,x D1 and y D

12

:

Example 7. Solve equation4x 2 Cx C4

x 2 C1 C x 2 C1x 2 Cx C1 D

31

6.

Solution Write the given equation in the form

4 C x

x 2 C1 C1

xx 2 Cx C1 D

5 C 16

;

then1

x

C 1

x

1

x

C 1

x C1

D 16

:

Let w Dx C 1x

, it follows that

1w

1w C1 D

16

;

w.w C1/ D6 ) w2

Cw 6 D0;) .w 2/.w C3/ D0; namely w D2 or w D 3:

(i) w D2 ) x C 1x D2, then .x 1/2

D0, so x 1 Dx 2 D1.(ii) w D 3 ) x C

1x D 3, then x

2

C3x C1 D0, so

x 3 D3 p 5

2 ; x4 D

3 Cp 52

:

Example 8. Solve equation x2

15 C 485x 2 D2x3 8x .Solution When both sides are multiplied by 15, the given equation becomes

x 2 C12x

2

D10 x 12x

:

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Let y Dx 12x

, by completing square it follows that

y 2 10y C24 D0;.y 4/.y 6/ D0; ) y D4 or 6:Then

y D4 ) x 12x D4 ) .x 6/.x C2/ D0; i.e. x1 D6; x 2 D 2:

y D6 ) x12x D6 ) x

2 6x 12 D0; i.e. x3 D3 p 21; x 4 D3 Cp 21:

Example 9. Solve equation 2x 2 3x C 2 3x

x 2 D1.Solution Since x 0, the given equation is equivalent to

2 x 2 C 1x 2

3 x C 1x D1:

Let y Dx C1x , then 2y 2 3y 4 D1, i.e., .2y 5/.y C1/ D0. Thus, y D52or 1.

(i) y D52 ) 2x C

2x D5 ) 2x

2 5x C2 D0 ) .2x 1/.x 2/ D0

) x 1 D 12

; x 2 D2.(ii) y D 1 ) x C

1x D 1 ) x

2

Cx C1 D0, no real solution.

Example 10. Solve equation x 2 C xx C12

D3.

Solution By completing the square on the left hand side, it follows that

x xx C1

2

C 2x 2

x C1 D3;

x 2

x C12

C2

x 2

x C1 3

D0:

Let y D x 2

x C1, then y 2 C2y 3 D0 ) .y C3/.y 1/ D0, so y D 3 or 1.

(i) y D 3 ) x 2

x C1 D3 ) x

2

C3x C3 D0, no real solution.

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(ii) y D 1 ) x 2

x C1 D 1 ) x

2 x 1 D 0 ) x1 D 1 p 5

2; x2 D

1 Cp 52 :

Testing Questions (A)

1: Solve the equation x3

C4x2

C2x 8x 2 C2x 3 D 2x

3

C5x2

C4x2x 2 Cx C1 .

2: Solve the equationx C1x C2 C

x C6x C7 D

x C2x C3 C

x C5x C6

.

3: Solve the equation1

.x 1/x C1

x.x C1/ C C1

.x C9/.x C10/ D11

12 .

4: Solve the equation1

x 2 C11x 8 C 1

x 2 C2x 8 C 1

x 2 13x 8 D0.

5: Solve the equation x 2

C

x 2

.x C1/2

D

5

4

.

6: Solve the equation3x 2 C4x 13x 2 4x 1 D

x 2 C4x C1x 2 4x C1

.

7: Find all the real solutions .x;y;z/ of the following system

9x 2

1 C9x 2 D 32

y; 9y 2

1 C9y 2 D 32

z; 9z 2

1 C9z 2 D 32

x:

8: (RUSMO/1993) Find all positive solutions of the system of fractional equa-

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tionsx 1

C

1

x2

D4;

x 2 C 1x 3 D1;

x 3 C 1x 4 D4;

x 4 C 1x 5 D1;:::

x 99 C 1x 100 D4;x 100 C

1x 1 D1:

9: Solve the equation in x :a Cxb Cx C

b Cxa Cx D

52

.

10: Solve the equation in x :a Cbb Cx C

a Ccx Cc D

2.a Cb Cc/x Cb Cc

(where a

Cb; a

Cc; b Cc; a Cb Cc are all not zero).11: Given that the equation

xx C1 C

x C1x D

4x Cax.x C1/

has only one real root,

nd the value of real number a .

Testing Questions (B)

1: Solve equationx 2 x C1

x 2 C2 Cx C1

x 2 x C1 D1.

2: Solve equationx 3 C7x

2

C24x C30x 2 C5x C13 D

2x 3 C11x2

C36x C452x 2 C7x C20

.

3: (RUSMO/2005) It is known that there is such a number s such that if realnumbers a; b; c ;d are all neither 0 nor 1, satisfying a Cb Cc Cd Ds and1a C 1b C1c C 1d Ds , then 11 a C 11 b C 11 c C 11 d Ds . Find s .

4: (VIETNAM/2007) Solve the system of equations

1 12y C3x D

2p x ; and 1 C

12y C3x D

6p y :

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5: (BELARUS/2005) Find all triples .x; y; z/ with x; y; z 2.0; 1/ satisfying

x C 12x 1 y C

12y 1 z C

12z 1

D 1 xy

z1

yzx

1 zx

y:

6: (NORTH-EUROPEAN/2006) Given that x; y; z are real numbers which arenot all equal, satisfying

x

C 1

y Dy

C 1

z Dz

C 1

x Dk;

where k is a real number. Find all possible values of k .

7: (GREECE/TST/2009) Find all real solutions .x;y;z/ of equation

.x C2/2

y Cz 2 C.y C4/

2

z Cx 4 C.z C6/

2

x Cy 6 D36;

giving x; y;z > 3 .

8: (BULGARIA/2004) Given the system of equations

8

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