50 AMC LECTURES

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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
1. SYSTEM OF LINEAR EQUATIONS

A group (two or more) of equations is called a system of equations. The solutions of a
system of equations should satisfy all the equations.

A system of two equations is as follows:
axby c (1)
1 1 1
a xb y c (2)
2 2 2
a b
Case I: When 1  1 , the system of equations has one unique solution.
a b
2 2
The followings show three methods to get the solutions:

Method 1 (Substitution method):

First solve either equation for one variable. Let’s solve (2) for x:
c b y
x 2 2 (3)
a
2
Now substitute the result for x in equation (1):
c b y
a  2 2 b y c
1 a 1 1
2
To solve for y, first multiply both sides of the equation by a to eliminate the
2
denominator:

a (c b y)a byca  ac a b yba yca
1 2 2 2 1 1 2 1 2 1 2 1 2 1 2
a b ya byac ca
1 2 2 1 1 2 1 2
ac ca
y(a b a b)ac ca  y  1 2 1 2 (4)
1 2 2 1 1 2 1 2 a b a b
1 2 2 1
Find x by substituting (4) in equation (3):

a c c a 
c b  1 2 1 2 
c b y 2 2 a b a b  c b c b
x 2 2  1 2 2 1  x 1 2 2 1
a a a b a b
2 2 1 2 2 1

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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
Method 2 (Elimination method)

We eliminate x and solve for y by first multiplying both sides of equation (2) by b and
1
both sides of equation (1) by b
2
ab xbb ycb (5)
1 2 1 2 1 2
a b xbb y c b (6)
2 2 1 2 2 1
(5) – (6):
cb c b
x(ab a b )cb c b  x 1 2 2 1 (7)
1 2 2 2 1 2 2 1 ab a b
1 2 2 21
ac ca
Then we get y: y  1 2 1 2
a b a b
1 2 2 1

Method 3 (Cramer’s rule):

a b
Case I: When 1  1 , the system of equations has one solution.
a b
2 2
 D
x  x

 D

D
y  y

 D
Where:
a b
D 1 1 a b b a
a b 1 2 1 2
2 2
c b
D  1 1 c b b c
x c b 1 2 1 2
2 2
a c
D  1 1 a c c a
y a c 1 2 1 2
2 2
a b c
Case II: When 1  1  1 , the system of equations has infinite many solutions.
a b c
2 2 2
a b c
Case III: When 1  1  1 , the system of equations has no solution.
a b c
2 2 2

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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
Example 1: (1954 AMC) A and B together can do a job in 2 days; B and C can do it in
2
four days; and A and C 2 days. The number of days required for A to do the job alone
5
is
(A) 1 (B) 3 (C) 6 (D) 12 (E) 2.8

Solution: (B).
Let x, y, z be the number of days it would take A, B, C to complete the job alone. We have
1 1 1
  (1)
x y 2
1 1 1
  (2)
y z 4
1 1 5
  (3)
z x 12
1 1 1 14 1 1 1 7
(1) + (2) + (3): 2(   )      (4)
x y z 12 x y z 12
1 4
(4) – (2):   x3
x 12

Example 2: Solve:
x2(y zu)19
(1)
y2(zu x)18
(2)
z2(u x y)17
(3)
u2(x y z)16
(4)

Solution: (x, y, z, u): (1, 2, 3, 4).

(1) + (2) + (3) + (4): 7(x yzu)70 x yzu 10
 2(x yzu)20 (5)
(5) – (1): x = 1
(5) – (2): y = 2
(5) – (3): z = 3
(5) – (4): u = 4.
The solutions are (x, y, z, u): (1, 2, 3, 4).
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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
Example 3: (1969 AMC) The number of points common to the graphs of
(x – y + 2)(3x + y – 4) = 0 and (x + y – 2)(2x – 5y + 7) = 0 is.
(A) 2 (B) 4 (C) 6 (D) 16 (E) infinite

Solution: (B).

Each of the graphs consists of a pair of nonparallel straight lines the first pair having
equations
I: x – y + 2 = 0 and II: (3x + y – 4) = 0
and the second pair having equations
III: x + y – 2 = 0 and IV: 2x – 5y + 7 = 0.
The intersections of line III with I and II are the two distinct points (0, 2) and (1, 1), and
13 29
those of line IV with I and II are two more distinct points (1, 1) and ( , ), giving a
17 17
total of four distinct points common to the graphs of the two given equations.

Example 4: (1986 AIME) Determine 3x 2x if x , x , x , x and x satisfy the system
4 5 1 2 3 4 5
of equations given below:
2x  x  x  x  x 6
1 2 3 4 5
x 2x  x  x  x 12
1 2 3 4 5
x  x 2x  x  x  24
1 2 3 4 5
x  x x 2x  x  48
1 2 3 4 5
x  x x  x 2x 96
1 2 3 4 5

Solution: 181.

Adding the five equations, then dividing both sides of the resulting equation by 6 we get:
x x x x x 31.
1 2 3 4 5
Subtracting the above equation from the given fourth and fifth equations, we get that x =
4
17 and x = 65. Thus 3x 2x 51130181.
5
3 5

2. SYSTEM OF NONLINEAR EQUATIONS

A system of equations is called nonlinear system of equations if at least one equation is
nonlinear.
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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
2.1. Solving The System Of Quadratic Equations With Two Unknowns

Type I:
Ax2 B xyC y2 DxE yF 0,
Ax2 B xyC y2 D xE yF 0,  1 1 1 1 1 1
 1 1 1 1 1 1  B yC
A2xB2yC2 0. x 2 A 2 .

2

 B yC B yC B yC
A( 2 2)2 B 2 2 yC y2 D 2 2 E yF 0,
 1 A 1 A 1 1 A 1 1
 2 2 2
B yC
x 2 2 .
 A

2

Example 5: (1971 AMC) If the line y = mx + 1 intersects the ellips x2 + 4y2 = 1 exactly
once, what is the value of m2?

Solution: 3/4.

y = mx + 1 (1)
x2 + 4y2 = 1 (2)

Substitute (1) into (2):
x2 + 4(mx + 1)2 = 1  x2 + 4(m2x2 +2 mx + 1) = 1  x2 + 4m2x2 +8mx + 4 = 1
 x2 (4m2 + 1) + 8mx + 3 = 0 (3)

Since these two lines intersect only once, (3) should have double roots, meaning that the
discriminant equals 0.
 = 0  (8m)2 – 4  (4m2 + 1)  3 = 0  64m2 – 48m2 – 12 = 0
 16m2 – 12 = 0  m2 = 12/16 = 3/4.

Example 6: Solve x and y for real numbers:
x2 y2
  1, (1)
16 9

3x4y 12. (2)

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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
Solution:
3
Rewrite (1) asy 3 x. Substituting it into (2) we get: x2 – 4x = 0.
4
 x = 0, x = 4.
1 2
And we get y = 3, y = 0.
1 2
So the solutions are: (0, 3), (4, 0).

Type II:

Ax2 BxyC y2 DxE yF 0,
 1 1 1 1 1 1
Ax2 B xyC y2 D xE yF 0.
2 2 2 2 2 2

Example 7: (1966 AMC) What is the number of distinct points common to the curve x2
+ 4y2 = 1 and 4x2 + y2 = 4?

Solution: 2.

x2 + 4y2 = 1 (1)
4x2 + y2 = 4 (2)
(1) + (2):
5x2 + 5y2 = 5  x2 + y2 = 1 (3)
(1) – (3): 3y2 = 0
So equation (1) becomes: x2 = 1  x = 1.
There are 2 points common: (1, 0) and (– 1, 0).

Example 8: (1991 AIME) Find x2 + y2 if x and y are positive integers such that
xy + x + y = 71 and x2 y+xy2 = 880

Solution:

Let m = x + y and n = xy.
Then we have
m + n = 71 (1)
mn = 880 (2)
Solving we get m = 16 and n = 55.
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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
So we have
x + y = 55 and x y = 16 (this system has no integer solutions).
or x + y = 16 and xy = 55  x = 5 and y = 11.
Therefore x2 + y2 = 52 + 112 = 146

Example 9: Solve x and y for real numbers:
6x2 3xy2y2 13y70, (1)

2x2 xy y2 6y10. (2)

Solution:

(1) – (2)  3: 5y2 – 5y – 10 = 0  (y – 2)(y + 1) = 0
Solving we get: y = 2, y = 1.
1 2
Substituting y = 2 into (2): 2x2 – 2x + 9 = 0.
1
This equation has no real solutions.
3
Substituting y = 1 into (2): 2x2 + x – 6 = 0. Solving we get: x 2, x  .
2 1 2 2
 3
x 2, x  ,
The system of equations has the solutions:  1  2 2
y1 1; y2 1.

Example 10: Solve x and y for real numbers:
3x2 7xy3y2 4x4y140,

5x2 13xy5y2 12x12y60.

Solution:

Let u = x + y, v = xy.
The given system of equations becomes:
3u2 v4u140, (1)

5u2 3v12u60. (2)
(1)  3  (2): 4u2 – 36 = 0.
 u = 3 or 3.
Then v = 1 or 25.
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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
Thus we get
x y 3 x y 3
 and 
xy 1 xy 25
So we only need to solve two equations:
t2 – 3t – 1 = 0 and t2 + 3t – 25 = 0
The solutions are:
 3 13  3 13  3 109  3 109
x  , x  , x  , x  ,
 1 2  2 2  3 2  4 2
   
 3 13  3 13  3 109  3 109
y  ; y  ; y  ; y  .
 1 2  2 2  3 2  4 2

2.2. Method To Solve The System Of Equations Degree Other Than 2

Example 11: Solve x and y for real numbers:
x4 x2y2  y4 91,

x2 xy y2 7.

Solution:

Let u = x + y, v = xy.
x4  y4 [(x y)2 2xy]2 2x2y2 (u2 2v)2 2v2,
x2  y2 (x y)2 2xy u2 2v,
The given system of equations can be written as
(u2 2v)2 2v2 v2 91,

u2 3v7.
Solving we get v = 3 and u =  4.
Then we have
x y 4 x y 4
 and 
xy 3 xy 3
The solutions are
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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
x 1, x 3, x 1, x 3,
 1  2  3  4
y 3; y 1; y 3; y 1.
   
1 2 3 4

Example 12: Solve x, y, and z for real numbers:
xyx y 1,

yz yz 5,

zxzx2.


Solution:

Add 1 to both sides of each equation we get:
(x1)(y1)2, (1)

(y1)(z1)6, (2)

(z1)(x1)3. (3)

Multiplying all of them together: [(x + 1)(y +1)(z + 1)]2 = 36,
 (x + 1)(y +1)(z + 1) =  6 (4)

Dividing (4) by (1), (2), (3) respectively:
z + 1 =  3, x + 1 =  1, y + 1 =  2.
The solutions are
x 0, x 2,
1 2
 
y 1, y 3,
1 2
 
z 2; z 4.
 
1 2

2.3. Solving System Of Radical Equations
x y xy 14,
Example 13: Solve for real x and y 
x2  y2 xy 84.

Solution:

Let u = x + y and v xy .
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50 AMC LECTURES (39) SYSTEMS OF EQUATIONS
uv14 uv14 u 10,
    
u2 v2 84 u2 v2 84 v4.
x y 10,
Therefore 
xy 16.
x 2, x 8,
Solving for x and y:  1  2
y 8; y 2.
 
1 2

Example 14: Solve for real x and y:

 x2  y2  x2  y2  y, (1)

x4  y4 144. (2)

Solution:

Squaring both sides of (1):
x2  y2 2 x4  y4 x2  y2  y2 2x2 2 x4 y4  y2 (3)
Substituting (2) into (3): y2 = 2x2 – 24 (4)
Substituting (4) into (2): x4 – 32x2 + 240 = 0
Solving we get: x2 5 orx2 3.
From (1) we know that y  0.
Substituting the values of x into (2), we get the solutions to the given system of
equations:
x2 5 x2 3,
 or  .
y 4; y 0.
We checked and know that all of them are the solutions:
{(2 5,4),(2 5,4),(2 3,0),(2 3,0)}.

2.4. Solving The System Of Rational Equations

Example 15: Alex's age is two times of the sum of Bob and Catherine. Alex's age was
four times of Bob's age two years ago. The ratio of Alex's age and Catherine's age will be
8 : 3 ten years from now. How old is Alex?
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