SIMULTANEOUS EQUATIONS
In lower forms, we have learnt the way to solve a system of two simultaneous linear equations
in two unknowns. In this chapter, we are going to learn ways to solve a system of one linear
equation and one quadratic equation in two unknowns and have a look at some of its applications.
1. Method of Substitution
Similar to a system of two simultaneous linear equations in two unknowns, we have an
analogous method of substitution to solve a system of one linear equation and one quadratic
equation in two unknowns. We illustrate this method by means of some examples.
Example 1.1.
Solve the following system of equations.
.......... 1
 yx

2
 y  x  4 x  2 ..........  2 
Solution.
Substituting (1) into (2), we get
x  x2  4 x  2
x 2  3x  2  0
 x  2  x  1  0
and so x  2 or x  1. In order to find the corresponding values for y, we substitute these values of
x into (1) to get y  2 when x  2 and y  1 when x  1. Equivalently, we may write
(x, y)  (2, 2) or (1, 1).
Example 1.2.
Solve the following system of equations.
.......... 1
 2 y  5x  9

2
 y  x  2 x  12 ..........  2 
Solution.
From (1), we have
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y
5
9
x  ..........  3 .
2
2
Substituting (3) into (2), we get
5
9
x   x 2  2 x  12
2
2
5 x  9  2 x 2  4 x  24
2 x 2  x  15  0
 2 x  5 x  3  0
and so x   52 or x  3. Substituting these values of x into (3), we get (x, y) 
  52 ,  434 
or (3, 3).
Example 1.3.
Solve the following system of equations.
.......... 1
y  x2

 2
2
 x  y  3x  2 y  43 ..........  2 
Solution.
Substituting (1) into (2), we get
x 2   x  2   3 x  2  x  2   43
2
x 2  x 2  4 x  4  3 x  2 x  4  43  0
2 x 2  3 x  35  0
 2 x  7  x  5   0
and so x  72 or x  5. Substituting these values of x into (1), we get (x, y) 
 72 , 112 
or (5, 3).
Exercise
1.
Solve the following system of equations.
 2x  y  2  0

2
2
 x y 4
2.
Solve the following system of equations.
 2 x  3 y  32

1
 4 3
 x  y  114

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2. Graphical Method
We may solve the system of equations graphically. We are going to show this by redoing the
above examples.
Example 2.1.
Solve the following system of equations.
.......... 1
 yx

2
 y  x  4 x  2 ..........  2 
Solution.
The graph of equations (1) and (2) is shown in Figure 1.
y
2
y x 2 4x 2
-4
-3
1
-2
x
-1
-1
-2
-3
y x
-4
Figure 1: The graph of y  x  4 x  2 and y  x.
2
Note that the coordinates of the points of intersection of the two lines are the required solutions
of the system. From Figure 1, the points of intersection are (1, 1) and (2, 2).
Example 2.2.
Solve the following system of equations.
.......... 1
 2 y  5x  9

2
 y  x  2 x  12 ..........  2 
Solution.
The graph of equations (1) and (2) are shown in Figure 2.
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y
y x 2 2x 12
10
5
2y 5x 9
-4
-2
2
4
x
-5
-10
-15
Figure 2: The graph of y  x  2 x  12 and 2y  5x – 9.
2
From Figure 2, the points of intersection occurs at about x  3 and x  2.5 and hence the
points of intersection is (3, 3) and (2.5, 10.75).
Example 2.3.
Solve the following system of equations.
.......... 1
y  x2

 2
2
 x  y  3x  2 y  43 ..........  2 

Note that (2) can be written as
x 2  y 2  3x  2 y  43
2
2
 2
3 
3
2
 x  3x       y  2 y  1     1  43
 2  
2

2
3
185
2

..........(3)
 x     y  1 
2
4

Any point lying on (3) is at a distance of
centered at
 32 ,  1
with radius
185
4
185
4
from the point
 32 ,  1 . Hence (3) is a circle
.
Solution.
The graph of equations (1) and (2) are shown in Figure 3.
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y
7.5
y x 2
5
2.5
x
-6
-4
-2
2
4
6
8
-2.5
x 2 y 2 3x 2y 43
-5
-7.5
Figure 3: The graph of x  y  3x  2 y  43 and y  x  2 .
2
2
From Figure 3, the answers are (3.5, 5.5) and (5, 3).
3. Applications in Practical Problems
The system of one linear and one quadratic equation is quite useful when solving practical
problems. We will illustrate its use through some examples.
Example 3.1.
The area and perimeter of a rectangle are respectively 216 cm2 and 60 cm. Find its length and
width.
Solution.
Let x cm and y cm be its length and width respectively. Then
xy  216 .......... 1

.

 2  x  y   60 ..........  2 
Since the width of a rectangle cannot be zero, by (1),
216
x
..........  3 .
y
Substituting (3) into (2), we get
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 216

2
 y   60
 y

2
216  y  30 y
2
y  30 y  216  0
 y  12  y  18  0
So y  12 cm or y  18 cm. Note that the second answer is rejected since y  18 gives x 
216
18
 12 ,
which contradicts to the assumption that x is the length and y is the width. Now when y  12,
. Hence the length of the rectangle is 18 cm and the width of it is 12 cm.
x  216
12  18
Example 3.2.
During free fall, the distance fallen (s, in metres) of an object and the time during which the object
has been falling (t, in seconds) have the following relation
s  5t 2 .
At a certain instant, 3s  56 – 11t. Find the distance fallen at that instant.
Solution.
To find the distance fallen, it is equivalent to solve for s in the system
2
.......... 1
 s  5t
.

 3s  56  11t ..........  2 
Substituting (1) into (2), we get
3  5t 2   56  11t
15t 2  11t  56  0
 3t  7  5t  8   0
So t  85 sec or t   73 sec (rejected as t  0 ). Put t  85 into (1), we get s  5  85  
2
64
5
m.
Example 3.3.
A rectangular garden of area 616 m2 is surrounded by a rectangular path of width 1 m. If the area
of the path is 104 m2 , find the dimension of the garden.
Solution.
Let the dimension of the rectangle be x m by y m. Then we have xy  616.
On the other hand, (x + 2)(y + 2) – 616  104.
Hence we have to solve the following system
xy  616 ..........1

.

  x  2  y  2   616  104 .......... 2
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From (2), we have
xy  2 x  2 y  4  616  104
.
xy  2 x  2 y  716 ..........  3
By substituting (1) into (3), we have
616  2 x  2 y  716
x  y  50
.
y  50  x ..........  4 
By substituting (4) into (1), we have
x  50  x   616
x  50 x  616  0
2
 x  22  x  28  0
So x  22 m or x  28 m. Now when x  22, y  28 and when x  28, y  22. So both values of x
gives the same dimension. Hence the dimension of the garden is 22 m by 28 m.
Exercise
1. An isosceles triangle has perimeter 72 cm and base angle 60°. Find the length of its base and its
area (in surd form).
2. A plate of area 369 cm2 is cut and bent to form two spheres. The sum of the diameters of the
two spheres is 27 cm. Find the volume of the smaller sphere.
3. During free fall, an object is dropped freely from certain height. At any instant, the distance
fallen (s, in metre) and the instantaneous speed (v, in metre per second) is given by the
following relation
v 2  20 s .
When the object reaches the ground, its distance fallen is 125% more than its speed. Find the
height of the point when the object is released.
4. Let C be a unit circle on the Cartesian coordinate plane centered at the origin. Then the equation
of C is x 2  y 2  1. Let A be the point (4, 0). Find the coordinates of the point(s) on the circle
such that the point(s) is/are at a distance of 4 from A.
5. A two-digit number is given. If the sum of the digits is 14 and the sum of square of the digits is
32 larger than the number, find the number.
6. A hollow cone of radius r cm and height h cm is cut straightly along its slant height and
unfolded to a sector. Express the area A of the sector in terms of  r and h. Furthermore, given
that the square of h is larger than 14 times of r by 49 and A is 228 cm2 , find the volume of the
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cone in terms of  in surd form.
4. Solution to Selected Exercise
Method of Substitution
1. (x, y)  (0, 2) or
  85 ,  65  .
2. (x, y)  (7, 6) or
224
.
 128
15 , 45 
Applications in Practical Problems
1. Let x cm be half the length of the base and y cm be the length of the slant. So the perimeter of
the isosceles triangle is 2x + 2y  72. On the other hand,
x
y
 cos 60  12 . So we need to solve
the system of equations given below.
 2 x  2 y  72

x 1



y 2

The answer is (x, y)  (12, 24) and so the height is
is
 2 12 12
242  122  432  12 3 cm . So the area
3  2  144 3 cm2 .
2. Let r and s be the radii of the two spheres. Then 2(r + s)  27. On the other hand, since the
surface areas for the spheres are 4 r 2 and 4 s 2 , we have
4 r 2  4 s 2  369 .
Hence we need to solve the system of equations below.

2  r  s   27

2
2
 4 r  4 s  369
So we know that the radius of the smaller sphere is 6 cm and so the volume of the smaller
sphere is
4
3
 63  288 cm3 .
3. When the object reaches the ground, s  (1+125%)v  2.25v. Hence we need to solve for s in the
following system of equations.
 v 2  20s

 s  2.25v
So we found that s  2.25  45  101.25 m or
405
4
m.
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4. Let Q  (x, y) be the coordinates of the required point. Since Q lies on C, we have
x2  y 2  1.......... 1 .
In addition, QA  4. So we have, after simplification, x 2  y 2  8x  0 . Substituting (1) into this,
we have 1 – 8x  0 i.e. x  18 and so we have y  

1
8
,
3 7
8

and

1
8
63
8
  3 87 . Hence the required points are

,  3 87 .
5. Let x be the tens digit and y be the unit digit. Then the number is 10x + y. So we have x + y  14
and x2  y 2  32  10 x  y  . By solving, we get x  6 or x  12.5. But the later answer is
rejected as x should be an integer which satisfies 0  x  9 . So y  8 and so the number is 68.
6. A  the curved surface area of the cone   rl   r r 2  h2 .
Next, h 2  14r  49 and A  228. So by solving (note that r 2  14r  49   r  7  ), we have
2
r  12 cm or r  19 cm. But the later answer is rejected as r  0. So
h  14 12  49  217 cm and so the volume of the cone is 48 217 cm3 .
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