Lesson Solving the system of algebraic equations of degree 2 and degree 1

The most general form of the polynomial equation of degree 2 with two unknowns x and y is
, (1)
where coefficients a, b, c, d , e and f are given numbers.

In this lesson we will consider the solution procedure for the system of two equations with two unknowns.
One equation is of the form (1), while the other is the linear equation
                      (2)
with given coefficients g, h and i.

To solve the system (1), (2) you have to find values of x and y that satisfy to both equations simultaneously.

The procedure of solution is as follows:
1) express one of the variable x or y from equation (2) via the other variable;
2) substitute this expression to the quadratic form (1) and collect common terms to reduce it to the standard quadratic equation form of one variable;
3) solve the obtained quadratic equation and get the value of the selected variable;
4) substitute the found value of the variable to equation (2) and get the value of the second variable.

Please note that methods of solution of the standard quadratic equations are described in this site in lesson 1, lesson 2 and lesson 3.

Examples below help you understand how this procedure works.

Example 1.
Solve the system of two equations
               (3)
                          (4)
Equation (3) represents the circle of the radius 2 with the center at point (3,1). This circle is shown in Figure 1 below by the red line.
Equation (4) represents straight line shown in Figure 1 in blue color.


Figure 1

Let's express x from equation (4):
                             (5)
Substitute this to equation (3):
.
Collecting of common terms leads to
.                   (6)
Apply the quadratic formula
.
So, two roots of equation (6) are y=1 and y=2.6.
Substituting these values to (5) gives two solutions of the system (3), (4).
First solution is x=1, y=1, and the second one is x=4.2, y=2.6.
Note that these solutions represent exactly intersection points of the circle and the straight line.

Example 2.
Solve the system of two equations
                (7)
                    (8)
Equation (7) represents the same circle of the radius 2 with the center at point (3,1). This circle is shown in Figure 2 below by the red line.
Equation (8) represents straight line shown in Figure 2 in blue color.
Note that the straight line in this example is tangent to the circle.


Figure 2

Let's express x from equation (8)
                      (9)
Substitute this to equation (7):
.
Collecting of common terms leads to
. (10)
The discriminant of the quadratic equation (10) is equal to
.
So, equation (10) has only one root y=(5+4*sqrt(5)/5.
Substituting this values to (9) gives the solution of the system (7), (8):
x=(15-2sqrt(5))/5, y=(5+4sqrt(5))/4.
Note that this solution represents exactly the tangent point of the circle and the straight line.

Example 3.
Solve the system of two equations
                (11)
                           (12)
Equation (11) represents the same circle of the radius 2 with the center at point (3,1). This circle is shown in Figure 3 below by the red line.
Equation (12) represents straight line shown in Figure 3 in blue color.


Figure 3

Let's express x from equation (12)
                             (13)
Substitute this to equation (11):
.
Collecting of common terms leads to
.                   (14)
The discriminant of the quadratic equation (14) is equal to
.
Because the discriminant is negative, equation (14) has no root in real numbers.
Hence, the system (11), (12) has no solution in real numbers.
Note that in this example the circle and the straight line have no intersection.

Example 4.
Solve the system of two equations
                        (15)
                         (16)
Equation (15) represents the parabola. This parabola is shown in Figure 4 below by the red line.
Equation (16) represents straight line shown in Figure 4 in green color.


Figure 4

Let's express x from equation (16)
                             (17)
Substitute this to equation (15):
.
Collecting of common terms leads to
.                        (18)
Apply the quadratic formula
.
So, two roots of equation (18) are y=(1+sqrt(33))/8 and y=(1-sqrt(33))/8.
Substituting these values to (17) gives two solutions of the system (15), (16).
First solution is x=(9+sqrt(33))/4, y=(1+sqrt(33))/8, and the second one is x=(9-sqrt(33))/4, y=(1-sqrt(33))/8.
Note that these solutions represent exactly intersection points of the parabola and the straight line.

Example 5.
Solve the system of two equations
                                  (19)
                         (20)
Equation (19) represents the hyperbola y=1/x. This hyperbola is shown in Figure 5 below by the red line.
Equation (20) represents straight line shown in Figure 5 in green color.


Figure 5

Let's express x from equation (20)
                             (21)
Substitute this to equation (19):
.
Collecting of common terms leads to
.                      (22)
Apply the quadratic formula
.
So, two roots of equation (22) are y=(-1+sqrt(3))/2 and y=(-1-sqrt(3))/2.
Substituting these values to (21) gives two solutions of the system (19), (20).
First solution is x=1+sqrt(3), y=(-1+sqrt(3))/2, and the second one is x=1-sqrt(3), y=(-1-sqrt(3))/2.
Note that these solutions represent exactly intersection points of the hyperbola and the straight line.