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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 the 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 the 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 the lesson 1, lesson 2 and lesson 3.
Examples below help you understand how this procedure works.
Example 1Solve the system of two equations
(3)
(4)
Equation (3) represents the circle of the radius 2 with the center at the point (3,1).
This circle is shown in Figure 1 by the red line.
Equation (4) represents the straight line shown in Figure 1 in blue color.
Express x from equation (4):
(5)
Substitute this to equation (3):
.
Collect the common terms:
. (6)

Figure 1. To the Example 1

Apply the quadratic formula
.
So, the two roots of the equation (6) are y=1 and y=2.6.
Substitute these values to (5) to get two solutions of the system (3), (4).
First solution is x=1, y=1. The second solution is x=4.2, y=2.6.
Note that these solutions represent exactly intersection points of the circle and the straight line.
Example 2Solve the system of two equations
(7)
(8)
Equation (7) represents the same circle of the radius 2 with the center at the
point (3,1). This circle is shown in Figure 2 by the red line.
Equation (8) represents the straight line shown in Figure 2 in blue color.
Note that the straight line in this example is tangent to the circle.
Express x from equation (8)
(9)
Substitute this to equation (7):
.
Collect the common terms:
. (10)

Figure 2. To the Example 2

The discriminant of the quadratic equation (10) is equal to
.
So, the equation (10) has only one root .
Substitute this value to (9) to get the solution of the system (7), (8):
, .
Note that this solution represents exactly the tangent point of the circle and the straight line.
Example 3Solve the system of two equations
(11)
(12)
Equation (11) represents the same circle of the radius 2 with the center at the
point (3,1). This circle is shown in Figure 3 by the red line.
Equation (12) represents the straight line shown in Figure 3 in blue color.
Express x from equation (12)
(13)
Substitute this to equation (11):
.
Collect the common terms:
. (14)

Figure 3. To the Example 3

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 4Solve the system of two equations
(15)
(16)
Equation (15) represents the parabola. This parabola is shown in Figure 4 by the
red line.
Equation (16) represents the straight line shown in Figure 4 in green color.
Express x from equation (16)
(17)
Substitute this to equation (15):
.
Collect the common terms:
. (18)

Figure 4. To the Example 4

Apply the quadratic formula
.
So, the two roots of the equation (18) are and .
Substitute these values to (17) to get two solutions of the system (15), (16).
First solution is , . The second solution is , .
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 = .
This hyperbola is shown in Figure 5 by the red line.
Equation (20) represents the straight line shown in Figure 5 in green color.
Express x from equation (20)
(21)
Substitute this to equation (19):
.
Collect the common terms:
. (22)

Figure 5. To the Example 5

Apply (the quadratic formula
.
So, the two roots of equation (22) are and .
Substitute these values to (21) to get two solutions of the system (19), (20).
First solution is , . The second solution is , .
Note that these solutions represent exactly intersection points of the hyperbola and the straight line.
My other lessons in this site on systems of equations that are not linear:
 Solving algebraic equations of high degree and
 Solving the system of algebraic equations of degree 2
under the topic Systems of equations that are not linear of the section AlgebraI.
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