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

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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.

### Example 1

Solve 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

.

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 2

Solve 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 3

Solve 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 4

Solve 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

.
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

.
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  Algebra-I.

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