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Linear
Systems
School year 2014/2015
Systems
A system consists of a set of equations, for
which we ask what are the common solutions.
ax+by=c
a’x+b’y=c’
System Degree
It’s the product of the higher degrees of the
individual equations that make up the system.
EXAMPLE:
4x-y²=3
x5-y=10
2 * 5 = gr.10
So a linear system is of first degree.
How do you solve linear systems?
- Substitution method
- Comparison method
- Reduction method
- Cramer’s method
Substitution method
Derive a variable by an equation and replace it
in the other equation.
ESEMPIO:
2x-5y=7
x-3y=1
x=3y+1
2(3y+1)-5y=7
x=3y+1
6y+2-5y=7
x=16
y=5
Comparison method
Derive the same variable
from both equations
a’x + b’y = c’
canonical
form
ax + by = c
x - y = 3
x + y = 9
x = y + 3
x = -y + 9
x = y + 3
y + 3 = -y + 9
x = y + 3
2y = 6
x = 6
y = 3
Example:
Reduction method
Reduction Method
It’s to add or subtract
the corresponding terms of the two equations
to obtain an equation with only one unknow.
EXAMPLE: PROCEDURE:
1. Multiplie one or both of the equations for factors
non-zero, so that the coefficients of one of the
variables are equal to or opposite.
2. If the coefficients obtained in step 1 are equal,
subtract member to member the two equations; if
the coefficients are opposite, add member to
member; so we get an equation in one unknown.
3. Solve the equation in a single variable.
4. Replace the solution in one of the two original
equations.
Cramer’s method
The system must be written in canonical form:
ax+by=c
a'x+b' y=c'
Determinant calculation (D)
+
-
= +ab'-a'b
Multiply +a with b' and -a' with b
a b
D=
a' b'
Dx calculation
+
-
= +cb'-bc'
Multiply +c with b' and -c' with b
c b
Dx=
c' b'
-
Dy calculation
+
= +ac'-a'c
Multiply +a with c' and -a' with c
a c
Dy=
a' c'
If D ≠ 0 the
system is
determined
x=Dx/D
y=Dy/D
If D = 0
-the system is
indeterminate with
Dx and Dy=0
-the system is
impossible with Dx
and Dy≠0
Literal Systems
The literal systems are those where in
addition to the variables there are other
letters (parameters).
Example
Transform the system in canonical form.
2x= 2a-y 2x+y=2a
(a+1)x+ay=2a (a+1)x+ay=2a
For literal system the most used method is Cramer, calculating the determinant D, Dx, Dy.
D= = 2a-(a+1) = 2a-a-1= a-1 Dx= = 2a2 -2a=2a(a-1)
Dy= = 4a -2a(a+1)=4a-2a2 -2a=2a-2a2 =2a(1-a)
2 1
a+1 a
2a 1
2a a
2 2a
a+1 2a
...continuous example
The system is determined if D ≠ 0 ie if a-1≠0 a≠1.
● If a≠1 then
● If a=1 then D=0, Dx=0 and Dy=0 and the system is indeterminated.
x= Dx/D= 2a(a-1)/a-1= 2a
y=Dy/D= 2a(a-1)/a-1= -2a(a-1)/a-1= -2a
x= 2a
y= -2a
Linear Fractional Systems
When a system is fractional?
● Are those systems in which at least one of the equations that compose
it appears the unknown of first degree (x; y) in the denominator.
● Is solved with the methods we have already seen. (eg. the
replacement method; method of comparison; reduction method etc ...),
but it should be the
EXISTENCE CONDITION (E. C.)
steps shall be non-zero
all denominators that contain the unknown
Example
Found:
● l.c.m= 2xy
● E.C.: x≠0 U y ≠0
Transform the system in canonical form...
…and choose the most appropriate method to solve it.
Result:
Check if the solution of the system
satisfies the E.C.
Sistems with 3 equations and 3 variables
CANONICAL FORM
Solve operations in brackets
Order and simplify the terms like putting the system in CANONICAL FORM
CANONICAL FORM
Find the value of y will go out and replaced in the other two equations using the
method of substitution
So the coefficient of y of the first equation is equal to 1, derive the value of y
Solve in order to remove the brackets
Order and simplify the similar terms in the equations in which the value replaced
The first equation must be simplified for 5
Since the coefficient of z of the first equation is equal to 1, derive from it the
value of z
Found the value of z and replace in the other two equations using
the substitution method
Solve operations in brackets
Order and simplify opposite terms
Find the value of x in the first equation and substitute in last
The system solution is given by the triplet (4; 0 ; 5 ) that simultaneously
solves all of the system equations. The system is therefore DETERMINED .
WORK MADE BY THE CLASS 2nd A Afm
OF ITCG “CORINALDESI” - SENIGALLIA (AN) - ITALY
Team 1 - Breccia Martina, Franceschetti Sofia, Pinca Julia Andrea
Team 2 - Valentini Alessia, Esposto Giorgia, Biagetti Elena, Montironi Ilaria, Carletti
Lucia
Team 3 - Fabri Luca, Franceschini Simone, Bernardini Alessio, Urbinelli Riccardo,
Saramuzzi Mirko
Team 4 - Rossi Davide, Ventura Devid, Latini Angelo
Team 5 - Cervasi Michela, Casella Federica, Raccuja Ilaria
Team 6 - Zhang Qiuye , Zhang Ting, Xie Sandro
Team 7 - Trionfetti Sara, Carbonari Gloria, Borgacci Francesca, Avaltroni Alessia

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Linear Systems

  • 2. Systems A system consists of a set of equations, for which we ask what are the common solutions. ax+by=c a’x+b’y=c’
  • 3. System Degree It’s the product of the higher degrees of the individual equations that make up the system. EXAMPLE: 4x-y²=3 x5-y=10 2 * 5 = gr.10
  • 4. So a linear system is of first degree.
  • 5. How do you solve linear systems? - Substitution method - Comparison method - Reduction method - Cramer’s method
  • 7. Derive a variable by an equation and replace it in the other equation. ESEMPIO: 2x-5y=7 x-3y=1 x=3y+1 2(3y+1)-5y=7 x=3y+1 6y+2-5y=7 x=16 y=5
  • 9. Derive the same variable from both equations a’x + b’y = c’ canonical form ax + by = c
  • 10. x - y = 3 x + y = 9 x = y + 3 x = -y + 9 x = y + 3 y + 3 = -y + 9 x = y + 3 2y = 6 x = 6 y = 3 Example:
  • 12. Reduction Method It’s to add or subtract the corresponding terms of the two equations to obtain an equation with only one unknow.
  • 13. EXAMPLE: PROCEDURE: 1. Multiplie one or both of the equations for factors non-zero, so that the coefficients of one of the variables are equal to or opposite. 2. If the coefficients obtained in step 1 are equal, subtract member to member the two equations; if the coefficients are opposite, add member to member; so we get an equation in one unknown. 3. Solve the equation in a single variable. 4. Replace the solution in one of the two original equations.
  • 15. The system must be written in canonical form: ax+by=c a'x+b' y=c'
  • 16. Determinant calculation (D) + - = +ab'-a'b Multiply +a with b' and -a' with b a b D= a' b'
  • 17. Dx calculation + - = +cb'-bc' Multiply +c with b' and -c' with b c b Dx= c' b'
  • 18. - Dy calculation + = +ac'-a'c Multiply +a with c' and -a' with c a c Dy= a' c'
  • 19. If D ≠ 0 the system is determined x=Dx/D y=Dy/D If D = 0 -the system is indeterminate with Dx and Dy=0 -the system is impossible with Dx and Dy≠0
  • 21. The literal systems are those where in addition to the variables there are other letters (parameters).
  • 22. Example Transform the system in canonical form. 2x= 2a-y 2x+y=2a (a+1)x+ay=2a (a+1)x+ay=2a For literal system the most used method is Cramer, calculating the determinant D, Dx, Dy. D= = 2a-(a+1) = 2a-a-1= a-1 Dx= = 2a2 -2a=2a(a-1) Dy= = 4a -2a(a+1)=4a-2a2 -2a=2a-2a2 =2a(1-a) 2 1 a+1 a 2a 1 2a a 2 2a a+1 2a
  • 23. ...continuous example The system is determined if D ≠ 0 ie if a-1≠0 a≠1. ● If a≠1 then ● If a=1 then D=0, Dx=0 and Dy=0 and the system is indeterminated. x= Dx/D= 2a(a-1)/a-1= 2a y=Dy/D= 2a(a-1)/a-1= -2a(a-1)/a-1= -2a x= 2a y= -2a
  • 25. When a system is fractional? ● Are those systems in which at least one of the equations that compose it appears the unknown of first degree (x; y) in the denominator. ● Is solved with the methods we have already seen. (eg. the replacement method; method of comparison; reduction method etc ...), but it should be the EXISTENCE CONDITION (E. C.) steps shall be non-zero all denominators that contain the unknown
  • 27. Found: ● l.c.m= 2xy ● E.C.: x≠0 U y ≠0
  • 28. Transform the system in canonical form... …and choose the most appropriate method to solve it.
  • 29.
  • 31. Check if the solution of the system satisfies the E.C.
  • 32. Sistems with 3 equations and 3 variables
  • 35. Order and simplify the terms like putting the system in CANONICAL FORM
  • 37. Find the value of y will go out and replaced in the other two equations using the method of substitution So the coefficient of y of the first equation is equal to 1, derive the value of y
  • 38. Solve in order to remove the brackets
  • 39. Order and simplify the similar terms in the equations in which the value replaced
  • 40. The first equation must be simplified for 5
  • 41. Since the coefficient of z of the first equation is equal to 1, derive from it the value of z
  • 42. Found the value of z and replace in the other two equations using the substitution method
  • 44. Order and simplify opposite terms
  • 45. Find the value of x in the first equation and substitute in last
  • 46. The system solution is given by the triplet (4; 0 ; 5 ) that simultaneously solves all of the system equations. The system is therefore DETERMINED .
  • 47. WORK MADE BY THE CLASS 2nd A Afm OF ITCG “CORINALDESI” - SENIGALLIA (AN) - ITALY Team 1 - Breccia Martina, Franceschetti Sofia, Pinca Julia Andrea Team 2 - Valentini Alessia, Esposto Giorgia, Biagetti Elena, Montironi Ilaria, Carletti Lucia Team 3 - Fabri Luca, Franceschini Simone, Bernardini Alessio, Urbinelli Riccardo, Saramuzzi Mirko Team 4 - Rossi Davide, Ventura Devid, Latini Angelo Team 5 - Cervasi Michela, Casella Federica, Raccuja Ilaria Team 6 - Zhang Qiuye , Zhang Ting, Xie Sandro Team 7 - Trionfetti Sara, Carbonari Gloria, Borgacci Francesca, Avaltroni Alessia