Transcript 9.2 Systems of Linear Equations and Inequalities

ANSWERS
PRE
Lesson 9.2 Systems of Linear Equations & Inequalities
Objectives:
•To solve systems using substitution.
•To solve systems using elimination.
•To put systems into triangular form so they can
be solved with back substitution.
For a system of linear equations in two
variables, exactly one of the following is true.
◦ The system has exactly one solution.
◦ The system has no solution.
◦ The system has infinitely many solutions.
( x, y )
Where the lines intersect
3 ways to solve Linear Systems
1. By Graphing
2. By Substituting
y
10
9
8
7
6
5
4
3
2
1
x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
1
2
3
4
5
6
7
8
9 10
-2
-3
-4
-5
-6
-7
-8
-9
-10
3. Eliminating
 x  2 y  2

3x  y  16
-1st make sure one equation has a variable isolated
In this case, one already is: x = -2y + 2
-2nd Substitute the expression into other equation
Equation 1
x = -2y + 2
Equation 2
3 x + y = 16
-3rd Substitute answer 3(
)+ y = 16
into one of the equations -6y + 6 + y = 16
3x +(-2)
y = 16
-5y + 6 = 16
x=6
-5y = 10
(x
y)
6 ,-2
y = -2
-1st
make sure one
equation has a
variable isolated
-2nd
Substitute the
expression into
other equation
-3rd Substitute answer
into one of the equations
 y  4x  9

y  x 3
y
x
x
3x  9  3
9 9
3x  6
x2
 2, 1
 3x  y  9

+ 3x  2 y  12
3x + y = -9
3x -3 = -9
3x = -6
x = -2
0 - y = 3
-1y = 3
y = -3
(-2,-3)
-
3x + 2y
2 = 26
()
3x + 8 = 26
3x = 18
x=6
0
-6y =-24
-6y =-24
y =4
(6, 4)
Equation 1
3x – 3y = 21
-Multiply Equation 1 by 2
Equation 1
(2)3x – (2)3y = (2)21
6x – 6y = 42
Equation 2
8x + 6y = -14
3 - 3y = 21
3x
()
6 - 3y = 21
-3y = 15
y = -5
+
14x + 0 = 28
14x = 28
x =2
(2, -5)
SYSTEMS OF LINEAR
EQUATIONS
 Here are two examples of systems of linear
equations in three variables.
System of
System in
Linear Equations
Triangular Form
x  2y  z  1
 x  3 y  3z  4
2 x  3 y  z  10
x  2y  z  1
y  2z  5
z 3
-Start with the last equation
x  2y  z 1
& solve upwards.
y  2z  5
y  22(z3) 5 5
z3
yz31
x  2(
2 y1)z(3
1)  1
y x 
12  3  1
z  3 x 1  1
x  2, y  1 , z  3
 2, 1,3
-Start with the last equation
x  5 y  z  22
& solve upwards.
y  4z  8
y  44(z3) 8 8
z 3
yz34
x  5(
)  22
5 y4)z(322
y x20
4 3 1
z  3 x  17  22
x  5, y  4 , z  3
5, 4,3
Classwork: Book: pg. 657; 6-10 all
Classwork: Book: pg. 649; 7-25 odd