Download Your Title Here - Nutley Public Schools

Review 2.1-2.3
Ex: Check whether the ordered pairs
are solns. of the system.
x-3y= -5
-2x+3y=10
A. (1,4)
1-3(4)= -5
1-12= -5
-11 = -5
*doesn’t work in the 1st
eqn, no need to check
the 2nd.
Not a solution.
B. (-5,0)
-5-3(0)= -5
-5 = -5
-2(-5)+3(0)=10
10=10
Solution
Solving a System Graphically
1. Graph each equation on the same
coordinate plane. (USE GRAPH PAPER!!!)
2. If the lines intersect: The point (ordered
pair) where the lines intersect is the
solution.
3. If the lines do not intersect:
a. They are the same line – infinitely many
solutions (they have every point in common).
b. They are parallel lines – no solution (they
share no common points).
Ex: Solve the system graphically.
2x-2y= -8
2x+2y=4
(-1,3)
Ex: Solve the system graphically.
2x+4y=12
x+2y=6
st
 1 eqn:
x-int (6,0)
y-int (0,3)
 2ND eqn:
x-int (6,0)
y-int (0,3)
 What does this mean?
the 2 eqns are for the
same line!
 ¸ many solutions





Ex: Solve graphically: x-y=5
1st eqn:
2x-2y=9
x-int (5,0)
y-int (0,-5)
2nd eqn:
x-int (9/2,0)
y-int (0,-9/2)
What do you notice
about the lines?
They are parallel! Go
ahead, check the slopes!
No solution!
3-2: Solving Systems of
Equations
using Substitution
Solving Systems of Equations
using Substitution
Steps:
1. Solve one equation for one variable (y= ; x= ; a=)
2. Substitute the expression from step one into the
other equation.
3. Simplify and solve the equation.
4. Substitute back into either original equation to find
the value of the other variable.
5. Check the solution in both equations of the system.
Example #1:
y = 4x
3x + y = -21
Step 1: Solve one equation for one variable.
y = 4x
(This equation is already solved for y.)
Step 2: Substitute the expression from step one
into
the other equation.
3x + y = -21
3x + 4x = -21
Step 3: Simplify and solve the equation.
7x = -21
x = -3
y = 4x
3x + y = -21
Step 4: Substitute back into either original
equation to find the value of the
other
variable.
3x + y = -21
3(-3) + y = -21
-9 + y = -21
y = -12
Solution to the system is (-3, -12).
y = 4x
3x + y = -21
Step 5: Check the solution in both equations.
Solution to the system is (-3,-12).
y = 4x
-12 = 4(-3)
-12 = -12
3x + y = -21
3(-3) + (-12) = -21
-9 + (-12) = -21
-21= -21
Example x#2:
+y=
10
5xfor
– one
y=2
Step 1: Solve one equation
variable.
x + y = 10
y = -x +10
Step 2: Substitute the expression from step one
into
the other equation.
5x - y = 2
5x -(-x +10) = 2
x+y=
10
y=2
Step 3: Simplify and solve 5x
the–equation.
5x -(-x + 10) = 2
5x + x -10 = 2
6x -10 = 2
6x = 12
x=2
x+y=
10
5x –
y=2
Step 4: Substitute back into
either
original
equation to find the value of the
other
variable.
x + y = 10
2 + y = 10
y=8
Solution to the system is (2,8).
x + y = 10
5x – y = 2
Step 5: Check the solution in both equations.
Solution to the system is (2,
8).
x + y =10
2 + 8 =10
10 =10
5x – y = 2
5(2) - (8) = 2
10 – 8 = 2
2=2

Solve by substitution:
1. y  2x  2
2x  3y  10
2. 2a  3b  7
2a  b  5
Steps:
3-2: Solving Systems of
Equations
using Elimination
1. Place both equations in Standard Form, Ax + By = C.
2. Determine which variable to eliminate with Addition
Subtraction.
3. Solve for the variable left.
4. Go back and use the found variable in step 3 to find
second variable.
5. Check the solution in both equations of the system.
or
5x + 3y
EXAMPLE
#1:= 11
5x = 2y + 1
STEP1:
C
STEP 2:
Write both equations in Ax + By =
form.
5x + 3y =1
5x - 2y
=1
Use subtraction
to eliminate
5x.
5x + 3y =11
5x + 3y = 11
-(5x - 2y =1)
-5x + 2y = -1
Note: the (-) is distributed.
STEP 3:
Solve for the variable.
5x + 3y =11
-5x + 2y = -1
5y =10
y=2
5x + 3y = 11
5x = 2y + 1
STEP 4: Solve for the other variable by
substituting
into either equation.
5x + 3y =11
5x + 3(2) =11
5x + 6 =11
5x = 5
x=1
The solution to the system is
(1,2).
5x + 3y= 11
5x = 2y + 1
Step 5:
Check the solution in both equations.
The solution to the system is (1,2).
5x + 3y = 11
5(1) + 3(2) =11
5 + 6 =11
11=11
5x = 2y + 1
5(1) = 2(2) + 1
5=4+1
5=5
Solving Systems of Equations
using Elimination
Steps:
1. Place both equations in Standard Form, Ax + By = C.
2. Determine which variable to eliminate with Addition
Subtraction.
or
3. Solve for the remaining variable.
4. Go back and use the variable found in step 3 to find
second variable.
5. Check the solution in both equations of the system.
the
Example x#2:
+ y = 10
5x – y = 2
Step 1: The equations are already in standard
form:
x + y = 10
5x – y = 2
Step 2: Adding the equations will eliminate y.
x + y = 10
x + y = 10
+(5x – y = 2)
+5x – y = +2
Step 3:
Solve for the variable.
x + y = 10
+5x – y = +2
6x = 12
x=2
x + y = 10
5x – y = 2
Step 4:
Solve for the other variable by
substituting into either equation.
x + y = 10
2 + y = 10
y=8
Solution to the system is (2,8).
x + y = 10
5x – y = 2
Step 5:
Check the solution in both equations.
Solution to the system is (2,8).
x + y =10
2 + 8 =10
10=10
5x – y =2
5(2) - (8) =2
10 – 8 =2
2=2
NOW solve these using
elimination:
1.
2.
2x + 4y =1
x - 4y =5
2x – y =6
x+y=3
Using Elimination to Solve a
Word Problem:
Two angles are supplementary. The
measure of one angle is 10 degrees
more than three times the other.
Find the measure of each angle.
Using Elimination to Solve a
Word Problem:
Two angles are supplementary. The
measure of one angle is 10 more
than three times the other. Find the
measure of each angle.
x = degree measure of angle #1
y = degree measure of angle #2
Therefore x + y = 180
Using Elimination to Solve a
Word Problem:
Two angles are supplementary. The
measure of one angle is 10 more
than three times the other. Find the
measure of each angle.
x + y = 180
x =10 + 3y
Using Elimination to Solve a
Word Problem:
Solve
x + y = 180
x =10 + 3y
x + y = 180
-(x - 3y = 10)
4y =170
y = 42.5
x + 42.5 = 180
x = 180 - 42.5
x = 137.5
(137.5, 42.5)
Using Elimination to Solve a
Word Problem:
The sum of two numbers is 70
and their difference is 24. Find
the two numbers.
Using Elimination to Solve a
Word problem:
The sum of two numbers is 70
and their difference is 24. Find
the two numbers.
x = first number
y = second number
Therefore, x + y = 70
Using Elimination to Solve a
Word Problem:
The sum of two numbers is 70
and their difference is 24. Find
the two numbers.
x + y = 70
x – y = 24
Using Elimination to Solve a
Word Problem:
x + y =70
x - y = 24
2x = 94
x = 47
47 + y = 70
y = 70 – 47
y = 23
(47, 23)
Now you Try to Solve These
Problems Using Elimination.
Solve
1. Find two numbers whose sum is
18 and whose difference is 22.
2. The sum of two numbers is 128
and their difference is 114. Find
the numbers.
MATRIX: A rectangular
arrangement of
numbers in rows and
columns.
The ORDER of a matrix
is the number of the
rows and columns.
The ENTRIES are the
numbers in the matrix.
This order of this matrix
is a 2 x 3.
columns
rows
 6 2  1
 2 0 5 


8
0

 10
1 3 
0
2 
4  3
 2
1

 7
0
1
3
1
0

9
5 7
1x4
4
6
5 9
2
7
 1

2
(or square
matrix)
3x3
2x2
3
8
6




3x5
(or square
matrix)
0
(Also called a
column matrix)
  9
7
 
0
 
6
4x1
(Also called a
row matrix)
To add two matrices, they must have the same
order. To add, you simply add corresponding
entries.
 5
 3

 0
 3  2
4    3
7   4
1 
0 
 3
5  (2)  3  1 


  33
40 
 0  4
7  (3)
 3
  0
 4
 2
4 
4 
 8 0 1 3   1 7
 5 4 2 9    5 3

 
=
=


5 2

3  2
8  (1)
07
 1 5
3 2
 5 5
43
23
9  ( 2)
7
0
7
7
4
5
5
7


To subtract two matrices, they must have the same
order. You simply subtract corresponding entries.
 9 2 4   4 0 7   9 4
 5 0 6    1 5  4 

 
   5 1
 1 3 8   2 3 2  1  (2)

 5

 4
 3
20 47 

0  5 6  (4)
33
8  2 
2
5
0
 3

10 
6 
=

2
8

 1
4 3   0
1


0  7   3  1
5
0   4 2
2-0
-4-1
8-3
0-(-1) -7-1
1-(-4)
5-2
3-8
0-7

=
8
1
7




2 -5 -5
5 1 -8
5
3
-7

In matrix algebra, a real number is often called a SCALAR.
To multiply a matrix by a scalar, you multiply each entry in
the matrix by that scalar.
 2
4
 4
0 
4( 2)


 1
 4( 4)
 8

 16
4(0) 

4( 1) 
0 

 4
 1
 2 
 0
 2   4


3   6
 1 4
 2 
 0  6
-2


 
-3 3
6 -5
-2(-3) -2(3)
-2(6)
-2(-5)
5 


 8 
 2  5 


3  (8) 
 
6 -6
 -12
10

This powerpoint was kindly donated to
www.worldofteaching.com
http://www.worldofteaching.com is home to over a
thousand powerpoints submitted by teachers. This is a
completely free site and requires no registration. Please
visit and I hope it will help in your teaching.