Notes Over 4.5
Solving a Linear System
1
Use an Inverse Matrix to solve the linear system.
.

2
3 x x




4
 y
2
2

 y
3
1



3 2
0
1
1


A


 x y

1




1
1


2 1
3 2





1
0












2 1


1
0
0
1



 x y



2 ,


2

0
3

0
 





2
3




 






Notes Over 4.5
Solving a Linear System
Use an Inverse Matrix to solve the linear system.
2 .
3
4 x x


4
5 y y
15


4





16
7
3 4
5


4
A





1
1 x y 




1
1


5

4

4 3







4
7





 




 
5

4
4 3











1
0
0
1



 x y








16


21


8 , 5





5
8



Notes Over 4.5
Solving a Linear System
Use an Inverse Matrix to solve the linear system.
3 .
6
3 x x


5
2
 y y
12


6



3
3
15

3


5
2


A
3


 x

1 y 



1
3



2 5

3 6





3
3













3
1 2


 







1
0
0
1



 x y





3 ,

2 5

3


6
 





3
3



Notes Over 4.5
Writing and Using a Linear System
4. You can purchase peanuts for $3 per pound, almonds for $4 per pound and cashews for $8 per pound. You want to create 140 pounds of a mixture that costs $6 per pound. If twice as many peanuts are used than almonds, how many pounds of each type should be used?




Let p = peanuts, a = almonds, and c = cashews
3 p p

 a
4
 a p

2 a c

1
3
1

1
4
2
1
8
0





8 c
140



 a c p





140





 
1 40
8 40
0




3
A p

1
 p
B
4
 a
 c a

8 c


140
840 p

2 a






40
20
80




0 peanuts almonds cashews

Notes Over 4.5
Writing and Using a Linear System
5. A chemist wants to use three different solutions to create a 600-liter mixture containing 25% acid. The first solution contains 30% acid, the second 20%, and the third 15%. If the mixture is to contain 100 more liters of the 15% solution than the 20%, how many liters of each solution should be used?
.
3 a a
 c
 b
.
2 b

 b





1
0
30

1
1
1
20
1
15




Let a = 30%, b = 20%, and c = 15% c

.
15 c


100



 a b c
600
.
25
 
30









6
1
00
00
15,000 a




 a

20 b b


15 c c
A

1
B
 b
 c






380
60
160

 



600


100
15 , 000
30%
20%
15%

Notes Over 4.5