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Chapter 3: Lines, Parabolas, and Systems
3.4: Systems of Linear Equations
Dr. Raja Mohammad Latif
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1 3.4: Systems of Linear Equations
===Lecture Sec 3.4 Begins now===
Home Work:
26
;
28
;
29
;
34
;
37
;
39
;
41
Independent System
: Lines intersect at one point: exactly
one solution.
3
x
4
y
= 1
;
2
x
+ 3
y
= 12
:
The two lines intersect in exactly one point.
S.S.
=
f
(3
;
2
g
:
Dependent System
solutions.
4
x
+
y
= 2
;
8
x
2
y
=
4
:
intersect.
Inconsistent System
: Lines are parallel: no solutions.
6
x
+ 4
y
= 7
;
3
x
2
y
= 4
:
There is no point where the two lines intersect.
177BZ6Example.Diet. A woman wants to use milk and
orange juice to increase the amount of calcium and vitamin A
in her daily diet.
An ounce of milk contains 37 milligrams of calcium and
57 micrograms of vitamin A.
An ounce of orange juice contains 5 milligrams of calcium
and 65 micrograms of vitamin A. How many ounces of milk
and orange juice should the woman drink each day to provide
exactly 500 milligrams of calcium and 1200 micrograms of
Dr. Raja Latif. Math 131 (052)
Department of Mathematical Sciences, KFUPM
3
vitamin A?
x = Number of ounces of milk
y = Number of ounces of orange juice
Next, we summarize the given information in the table:
Milk
Orange Juice
Total Needed
Calcium
37
5
500
Vitamin A
57
65
1200
Next we use the information in the table to form equations
involving x and y
0
@
Calcium
in x oz
of milk
1
A
+
0
@
Calcium in
y oz of
orang juice
1
A
=
Total calcium
needed (mg)
±
37x +
5y =
500
0
@
Vitamin A
in x oz
of milk
1
A
+
0
@
Vitamin A
in y oz of
orang juice
1
A
=
Total calcium
needed (mg)
±
57x
+
65y
=
1200
Solve using elimination by addition:
481x
65y
=
6500
57x
+
65y
=
1200
424x
=
5300
=
)
x = 12.5
Dr. Raja Latif. Math 131 (052)
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37(12.5) + 5y = 500
=
)
5y =37.5
=
)
y = 7.5.
Drinking 12.5 ounces of milk and 7.5 ounces of orange
juice each day will provide the required amounts of calcium
and vitamin A.
=========================
152AL6Example. (Mixture) The Britannia Store, which
specializes in selling all kinds of nuts, sells peanuts at $0.70
per pound and cashews at $1.60 per pound.
peanuts are not selling well and decides to mix peanuts and
cashews to make a mixture of 45 pounds, which could sell for
$1.00 per pound.
How many pounds of peanuts and cashews should be
mixed to keep the same revenue?
Solution. Let the mixture contain x pounds of peanuts
and y pounds of cashews.
Since the total mixture is 45 pounds,
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This note was uploaded on 12/17/2009 for the course MATH MATH131 taught by Professor Dr.rajalatif during the Spring '09 term at King Fahd University of Petroleum & Minerals.
 Spring '09
 Dr.RajaLatif
 Math, Linear Equations, Equations, Systems Of Linear Equations

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