Download Math 103 College Algebra Lecture Notes on Chapter 5.1-5.2

College Algebra Worksheets
By O. Pauline Chow
HACC
Math 103 College Algebra
Lecture Notes on Chapter 5.1-5.2 (audio dated 9/28/07 at 6:40 pm, 32 minutes long)
Graphing method
Substitution method
Addition method
Applications
3x3 systems
Systems of Linear Equations
In this chapter, you will review the methods of solving 2x2 and 3x3 systems of linear equations. For 2x2 systems, you will solve
using graphing, substitution, and addition method. For 3x3 systems, you use the addition method.
Graphing Method
Ex 1: Solve the 2x2 system using graphing method.
y = –2x + 6
4x + 2y = 4
For each equation, you need to find at least two points to graph.
The graphs of the two linear equations are two parallel lines. Since they do not intersect, there is no solution to the system and we
say the system is inconsistent.
Ex 2: Solve the 2x2 system using graphing method.
2x – y = 4
x + 2y = –3
Note: When two lines intersect at a point, the point is called the point of intersection and we say the system is consistent.
When two lines overlap each other, every point on the two lines is a point of intersection. We say the system is dependent and
there is an infinite number of solutions.
College Algebra Worksheets
By O. Pauline Chow
HACC
Substitution Method
Ex 1: Solve the 2x2 system using substitution method.
–2x + 5y = 14
7x + 6y = –2
(1)
(2)
You need to solve for one of the two variables using one of the two equations, say, y in the first equation.
–2x + 5y = 14
5y = 2x + 14
Add 2x to each side.
y = 2 x + 14
(3)
5
Now substitute equation (3) into equation (2).
7x + 6 (2x + 14) = –2
5
35x + 6(2x + 14) = –10
35x + 12x + 84 = –10
47x + 84 = –10
47x = –94
x = –2
Multiply each side by 5.
Combine like terms on the left side.
Subtract 84 from each side.
Divide each side by 47
Now substitute –2 for x into equation (3).
y = 2(–2) + 14 = 10 = 2
5
5
So, the solution to the system is (–2, 2). The system is consistent.
Ex 2: Solve the 2x2 system using substitution method.
x + 3y = –3
5x – 6y = 27
(1)
(2)
Solve equation (1) for x.
x = –3 – 3y (3)
Substitute equation (3) into (2).
5(–3 – 3y) – 6y = 27
–15 – 15y – 6y = 27
–21y = 42
y = –2
Substitute –2 for y into equation (3)
x = –3 – 3(–2) = –3 + 6 = 3
The solution to the system is (3, –2).
Addition Method
Ex 1: Solve the 2x2 system using addition method.
–2x + 5y = 14
7x + 6y = –2
(1)
(2)
College Algebra Worksheets
By O. Pauline Chow
HACC
You need to multiply an appropriate number to each equation, so that the coefficients of one of the two variables will add up to
zero. So, you multiply 7 to equation (1) and 2 to equation (2).
7(–2x + 5y = 14)
2(7x + 6y = –2)
–14x + 35y = 98
14x + 12y = –4
Now you add the two equations.
–14x + 35y = 98
+ 14x + 12y = –4
47y = 94 → y = 2
Now substitute 2 for y in one of the two equations, say (1) to solve for x.
–2x + 5(2) = 14
–2x + 10 = 14
–2x = 4
x = –2
So, the solution to the system is (–2, 2).
Ex 2: Solve the 2x2 system using addition method.
–4x + 3y = –31
3x – 2y = 22
(1)
(2)
We will multiply 3 to equation (1) and 4 to equation (2) and add.
3(–4x + 3y = –31)
4(3x – 2y = 22)
–12x + 9y = –93
12x – 8y = 88
y = –5
Now substitute –5 for y into equation (2) to solve for x.
3x – 2(–5) = 22
3x + 10 = 22
3x = 12, x = 4
The solution to the system is (4, –5).
Ex 3: Solve the 2x2 system using addition method.
2x – y = –3
6x – 3y = 1
(1)
(2)
We will multiply –3 to equation (1) and add to equation (2).
–3(2x – y = –3)
6x – 3y = 1
–6x + 3y = 9
6x – 3y = 1
0 = 10, impossible
College Algebra Worksheets
By O. Pauline Chow
HACC
There is no solution. The system is inconsistent.
Applications
Ex 1: Tanya got 6 paperbacks and 3 hardbacks for $8.25. Gretta got 4 paperbacks and 5 hardbacks for $9.25. Find the price for
each. What is the total price for 7 paperbacks and 9 hardbacks?
Let the price of one paperback be $x and that of hardback be $y.
Since Tanya got 6 paperbacks and 3 hardbacks for $8.25, the first equation is
6x + 3y = 8.25
Since Gretta got 4 paperbacks and 5 hardback for $9.25, the second equation is
4x + 5y = 9.25
6x + 3y = 8.25
4x + 5y = 9.25
(1)
(2)
Multiply –2 to equation (1) and add to 3 times equation (2).
–12x – 6y = –16.5
12x + 15y = 27.75
9y = 11.25
y = 1.25, price of one hardback is $1.25
6x + 3(1.25) = 8.25
6x + 3.75 = 8.25
6x = 4.5
x = 0.75, price of one paperback is $0.75
Cost of 7 paperbacks and 9 hardbacks is 7(0.75) + 9(1.25) or $16.5.
Ex 2: In chemistry lab, Stephanie has to make a 35 milliliter solution that is 12% HCL. All that is in the lab is 8% and 15% HCL.
How many milliliters of each solution should she use to obtain the desired mix?
% of HCL
Volume of solution
Volume of HCL
8%
x ml
0.08x ml
15%
y ml
0.15y ml
12%
35 ml
0.12(35) ml
x + y = 35
x = 35 – y
0.08x + 0.15y = 0.12(35) 8x + 15y = 420
Substitute equation (1) into equation (2).
8(35 – y) + 15y = 420
280 – 8y + 15y = 420
7y = 140, y = 20
(1)
(2)
x = 35 – 20 = 15
Stephanie should mix 15 milliliters of 8% HCL and 20 milliliters of 15% HCL.
Ex 3 A private pilot flies a Cessna 172 on a trip that is 480 miles each way. It takes her 3 hours to get there with tailwind and 4
hours to return with headwind. What are the speed of the Cessna in still air and the speed of the wind?
First you let x mph be the speed of the Cessna in still air and y mph be the speed of the wind.
tail wind
head wind
distance
480 miles
480 miles
speed
(x + y) mph
(x – y) mph
time
3 hours
4 hours
College Algebra Worksheets
By O. Pauline Chow
HACC
Now apply the formula: distance = time * speed
3(x + y) = 480 x + y = 160
4(x – y) = 480 x – y = 120
x + y = 160
x – y = 120
2x = 280, x = 140
y = 160 – 140 = 20
Speed of the Cessna is 140 mph in still air and speed of the wind is 20 mph.
3x3 System
To solve a 3x3 system, you need to use the addition method repeatedly to eliminate one of the 3 variables in the system and reduce
the 3x3 system to a 2x2 system.
Ex 1: Solve a 3x3 system.
2x – 6y + 4z = 8
3x – 9y + 6z = 12
5x – 15y + 10z = 20
(1)
(2)
(3)
In order to keep track of the steps, starting with the first variable, say x. Use equations (1) and (2) to eliminate x by multiplying
equation (1) by 3 and add to (–2) times equation (2). Use equations (1) and (3) to eliminate x by multiplying equation (1) by 5 and
add to (–2) times equation (3).
3* eqt(1) + (–2)*eqt(2):
5*eqt(1) + (–2)*eqt(3):
6x – 18y + 12z = 24
–6x + 18y – 12z = –24
0=0
10x – 30y + 20z = 40
–10x + 20y – 20y = –40
0=0
Since both give an identity, the system is dependent, i.e. all three equations are equivalent to equation (1). The solution set is {2x
– 6y + 4z = 8, where x, y and z are any real numbers}
Ex 2:
–2x + y – 3z = 6
4x – y + z = 2
2x – y + 3z = 1
(1)
(2)
(3)
2* (1) + (2):
–4x + 2y – 6z = 12
+ 4x – y + z = 2
y – 5z = 14
(1) + (3):
–2x + y – 3z = 6
+ 2x – y + 3z = 1
0 = 7 impossible
The system is inconsistent and there is no solution.
Ex 3:
2x + 3y + 5z = –14
3x – 2y + z = 5
–5x + y – 3z = –2
–3* eqt(1) + (2)*eqt(2):
–6x – 9y – 15z = 42
+ 6x – 4y + 2z = 10
–13y – 13z = 52
y + z = -4 (4)
(1)
(2)
(3)
5*eqt(1) + (2)*eqt(3):
10x + 15y + 25z = –70
+ –10x + 2y – 6z = –4
17y + 19z = –74 (5)
College Algebra Worksheets
By O. Pauline Chow
HACC
–17* eqt(4) + eqt(5)
–17y – 17z = 68
17y + 19z = –74
2z = –6, z = –3
Substitute –3 for z into equation (4).
y + (–3) = -4
y = –1
Substitute –3 for z and –1 for y into equation (1) to solve for x.
2x + 3y + 5z = –14
2x + 3(–1) + 5(–3) = –14
2x – 3 – 15 = –14
2x = 4, x = 2
The solution to the system is (2, –1, –3).
Ex 4:
100x + 200y + 500z = 47
350x + 5y + 250z = 33.9
200x + 80y + 100z = 23.4
(1)
(2)
(3)
7* eqt(1) + (–2)*eqt(2):
700x + 1400y + 3500z = 329
+ –700x – 10y – 500z = –67.8
1390y + 3000z = 261.2 (4)
–2*eqt(1) + eqt(3):
–200x – 400y – 1000z = –94
+ 200x + 80y + 100z = 23.4
–320y – 900z = –70.6
Next you will need to eliminate y.
32*eqt(4) + 139*eqt(5):
44480y + 96000z = 8358.4
+ –44480y – 125100z = –9813.4
–29100z = –1455
z = 0.05
Now substitute 0.05 for z in either (4) or (5) to solve for y.
1390y + 3000(0.05) = 261.2
1390y + 150 = 261.2
1390y = 111.2
y = 0.08
Substitute both values of y and z into (1), (2), or (3) to solve for x.
100x + 200(0.08) + 500(0.05) = 47
100x + 16 + 25 = 47
100x = 6
x = 0.06
The solution to the system is (0.06, 0.08, 0.05). Be sure to check the solution using (2) and (3).
(5)