Contemporary business mathematics with canadian applications

Contemporary Business Mathematics with
Canadian Applications
Eleventh Canadian Edition
Chapter 2
Linear Systems
Copyright © 2018 Pearson Canada Inc.

Linear Systems (1 of 2)
In many types of problems, the relationship between two or more
variables can be represented by setting up linear equations. Algebraic as
well as graphical techniques are available to solve such problems.

Linear Systems (2 of 2)
Solving a system of two linear equations by algebraic elimination

Solving Linear Systems by Algebraic
Elimination
• If the coefficients (numbers in front of variable) of one variable are the
same in both equations;
– Reduce to one equation by addition or subtraction as follows.
a) If the coefficients of one variable are numerically equal but opposite
in sign
– Addition will eliminate the variable.
b) If the coefficients of one variable are numerically equal and have
the same sign,
– Subtraction can eliminate the variable.
– Alternatively, one equation can be multiplied by −1; addition can
then be used.

Ex
ample W here the Coef
f
icients are Eq
ual
• Solve 5x + 4y =7
3x − 4y = 17
5x + 4y = 7 equation 
3x − 4y = 17 equation 
8x =24 add  and  to eliminate y
x = 3

Ex
ample (2 of 2)
• Substitute x=3into any one of the equations
5(3) +4y = 7substitute 3f
or xin 
15+4y = 7
4y = −8
y = −2
x = 3, y = -2 is the solution

Ex
f
icients are Not
Numerically Eq
ual (1 of 2)
• Solve x + 4y = 18
2x + 5y = 24
T
o eliminate the term in x, multiply  by 2.
2x + 8y = 36  multiplied by 2
−2x − 5y = −24 multiplied by −1
3y = 12 add
y = 4

Ex
f
icients are Not
Numerically Eq
ual (2 of 2)
• Substitute x=4into any one of the equations
x + 4(4) =18substitute 4f
or y in 
x + 16=18
x = 2
x = 2, y = 4is the solution

Solving Linear Systems in Two Variables
I
nvolving Decimals or Fractions
1. W hen one or both equations contain decimals, it is best to multiply
the equation(s) by an appropriate factor of 10to eliminate the
decimal.
2. If one or both equations contain fractions, multiply each equation by
its Lowest Common Denominator (LCD) to eliminate the fraction.
Then solve as shown previously.

I
nvolving Decimals
• It is best to eliminate the decimals or fractions by multiplying
Solve (1) 1.5x+0.8y =1.2 and (2) 0.7x+1.2y =−4.4
- Multiply (1) by factor of 10to get :15x+8y =12
- Multiply (2) by factor of 10to get:7x+12y =-
44
15x+8y =12  x3 45x+24y =36
7x+12y =-
44 x-
2   -
14x– 24y =88
Subtract: 31x=124
x=4(rounded)
Substitute into either of the equations:
15(4) +8y =12
y =-
6

I
nvolving Fractions
• Solve (1)
1
4
+
2
3
=5 and (2)
1
3
+
1
2
=4
– Multiply (1) by the LCD of 12 to get 3x + 8y =60
– Multiply (2) by the LCD of 6to get 2x + 3y =24
3x+8y = 60 → ×2 → 6x+16y = 120
2x+3y =24 → ×-
3→ -
6x-9y =-
72___
Subtract 7y =48
y =6.86(rounded)
substitute into either the of equations
3x+8(6.86) =60
x=1.71
The solution is approx
imately (1.71,6.86)

Systems ofTwo Linear Eq
uations with No
Solution or Many Solutions
• Some systems of linear equations are inconsistent, meaning that there
is no solution for them. Other systems are consistent and dependent
with many solutions.
• How can you tell?
– If you get a result that is a contradiction and impossible this means
there is no solution. These systems are called inconsistent linear
equations.
 For ex
ample, after solving you may end up with 0=8.
– If you get a result that says 0=0this means that the system is
consistent and dependent. There are an infinite number of
solutions. The original systems are equivalent.

Graphing Linear Eq
uations (1 of 4)
• The horizontal line is called the X ax
is, while the vertical line is called
the Y ax
is.
• The point of intersection of the two ax
es is called the origin.
• The position of any point relative to the pair of ax
es is defined by an
ordered pair of numbers (x, y).
– They represent the distance from the origin on the horiz
ontal and
vertical ax
is respectively.

Graphing Linear Eq
uations (2 of 4)

Graphing Linear Eq
uations (3 of 4)

Graphing Linear Eq
uations (4 of 4)
• Graph x+y =4
• Rearrange equation, choose values for xand find the corresponding
values for y
y =4− x
T
able ofvalues
x 0 4 2
Y 4 0 2

The Slope–
Y-
I
ntercept Form ofa Linear
Eq
uation (1 of 3)
• Every line has two important characteristics:its steepness, called the
slope (rise over run), and a point where the line intersects with the Y
ax
is, called the Y-
intercept.

The Slope–
Y-
I
Eq
uation (2 of 3)
• Graph the line 2x− y =4
−y =−2x+4
y =2x−4
– slope m =2, y-
intercept =−4
– note slope can be either
/ =
+2/
+1
=
−2/
−1
Blank Table ofvalues Blank
x 3 2
y 2 0
point (3,2) (2,0)

The Slope–
Y-
I
Eq
uation (3 of 3)
2x− y =4
• Notice the slope is
the same
everywhere on the
line.
Graph

The Slope–
Y-
I
ntercept Form Special Case 1
(1 of 2)
Lines parallel to the X-
ax
is
• Lines parallel to the X-
ax
is.
– There is no mx in the
equation y =b.
– This is a line parallel to
the X ax
is that crosses
the Y ax
is at point (0, b)
and has a slope of 0.
Graph

Special Cases Lines Parallel to the X Ax
es
• Graph
i. y =3and
ii. y =−2
(i) The line represented by y =3is a line parallel
to the X ax
is and three units above it.
(ii) The line represented by y =−2 is a line parallel
to the X ax
is and two units below it.

The Slope–
Y-
I
ntercept Form Special Case 2
(2 of 2)
Lines parallel to the Y-
ax
is
• Lines parallel to the Y-
ax
is.
– X =a
– The slope of the line is
undefined.
Graph

Special Cases Lines Parallel to the Y Ax
es
• Graph
i. x=3and
ii. x=−5
(i) The line represented by x =3is a line parallel
to the Y ax
is and three units to the right of it.
(ii) The line represented by x =−5is a line parallel
to the Y ax
is and five units to the left of it.

Dif
f
erent Cases f
or the Slope
Positive Slope As xincreases,y increases
Negative Slope As xincreases, y decreases
0Slope Horiz
ontal line, as xincreases,
y remains the same
Undefined Slope Vertical line, xremains the
same as y increases

Graph a System ofLinear Eq
uations and
Solve (1 of 2)
• Equation #1
x=y
• Equation #2
x− 2y +2000=0

uations and
Solve (2 of 2)
• x =y (first equation)
– b =0, the y-
intercept is 0(point is 0,0)
– m =1, the slope is 1 (rise/
run =rise 1, run 1)
• x − 2y +2000=0(second equation)
x − 2y =−2000
−x +2y =+2000
2y =x +2000
y =½ x +1000
– b =1000, the y-
intercept is 1000
– m =½ (the slope)

uations and
Solve
• Now substitute x =y into
the second equation
y =½ y +1000
½ y =1000
y =2000
• The solution is the point of
intersection S (2000,2000).

Problem Solving (1 of 3)
• The Clark
son Soccer League has set a budget of $3840for soccer
balls. High-
quality game balls cost $36each while lower-
quality
practice balls cost $20each. If 160balls are to be purchased, how
many balls of each type can be purchased to ex
actly use up the
budgeted amount?

• Let xrepresent the number of game balls.
• Let y represent the number of practice balls.
T
otal number of balls is x+y , so x+y =160
Value of the game balls is $36x
Value of the practice balls is $20y
T
otal value of the balls is $(36x, 20y)

x+y =160 
36x+20y =3840 
−20x− 20y =−3200  multiplied by -
20
16x =640 add
x=40, substitute x=40into 
40+y =160
y =120
• 40game balls and 120practice balls can be bought

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