Download Chapter 4 Linear Equations and Inequalities in Two Variables

Chapter one
Linear Equations
Linear Equations
Definition of a Linear Equation
A linear equation in one variable x is an
equation that can be written in the form
ax + b = 0, where a and b are real numbers and
a is not equal to 0.
An example of a linear equation in x is 4x + 2 = 6. Linear
equations in x are first degree equations in the variable x.
LINEAR EQUATION IN TWO VARIABLES
LINEAR EQUATION IN TWO VARIABLES
Definition of an equation:
an equation is any expression containing an “equals” sign.
Consider the following equation involving the two quantities, x
and y:
y = -3 + 2x
The equation can easily be used to work out the value of y for
any given value of x. All you need to do is insert your given
value for x, and then evaluate the expression on the right hand
side of the equation – this will be the corresponding value of y.
LINEAR EQUATION IN TWO VARIABLES
Repeating this for other values of x we get:
x
y
-2
-3 + 2(-2) = -7
-1
-3 + 2(-1) = -5
0
-3 + 2(0) = -3
1
-3 + 2(1) = -1
2
-3 + 2(2) = 1
3
-3 + 2(3) = 3
4
-3 + 2(4) = 5
LINEAR EQUATION IN TWO VARIABLES
Presenting the concept of the equation of a straight line
To plot y against x on a graph as illustrated in Figure by plotting the points
-8 -7 -6 -5 -4 -3 -2 -1
-7 -6 -5 -4 -3 -2 -1
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given by the pairs of co-ordinates (e.g. (-2, -7) and (4,5)) in the table above:
1 2 3 4 5 6 7 8
System of equations
A pair of linear equations in two variables
is said to form a system of simultaneous
linear equations.
For Example, 2x – 3y + 4 = 0
x + 7y – 1 = 0
Form a system of two linear equations in
variables x and y.
general form of a linear equation in two variables
• The general form of a linear equation in two
variables x and y is:
ax + by + c = 0 , a not = 0, b not=0, where
a, b and c being real numbers.
• A solution of such an equation is a pair of values,
one for x and the other for y, which makes two sides
of the equation equal.
• Every linear equation in two variables has infinitely
many solutions which can be represented on a
certain line.
GRAPHICAL SOLUTIONS OF A LINEAR
EQUATION
• Let us consider the following system of two
simultaneous linear equations in two variable.
• 2x – y = -1
• 3x + 2y = 9
Here we assign any value to one of the two
variables and then determine the value of the
other variable from the given equation.
GRAPHICAL SOLUTIONS OF A LINEAR EQUATION
For the equation
X
0
2
Y
1
5
2x –y = -1 ---(1)
2x +1 = y
Y = 2x + 1
3x + 2y = 9 --- (2)
2y = 9 – 3x
9- 3x
Y = ------2
X
3
-1
Y
0
6
GRAPHICAL SOLUTIONS OF A LINEAR EQUATION
Y
(-1,6)
(2,5)
(0,3)
(0,1)
X
X’
Y’
X= 1
Y=3
ALGEBRAIC METHODS OF SOLVING
SIMULTANEOUS LINEAR EQUATIONS
The most commonly used algebraic methods of solving simultaneous
linear equations in two variables are
1. Method of elimination by substitution
2. Method of elimination by equating the coefficient
3. Method of Cross- multiplication
ELIMINATION BY SUBSTITUTION
Let us take an example
x + 2y = -1 ------------------ (i)
2x – 3y = 12 -----------------(ii)
SUBSTITUTION METHOD
x + 2y = -1
x = -2y -1 ------- (iii)
Substituting the value of x in equation (ii), we get:
2x – 3y = 12
2 ( -2y – 1) – 3y = 12
- 4y – 2 – 3y = 12
- 7y = 14 , y = -2 ,
SUBSTITUTION METHOD
Putting the value of y in eq (iii), we get
x = - 2y -1
x = - 2 x (-2) – 1
=4–1=3
Hence the solution of the equation is
( 3, - 2 )
ELIMINATION METHOD
• Example: we want to solve,
3x + 2y = 11
2x + 3y = 4
ELIMINATION METHOD
Let 3x + 2y = 11 --------- (i)
2x + 3y = 4 ---------(ii)
Multiply 3 in equation (i) and 2 in equation (ii) and
subtracting eq iv from iii, we get
9x + 6y = 33 ------ (iii)
4x + 6y = 8 ------- (iv)
5x = 25
=> x = 5
ELIMINATION METHOD
• putting the value of y in equation (ii) we get,
2x + 3y = 4
2 x 5 + 3y = 4
10 + 3y = 4
3y = 4 – 10
3y = - 6
y=-2
Hence, x = 5 and y = -2
Graphing Linear Equations in Two Variables
The Rectangular Coordinate System
y-axis
2nd quadrant
1st quadrant
x-axis
3rd quadrant
4th quadrant
EXAMPLE
Plot the points (3,2) and (-2,-4).
SOLUTION
 (3,2)
(-2,-4) 
The Graph of an Equation
The graph of an equation in two variables is the set of
points whose coordinates satisfy the equation.
An ordered pair of real numbers (x,y)
is said to satisfy the equation when substitution of the x and y
coordinates into the equation makes it a true statement.
For example, in the equation
y = 2x + 6, the ordered pair (1,8) is a solution. When we substitute
this point the sentence reads 8 = 8, which is true.
The ordered pair (2,3) is not a solution. When we substitute this
point, the sentence reads 3 = 10, which is not true.
Solution of an Equation in Two Variables
Example:
Given the equation 2x + 3y = 18, determine if the
ordered pair (3, 4) is a solution to the equation.
We substitute 3 for x and 4 for y.
2(3) + 3 (4) ? 18
6 + 12 ? 18
18 = 18 True.
Therefore, the ordered pair (3, 4) is a solution to the
equation 2x + 3y = 18.
Finding Solutions of an Equation
Find five solutions to the equation y = 3x + 1.
Start by choosing some x values and then computing the
corresponding y values.
If x = -2, y = 3(-2) + 1 = -5.
Ordered pair (-2, -5)
If x = -1, y = 3(-1) + 1 = -2.
Ordered pair ( -1, -2)
If x =0, y = 3(0) + 1 = 1.
Ordered pair (0, 1)
If x =1, y = 3(1) + 1 =4.
Ordered pair (1, 4)
If x =2, y = 3(2) + 1 =7.
Ordered pair (2, 7)
Graph of the Equation
Plot the five ordered pairs to obtain the graph of y = 3x + 1
(2,7)
(1,4)
(0,1)
(-1,-2)
(-2,-5)
Linear equations in two variables
Equations of the form ax + by = c are called
linear equations in two variables.
y
This is the graph of the
equation 2x + 3y = 12.
(0,4)
(6,0)
-2
The point (0,4) is the y-intercept.
The point (6,0) is the x-intercept.
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2
x
Slope of a line
Slope of a line
The slope of a line is a number, m, which measures its
steepness.
y
m is undefined
m=2
1
m=
2
m=0
x
2
-2
m=-
27
1
4
Slope of a line
The slope of the line passing through the two points
(x1, y1) and (x2, y2) is given by the formula
y2 – y1
, (x1 ≠ x2 ).
m=
x2 – x1
The slope is the
change in y divided
by the change in x as
we move along the
line from (x1, y1) to
(x2, y2).
y
(x2, y2)
y2 – y1
change in y
(x1, y1)
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x2 – x1
change in x
x
Slope formula
Example: Find the slope of the line passing through
the points (2, 3) and (4, 5).
Use the slope formula with x1= 2, y1 = 3, x2 = 4, and y2 = 5.
y2 – y1
5–3
m=
=
x2 – x1
4–2
2
= =1
2
y
(4, 5)
2
(2, 3)
2
x
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Slope-intercept form
Example: Graph the line y = 2x – 4.
1. The equation y = 2x – 4 is in the slope-intercept form. So,
m = 2 and b = - 4.
y
2. Plot the y-intercept, (0, - 4).
x
3. The slope is 2. m =
change in y
2
=
1
change in x
4. Start at the point (0, 4).
Count 1 unit to the right and 2 units up
to locate a second point on the line.
The point (1, -2) is also on the line.
5. Draw the line through (0, 4) and (1, -2).
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(1, -2)
2
(0, - 4)
1
Slope-intercept form
A linear equation written in the form y – y1 = m(x – x1)
is in point-slope form.
The graph of this equation is a line with slope m
passing through the point (x1, y1).
Example:
y
The graph of the equation
8
m=-
y – 3 = - 1 (x – 4) is a line
4
2
of slope m = - 1 passing
2
1
2
(4, 3)
x
through the point (4, 3).
4
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8
Slope-intercept form
Example: Write the slope-intercept form for the equation of
the line through the point (-2, 5) with a slope of 3.
Use the point-slope form, y – y1 = m(x – x1), with m = 3 and
(x1, y1) = (-2, 5).
y – y1 = m(x – x1)
Point-slope form
y – y1 = 3(x – x1)
Let m = 3.
y – 5 = 3(x – (-2))
Let (x1, y1) = (-2, 5).
y – 5 = 3(x + 2)
Simplify.
y = 3x + 11
Slope-intercept form
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Slope-intercept form
Example: Write the slope-intercept form for the
equation of the line through the points (4, 3) and (-2, 5).
m= 5–3 =- 2 =- 1
-2 – 4
6
Calculate the slope.
3
y – y1 = m(x – x1)
Point-slope form
1
(x – 4)
3
y = - 1 x + 13
3
3
1
Use m = - and the point (4, 3).
3
y–3=-
Slope-intercept form
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Slope-intercept form
Two lines are parallel if they have the same slope.
If the lines have slopes m1 and m2, then the lines are
parallel whenever m1 = m2.
y
(0, 4)
Example:
The lines y = 2x – 3
y = 2x + 4
and y = 2x + 4 have slopes
m1 = 2 and m2 = 2.
x
y = 2x – 3
The lines are parallel.
(0, -3)
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Slope-intercept form
Two lines are perpendicular if their slopes are
negative reciprocals of each other.
If two lines have slopes m1 and m2, then the lines are
perpendicular whenever
y
1
m2= or m1m2 = -1.
y = 3x – 1
m1
(0, 4)
Example:
The lines y = 3x – 1 and
1
y = - x + 4 have slopes
3
1
m1 = 3 and m2 = - .
3
x
(0, -1)
The lines are perpendicular.
Reciprocals = ‫مقلوب‬
1
y=- x+4
3
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Example
A certain factory has daily fixed overhead expenses of $2000, while each item produced
Costs $100. Find an equation that relates the daily cost C to the number X of items
Produced each day.
Solution: The fixed overhead expense of $2000 represents the fixed cost,
the cost incurred no matter how many items are produced. Since each item
C y
produced costs $100, the variable cost of producing
production is
C = 100x + 2000
The graph of this equation is given by the line
Dollars
X items is 100X. Thus the total daily cost C of
2200
2100
2000
x
Shown. Notice that the fixed cost $2000 is
Represented by the y-intercept, while the $100
Cost of producing each item is the slope. Also
Notice that a different scale is used on each36 axis.
1 2 3 4 5
No. of items
Solved problems
1- Check whether the graphs of the equations given below represent parallel lines.
Explain your answer.
line a: y = –5x + 3
line b: y + 5x = –2
Solution
line a: y = –5x + 3
line b: y = –5x – 2
Identify the slope of each equation.
For line a: y = –5x + 3,
Slope = –5
For line b: y = –5x – 2
Slope = –5
Line a and line b have equal slope, –5.
Parallel lines have equal slope. So, line a and line b are parallel lines.
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Solved problems
2- Check whether the graphs of the equations given below represent parallel lines.
Explain your answer.
line a: y = x + 7
line b: x – y = –2
Solution
line a: y = x+ 7
line b: y = x + 2
Identify the slope of each equation.
For line a: y = x + 7,
Slope = 1
For line b: y = x + 2
Slope = 1
Line a and line b have equal slope that is 1.
Parallel lines have equal slope. So, line a and line b are parallel lines.
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Solved problems
3- Find the slope of the line that passes through the points (-1 , 0) and (3 , 8).
Solution : The slope m is given by
m = (y2 - y1) / (x2 - x1) = (8 - 0) / (3 - -1) = 2
4- Find the slope of the line that passes through the points (2 , 0) and (2 , 4).
Solution : The slope m is given by
m = (y2 - y1) / (x2 - x1) = (4 - 0) / (2 - 2) = 4 / 0
Division by zero is not allowed in math. Therefore the slope of the line defined by the
points (2 , 0) and (2 , 4) is undefined. The line through the points (2 , 0) and (2 , 4) is
perpendicular to the x axis.
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Solved problems
5- Find the slope of the line that passes through the points (7 , 4) and (-9 , 4).
Solution : The slope m is given by
m = (y2 - y1) / (x2 - x1) = (4 - 4) / (-9 - 7) = 0 / -16 = 0
The line defined by the points (7 , 4) and (-9 , 4) is parallel to the x axis and its slope is
equal to zero.
6- Are the three points A(2 , 3) , B(5 , 6) and C(0 , -2) collinear?
Solution : We first find the slope defined by the points A(2 , 3) and B(5 , 6).
m(AB) = (y2 - y1) / (x2 - x1) = (6 - 3) / (5 - 2) = 3 / 3 = 1
We next find the slope defined by the points B(5 , 6) and C(0 , -2).
m(BC) = (y2 - y1) / (x2 - x1) = (-2 - 6) / (0 - 5) = -8 / -5 = 8 / 5
The two slopes are not equal therefore the40points are not collinear.
Solved problems
7- What is the slope of the line perpendicular to the line whose equation is given by
- 2y = - 8 x + 9?
Solution : The equation of the given line is
- 2y = - 8 x + 9
Rewrite in slope intercept form.
y = 4 x - 9/2
The slope of the given line is 4. Two perpendicular lines have slopes m1 and m2
related by:
m1*m2 = -1
If we set m1 = 4 then m2, the slope of the line perpendicular to the given line, is
equal to -1/4.
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Solved problems
8- Is the triangle whose vertices are the points A(0 , -1) , B(2 , 1) and C(-4 , 3) a right
triangle?
Solution : We first find the slope of the line defined by points A and B
m(AB) = (y2 - y1) / (x2 - x1) = (1 - -1) / (2 - 0) = 1
We next find the slope of the line defined by points A and C
m(AC) = (3 - -1) / (-4 - 0) = 4 / -4 = -1
The product of the slopes m(AB) and m(AC) is equal to -1 and this means that the
lines defined by A,B and A,C are perpendicular and therefore the triangle whose
vertices are the points A, B and C is a right triangle.
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Solved problems
9- What is the slope of the line -7y + 8x = 9
Solution : The given equation
-7y + 8x = 9
Rewrite the equation in slope intercept form.
-7y = -8x + 9
y = (8/7) x - 9/7
The slope of the given line is 8/7
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Solved problems
10- What is the slope of the line y = 9?
Solution : The given equation
y=9
The given equation is a line parallel to the x axis therefore its slope is equal to 0.
11- What is the slope of the line x = -5?
Solution : The given equation
x = -5
The given equation is perpendicular to the x axis therefore its slope is undefined.
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Solved problems
12- Which of the following equations passes through the points (2,4) and (-3,-6)?
Choose:
y = (1/2)x - 2
y = 2x
y = 2x + 4
y = (-1/2)x + 2
13- What is the slope of a line perpendicular to 2y = -6x - 10?
Choose:
-3
3
-1/3
1/3
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Solved problems
14- Given 3y - 4x = 5 and 4y + 6 = 3x.
Are these lines parallel, perpendicular or neither?
Choose :
parallel
perpendicular
neither
15- What is the equation of a line that passes through the point (4,-5) and is
parallel to 3x + 2y = 12?
Choose :
y = -3x + 6
y = (3/2)x + 1
y = 3x + 1
y = (-3/2)x + 1
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Solved problems
16- What is the equation of a line that passes through the point (-1,-2) and is
perpendicular to -5x = 6y + 18?
Choose :
y = (6/5)x - (4/5)
y = (-6/5)x + (6/5)
y = (6/5)x + (4/5)
y = (-6/5)x - (4/5)
17- Given 4y - 2x = 10 and -6y - 6 = -3x.
Are these lines parallel, perpendicular or neither?
Choose :
parallel
perpendicular
neither
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