Chapter one Linear Equations. Linear Equations Definition of a Linear Equation A linear equation in...

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


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.


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


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

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

1 2

3 4

5 6

7

-8 -7 -6 -5 -4 -3 -2 -1

-7 -

6 -5

-4

-3 -

2 -

1

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.

• The general form of a linear equation in two variables x and y is:

ax + by + c = 0 , a not = 0, b not=0, wherea, 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.

general form of a linear equation in two variables

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 = 9Here we assign any value to one of the two

variables and then determine the value of the other variable from the given equation.

For the equation

2x –y = -1 ---(1) 2x +1 = y Y = 2x + 1

3x + 2y = 9 --- (2)2y = 9 – 3x

9- 3xY = ------- 2

X 0 2

Y 1 5

X 3 -1

Y 0 6


XX’

Y

Y’

(2,5)(-1,6)

(0,3)(0,1)

X= 1Y=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• Method of elimination by equating the coefficient• Method of Cross- multiplication

ELIMINATION BY SUBSTITUTION

Let us take an example x + 2y = -1 ------------------ (i)

2x – 3y = 12 -----------------(ii)

SUBSTITUTION METHOD

x + 2y = -1x = -2y -1 ------- (iii)

Substituting the value of x in equation (ii), we get:

2x – 3y = 122 ( -2y – 1) – 3y = 12- 4y – 2 – 3y = 12- 7y = 14 , y = -2 ,

SUBSTITUTION METHOD

Putting the value of y in eq (iii), we getx = - 2y -1x = - 2 x (-2) – 1

= 4 – 1 = 3Hence 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 = 42 x 5 + 3y = 410 + 3y = 43y = 4 – 103y = - 6y = - 2

Hence, x = 5 and y = -2

Graphing Linear Equations in Two Variables

The Rectangular Coordinate System

1st quadrant2nd quadrant

3rd quadrant 4th quadrant

x-axis

y-axis

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Plot the points (3,2) and (-2,-4).

(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 equationy = 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.

Example:Given the equation 2x + 3y = 18, determine if theordered pair (3, 4) is a solution to the equation.

We substitute 3 for x and 4 for y.2(3) + 3 (4) ? 186 + 12 ? 1818 = 18 True.

Therefore, the ordered pair (3, 4) is a solution to the equation 2x + 3y = 18.

Solution of an Equation in Two Variables

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)

Finding Solutions of an Equation

Plot the five ordered pairs to obtain the graph of y = 3x + 1

(2,7)

(1,4)

(0,1)

(-1,-2)

(-2,-5)

Graph of the Equation

25

Equations of the form ax + by = c are called linear equations in two variables.

The point (0,4) is the y-intercept.

The point (6,0) is the x-intercept.

x

y

2-2

This is the graph of the equation 2x + 3y = 12.

(0,4)

(6,0)

Linear equations in two variables

Slope of a line

27

y

x2-2

The slope of a line is a number, m, which measures its steepness.

m = 0

m = 2m is undefined

m =1

2

m = -1

4

Slope of a line

28

x

y

x2 – x1

y2 – y1

change in y

change in x

The slope of the line passing through the two points (x1, y1) and (x2, y2) is given by the formula

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).

y2 – y1

x2 – x1

m = , (x1 ≠ x2 ).

(x1, y1)

(x2, y2)

Slope of a line

29

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

x2 – x1

m = 5 – 3

4 – 2= =

22

= 1

2

2(2, 3)

(4, 5)

x

y

Slope formula

30

1

Example: Graph the line y = 2x – 4.

2. Plot the y-intercept, (0, - 4).

1. The equation y = 2x – 4 is in the slope-intercept form. So, m = 2 and b = - 4.

3. The slope is 2.

The point (1, -2) is also on the line.

1= change in y

change in xm = 2

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.

2

x

y

5. Draw the line through (0, 4) and (1, -2).

(0, - 4)

(1, -2)

Slope-intercept form

31

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:

The graph of the equation

y – 3 = - (x – 4) is a line

of slope m = - passing

through the point (4, 3).

1

2 1

2

(4, 3)

m = -1

2

x

y

4

4

8

8


32

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


33

Example: Write the slope-intercept form for the equation of the line through the points (4, 3) and (-2, 5).

y – y1 = m(x – x1) Point-slope form

Slope-intercept formy = - x + 13

31

3

2 1 5 – 3 -2 – 4

= - 6

= - 3

Calculate the slope.m =

Use m = - and the point (4, 3).y – 3 = - (x – 4)1

3 3

1


34

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.

Example: The lines y = 2x – 3 and y = 2x + 4 have slopes m1 = 2 and m2 = 2.

The lines are parallel.

x

y

y = 2x + 4

(0, 4)

y = 2x – 3

(0, -3)


35

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

The lines are perpendicular.

1m1

m2= - or m1m2 = -1. y = 3x – 1

x

y

(0, 4)

(0, -1)

y = - x + 41

3Example:

The lines y = 3x – 1 and

y = - x + 4 have slopes

m1 = 3 and m2 = - .

1

3 1

3


Reciprocals = مقلوب

36

x

C y

ExampleA 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

produced costs $100, the variable cost of producing

X items is 100X. Thus the total daily cost C of

production is C = 100x + 2000

The graph of this equation is given by the line

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 each axis.

1 2 3 4 5

220021002000D

olla

rs

No. of items

37

Solved problems

1- Check whether the graphs of the equations given below represent parallel lines. Explain your answer.line a: y = –5x + 3line b: y + 5x = –2 Solution

line a: y = –5x + 3line b: y = –5x – 2 Identify the slope of each equation.For line a: y = –5x + 3,Slope = –5For line b: y = –5x – 2Slope = –5 Line a and line b have equal slope, –5.Parallel lines have equal slope. So, line a and line b are parallel lines.

38

Solved problems

2- Check whether the graphs of the equations given below represent parallel lines. Explain your answer.line a: y = x + 7line b: x – y = –2

Solution

line a: y = x+ 7line b: y = x + 2 Identify the slope of each equation.For line a: y = x + 7,Slope = 1For line b: y = x + 2Slope = 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.

39

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).


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.

40

Solved problems

5- Find the slope of the line that passes through the points (7 , 4) and (-9 , 4).


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 the points are not collinear.

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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.

Solved problems

42

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.

Solved problems

43

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?


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?


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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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

Solved problems

47

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

Solved problems

Chapter one Linear Equations. Linear Equations Definition of a Linear Equation A linear equation in...

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