Linear equations Lecture for Maths1.pptx

Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 2.1 - 1

Applied Mathematics for
Business and Social
Sciences (8405)

Chapter No. 2
Linear Equations

Linear Equations in One Variable
Objectives
1. Decide whether a number is a solution of a linear
equation.
2. Solve linear equations using the addition and
multiplication properties of equality.
3. Solve linear equations using the distributive
property.
4. Solve linear equations with fractions or decimals.
5. Linear equation in two variables.
6. Solving simultaneous equations.
7. Slope and intercept.
8. Representing lines in slope intercept form.

Using Linear Equations of One Variable
Algebraic Expressions vs. Equations
algebraic expressions: are the statements without any
sign of =, > or <
– 9y + 5, 10k, and 2 5
7
a
b c
-
Equations are statements that two algebraic
expressions are equal:
3x – 13 = 29, 2 + y = – 11, and 3m = 4m – 2
An equation always contains an equals sign, but an
expression does not.

Linear Equation in One Variable
A linear equation is also called a first-degree
equation since the greatest power on the
variable is one.
5x + 10 = 13
A linear equation in one variable can be written in the
form ax + b = c
where A, B, and C are real numbers, with A = 0.
/

Determine whether the following equations are
linear or nonlinear.
8x + 3 = –9
9x3
– 8 = 15
x
7 = –12
4 16
 
x
Yes, x is raised to the first power.
No, x is not raised to the first
power.
power.
power.

2.1 Using Linear Equations of One Variable
Deciding Whether a Number is a Solution
If a variable can be replaced by a real number that
makes the equation a true statement, then that
number is a solution of the equation, x – 10 = 3.
x – 10 = 3
13
13 – 10 =
3
x – 10 = 3
8
8 – 10 = 3
(true) (false)
13 is a solution 8 is not a solution

Finding the Solution Set of an Equation
An equation is solved by finding its solution set
– the set of all solutions.
The solution set of x – 10 = 3
is {13}.
Equivalent equations are equations that have
the same solution set. These are equivalent
equations since they all have solution set {–3}.
3x + 5 = –4 3x = –9 x = –3

Solving Linear Equations
An equation is like a balance scale, comparing
the weights of two quantities.
Expression-1 Expression-2
We apply properties to produce a series of simpler
equivalent equations to determine the solution set.
Variable Solution
=
=

C
Addition Property of Equality
The same number may be added to both sides of
an equation without changing the solution set.
A =
=
C
+
A = B
+
A B
B

C
Multiplication Property of Equality
Each side of an equation may be multiplied by
the same nonzero number without changing the
solution set.
A =
=
C
A = B
A B
B

Addition and Multiplication Properties of
Equality
For all real numbers A, B, and C, the equation
A = B and A + C = B + C
are equivalent.
Addition Property of Equality
For all real numbers A, B, and for C = 0, the equation
A = B and A C = B C
are equivalent.
Multiplication Property of Equality
/

Addition and Multiplication Properties of
Equality
Because subtraction and division are defined in
terms of addition and multiplication,
we can extend the addition and multiplication
properties of equality as follows:
The same number may be subtracted from each side of an
equation, and each side of an equation may be divided by
the same nonzero number, without changing the solution
set.

Solving Linear Equations in One Variable
Step 1 Clear fractions. Eliminate any fractions by multiplying
each side by the least common denominator.
Step 2 Simplify each side separately. Use the distributive
property to clear parentheses and combine like terms
as needed.
Step 3 Isolate the variable terms on one side. Use the
addition property to get all terms with variables on one
side of the equation and all numbers on the other.
Step 4 Isolate the variable. Use the multiplication property to
get an equation with just the variable (with coefficient
of 1) on one side.
Step 5 Check. Substitute the proposed solution into the
original equation.

Solve 3x + 2 = 10.
3x + 2 = 10
3x + 2 – 2 = 10 – 2
3x = 8
Subtract 2.
Combine like terms.
Divide by 3.
Proposed solution.
3 8
3 3

x
8
3

x

3x + 2 = 10
3 • + 2 = 10
3
8 Check by substituting the proposed
solution back into the original equation.
8 + 2 = 10
Since the value of each side is 10, the
proposed solution is correct.
The solution set is
8
3
 
 
 
.

Solve 2x – 5 = 5x – 2.
2x – 5 = 5x – 2
2x – 5 – 5x = 5x – 2 – 5x
–3x – 5 = –2
Subtract 5x.
Combine like terms.
Add 5.
Divide by –3.
–3x – 5 + 5= –2 + 5
–3x = 3 Combine like terms.
x = –1 Proposed solution.
3 3
3 3



x

2x – 5 = 5x – 2
Check by substituting the
proposed solution back into
the original equation.
–2 – 5 = –5 – 2
Since the value of each
side is –7 , the proposed
solution is correct.
The solution set is {–1}.
2(–1) – 5 = 5(–1) – 2
–7 = –7

Solve 5(2x + 3) = 3 – 2(3x –
5).
5(2x + 3) = 3 – 2(3x – 5)
10x + 15 = 3 – 6x + 10
10x + 15 – 15 = 3 – 6x + 10 –
15
10x = – 6x – 2
10x + 6x = –6x – 2 + 6x
16x = –2
Distributive Prop.
Add –15.
Collect like terms.
Add 6x.
Collect like terms.

Sec 2.1 - 21
Divide by 16.
1
8

x Proposed solution.
16x = –2
16 2
16 16


x

Check proposed solution:
   
5 2 3 3 2 3 5 22 43
5 3 2
8 8
110 86
5 2 3 3 2 3 5 3
8
110 110
Checks
8 8
The solution set se
8
2 3 110 24 86
5 3 3 2 5
8 8 8 8
1 1
8 8
t is
8
2 24 3 40
5 3 2
8 8 8 8
       
  
   
   
   
   
     
   
   
   
   
   
       
   
   
   
     
   
   

 
x x
1
.
8
 

 
 

Solving Linear Equations with Fractions
Solve 2 1 1 3
2 3 4
 
 
x x
.
   
Clear fractions.
Distributive property
2 1 1 3
12 12
2 3 4
6 2 1 4 3 3
1
.
Distributive property.
A
2 6 4 3 9
12 dd 3 .
6 4 3
3 3
9
 
   
 
   
   
   
   
    
 
x x
x x
x x
x
x x x
x

Solving Linear Equations with Fractions
Collect like terms.
Add 6.
Coll
12 6 4 3 9 3
9
ect like terms
Divide by 9.
Proposed solu
6 5
9 6 5
9 1
tion
1
9 11
11
6 6
9
9
.
9
     
 
 

 




x x x x
x
x
x
x
x
continued

Solving Linear Equations with Decimals
Solve ( )
1.5 2 2.8
x x
+ = + .
 
  Multiply by 10.
Add 10 .
Collect like terms.
Add 30.
1.5 2 2.8
15 2 28 10
15 30 28 10
15 30 28 10
5 30 28
10
5 30
1
8
30 3
2
0
0
  
  
  
   

 
 
 


x x
x x
x x
x
x
x
x
x
x
x

Solving Linear Equations with Decimals
Collect like terms.
Divide by 5.
Proposed solution.
30
2
The
5 30 28
5 2
5 2
2
solution set is
5
30
5 5
.
5
 

 
 
 




 

x
x
x
x
continued

Conditional, Contradiction, and Identity
Equations
Linear equations can have exactly one solution,
no solution, or an infinite number of solutions.
Type of Linear
Equation
Number of Solutions Indication When
Solving
Conditional One Final results is
x = a number.
Identity Infinite; solution set
{all real numbers}
Final line is true,
such as 5 = 5.
Contradiction None; solution set is Final line is false,
such as –3 = 11.
.


Equations
A contradiction has no solutions.
Adding 7.
Collecting like terms
Sol 7 2.
7 2
7 2
.
Add .
Col
5
5
0 5 lecting like term .
7
s
ve
7

  
 

 

  
 






x x
x x
x x
x x
x x
x
x
x
Since 0 = –5 is never true, and this equation is
equivalent to x + 7 = x + 2, the solution set is
empty.

Equations
An identity has an infinite number of solutions.
 
 
Adding 2.
Collecting like terms.
Adding 2 .
Collecting like
2
Solv
te
e 2 2 1 .
2 2 2 1
2 2 2 2
2 2 2 2
2
2
2
2 2
rms.
2
0 0
2 2
 

  
  


 
  
 



x x
x x
x x
x x
x x
x x
x
x x
Since 0 = 0 is always true, and this equation is
equivalent to 2x + 2 = 2(x + 1), the solution set is
all real numbers.

LINEAR EQUATION: A Linear Equation is
an algebraic equation in which terms are
a constants or the product of a constants
and variables. Linear Equations can have
one or more variables.Ex:2x-3=5(linear
equation in one variable)
Ex:2x+3y=7(linear equation in two
variables)

Representing a
linear equation
Y=mx + c

Slope of a Line (m)
Slope basically describes the
steepness of a line

If a line goes up from left to right, then
the slope has to be positive
Conversely, if a line goes down from left
to right, then the slope has to be negative

Definitions of Slope
Slope is simply the change in the
vertical distance over the change in
the horizontal distance
1
2
1
2
x
x
y
y
x
y
run
rise
m
slope









1
2
1
2
x
x
y
y
m



The formula above is the one which we
will use to find the slope of specific lines
In order to use that formula we need to
know, or be able to find 2 points on the
line

If a line is in the form Ax + By = C,
we can use the following formula to
find the slope:
B
A
m 


Examples
  
 
3
1
6
2
1
5
4
6
6
,
5
,
4
,
1







m
m
3
2
5
3
2




m
y
x

Horizontal lines have a slope of zero
while vertical lines have no slope
Horizontal
y=
Vertical
x=
m = 0
m = no
slope

The World Of Linear
Equations
Writing Linear Equations
In Slope-Intercept Form
y = mx + b

If you are given:
The slope and y-intercept
 Finding the equation of
the line in y= mx + b
form. Given: slope
and y-intercept. Just
substitute the “m” with
the slope value and the
“b” with the y-intercept
value.
 Slope = ½ and
 y-intercept = -3
y= mx + b
½
-3
y= ½x – 3

If you are given: A Graph
Find the:
 y – intercept = b = the point
where the line crosses the y
axis.
 Slope = = m =
run
rise
s
x'
in
change
s
y'
in
change
 y – intercept = b = -3

Slope = = m = ½
y= mx + b
2
over
1
up ½
-3
y= ½x – 3

If you are given:
The slope and a point
 Given: slope (m)
and a point (x,y).
To write equations
given the slope and
a point using Point-
Slope Form.
 Slope =½ and point (4,-1)
½ 4-1
y= ½x – 3
Point-Slope Form
1 1
y y m(x x )
  
1 1
y y m(x x )
  
   
4
2
1
1 


 x
y
 
4
2
1
1 

 x
y
2
2
1
1 

 x
y
-1 -1

If you are given:
Two points
 Finding the equation
of the line in y= mx +
b form. Given: Two
points. First find the
slope (m) and then
substitute one of the
points x and y values
into Point-Slope Form.
1 1
y y m(x x )
  
Point-Slope Form
Point (-2, -4) & Point (2, -2)
Find the:
 Slope = = m =
s
x'
in
change
s
y'
in
change
run
rise
 
 






2
2
4
2




2
2
4
2
2
1
4
2

 Slope =½ and point (2, -2)
1 1
y y m(x x )
  
½ -2
2
   
2
2
1
2 


 x
y
1
2
1
2 

 x
y
-2 -2
y= ½x – 3

Write the equation of a line that has
a y-intercept of -3 and a slope of -4.
1. y = -3x – 4
2. y = -4x – 3
3. y = -3x + 4
4. y = -4x + 3

Write an equation of the line that goes
through the points (0, 1) and (1, 4).
1. y = 3x + 4
2. y = 3x + 1
3. y = -3x + 4
4. y = -3x + 1

To find the slope and y-intercept of an
equation, write the equation in slope-
intercept form: y = mx + b.
Find the slope and y-intercept.
1) y = 3x – 7
y = mx + b
m = 3, b = -7

2) y = x
y = mx + b
y = x + 0
3) y = 5
y = mx + b
y = 0x + 5
2
3
m =
b = 0
2
3
2
3
m = 0
b = 5

-3 -3 -3
4) 5x - 3y = 6
Write it in slope-intercept form. (y = mx + b)
5x – 3y = 6
-3y = -5x + 6
y = x - 2
5
3
m =
b = -2
5
3

Write it in slope-intercept form. (y = mx + b)
2y + 2 = 4x
2y = 4x - 2
y = 2x - 1
5) 2y + 2 = 4x
2 2 2
m = 2
b = -1

Find the slope and y-intercept of
y = -2x + 4
1. m = 2; b = 4
2. m = 4; b = 2
3. m = -2; b = 4
4. m = 4; b = -2

Linear equations Lecture for Maths1.pptx

Linear equations Lecture for Maths1.pptx

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Linear equations Lecture for Maths1.pptx