Learning Objectives in Mathematics Education
Identify key words and phrases, translate sentences to
mathematical equations, and develop strategies to
solve problems.
Solve word problems involving relationships between num
bers.
Solve geometric problems involving perimeter.
Solve percentage and money problems including simple int
erest.
Set up and solve uniform motion problems.
It’s
a
Mathematical
Magic
Ok, So Today we will learn about some more mathematical
tricks….
6
A= 5
B= 8
X= 12
A= #
B= ##
X= *#
I
I
I
There are two circles, In first circle the value of A,B and X are specifi
ed but in second circle it
is
unknown
.
A letter or symbol that stands
for a known number is
called
Constant
.
EXAMPLE OF LINER EQUATIONS :
x + 4 = - 2
2x + 5 = 10
5 – 3x = 8
1)
INVERSE OPERATION METHOD
2)
TRANSPOSITION METHOD
LINEAR EQUATION FORMATION ACTIVITY VIDEO
INVERSE OPERATION METHOD
INVERSE OPERATION METHOD VIDEO
The linear equations in one variable can be solved
mathematically in a systematic method by the inverse operations.
•
In this method, both sides of the linear equation are
balanced by the basic mathematical operations
inversely for making variable to appear at one side
and its equivalent quantity to appear at the other Side
of the equation.
The inverse operations method is always recommendable and
better than the other method.
Addition form
If a variable is connected to a number by addition at one side
of the equation, then use opposite operation subtraction in
both sides of the linear equation for eliminating the number
from one side of the equation completely.
x+9=14
In the left-hand side expression, the number 9 is added to
variable x. It should be eliminated from this expression to
find the value of x. So, let us subtract 9 from the left-hand
side expression . But it makes the right-hand side expression
imbalanced. Therefore, subtract both sides of the equation
by 9 for making the equation systematic and balanced.
⟹
x+9
−
9=14
−
9
⟹
x+9
−
9=5
∴
x=5
Subtraction form
If a variable is connected to a number by subtraction at one
side of the equation, then use its inverse operation addition at
both sides of the
linear equation for
eliminating
the number
completely from one side of
the equation.
x
−
5=11
The number 5 is subtracted from variable x. In order to
solve x, it must be eliminated from left-hand side expression
and it is possible by adding 5 to the expression but the
expression in the right-hand side become imbalanced due to
this operation. However, It can be balanced
by adding 5 to both sides of the equation.
⟹
x
−
5+5=11+5
⟹
x
−
5+5=16
∴
x=16
Multiplication form
If a variable is connected to a number by multiplication at
one side the linear equation, then try its inverse operation
division at both left and right-hand sides of the equation for
solving the variable.
4x=24
In this example, the number 4 is multiplied to variable x.
For finding the value of x, it’s essential to eliminate the
number 4 from the expression. It is usually done by the
division but it imbalances the right-hand side of the
equation. So, divide both sides the equation by the
coefficient of the variable.
⟹
4x/4=24/4
⟹
4x/4=24/4
∴
x=6
Division form
If a variable is connected to a number by division in one
side of the linear equation, then apply its inverse operation
multiplication in both the sides of the mathematical
equation for evaluating the variable.
x/8=2
The number 8 divides the variable x in this example. To
solve for the value of x, it is
necessary
to eliminate the
number 8
from the expression. It can be done by the
multiplication and it
im
balances the right-hand side
expression. Therefore, it is essential to multiply 8 to both
sides the equation
.
⟹
8
×
x/8=8
×
2
⟹
8x/8=16
⟹
8x/8=16
∴
x=16
TRANSPOSITION METHOD
T
R
A
N
S
P
O
S
E
M
E
T
H
O
D
P
D
F
By transposing a term from one side to ano
ther side, we mean changing its sign and c
arrying it to the other side. In transpositio
n, the plus sign of the term changes into mi
nus sign on the other side and vice –versa.
The transposition method involves the follo
wing steps:
Obtain the linear equation.
TRANSPOSE METHOD VIDEO
•
Identify the variable and constants.
•
Simplify the L.H.S. and R.H.S. to their simplest forms
by removing bracket.
•
Transpose all terms containing variable on L.H.S. an
d constant term on R.H.S.
•
•
Note that the sign of the terms will change in
shifting them from L.H.S. to R.H.S. and vice-versa.
•
Simplify L.H.S and R.H.S. in the simplest form so
that each side contains just one term.
Solve the equation obtained in Step V by
dividing both sides by the coefficient of the variable o
n L.H.S.
EXAMPLE OF
TRANSPOSITION METHOD
CHECKING OF LINEAR EQUATION
CHECKING OF LINEAR EQUATION
WORD PROBLEMS
Example 1:
Translate: Four less than twice some
n
u
m
b
e
r
i
s
1
6
.
Solution: First, choose a variable for the unknown
n
u
m
b
e
r
a
n
d
i
d
e
n
t
i
f
y
t
h
e
k
e
y
w
o
r
d
s
a
n
d
p
h
r
a
s
e
s
.
Let
x
represent the unknown indicated by
“some number.”
2
x
-
4
=
1
6
Remember that subtraction is not commutative.
For this reason, take care when setting up
differences.
In this example,
4−2x=16
is an incorrect translation.
Answer:
2x−4=16
Example 2:
The difference between two integers is 2. The larger
integer is
6 less than twice the smaller. Find the integers.
Solution:
Use the relationship between the two integers in the second
sentence, “The larger integer is
6 less
than
twice the smaller
,”
to identify the unknowns in terms of one variable.
Let x represent the smaller integer
Let 2x-6 represent the larger integer
Since the difference is positive, subtract the smaller integer from
the larger.
(2x-6)-x=2
Solve.
2x-6-x=2
X-6=2
X-6+6=2+6
X=8
Use 2
x
− 6 to find the larger integer.
2x-6=2(8)-6=16-6=10
Answer: The two integers are 8 and 10. These integers clearly solve
the problem.
E
x
a
m
p
l
e
-
3
T
h
e
d
i
f
f
e
r
e
n
c
e
b
e
t
w
e
e
n
t
h
e
t
w
o
n
u
m
b
e
r
s
i
s
4
8
.
T
h
e
r
a
t
i
o
o
f
t
h
e
t
w
o
n
u
m
b
e
r
s
i
s
7
:
3
.
W
h
a
t
a
r
e
t
h
e
t
w
o
n
u
m
b
e
r
s
?
S
o
l
u
t
i
o
n
:
L
e
t
t
h
e
c
o
m
m
o
n
r
a
t
i
o
b
e
x
.
Their difference = 48
According to the question,
7x - 3x = 48
⇒
4x = 48
⇒
x = 48/4
⇒
x = 12
Therefore, 7x = 7 × 12 = 84
3x = 3 × 12 = 36
Simplify the process of solving real-world problems by creating
mathematical models that describe the relationship among unknowns
se algebra to solve the resulting equations.
Guessing and checking for solutions is a poor practice. This
technique might sometimes produce correct answers, but is unreliable,
especially when problems become more complex.
Read the problem several times and search for the key words
and phrases. Identify the unknowns and assign variables or expressions to
the unknown quantities. Look for relationships that allow you to use only
one variable. Set up a mathematical model for the situation and use
algebra to solve the equation. Check to see if the solution makes
sense and present the solution in sentence form.
Do not avoid word problems: solving them can be fun and rewarding
. With lots of practice you will find that they really are not so bad after all.
Modelling and solving applications is one of the major reasons to study
algebra.
Do not feel discouraged when the first attempt to solve a word
problem does not work. This is part of the process. Try something different
and learn from incorrect attempts.
ADVICE FOR STUDENTS
The learning objectives in this mathematics course include identifying key words, translating sentences into mathematical equations, and developing problem-solving strategies. Students will solve word problems involving relationships between numbers, geometric problems with perimeter, percentage and money problems, and uniform motion problems. The content covers the use of variables and constants in equations, and examples of linear equations are provided. The course also introduces mathematical tricks and concepts like symbolizing unknown numbers with variables.
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Presentation Transcript
SHARIQUE MINHAZ PRT MATHEMATICS
LEARNING OBJECTIVES LEARNING OBJECTIVES Identify key words and phrases, translate sentences to Identify key words and phrases, translate sentences to mathematical equations, and develop strategies to mathematical equations, and develop strategies to solve problems. solve problems. Solve word problems involving relationships between num Solve word problems involving relationships between num bers. bers. Solve geometric problems involving perimeter. Solve geometric problems involving perimeter. Solve percentage and money problems including simple int Solve percentage and money problems including simple int erest. erest. Set up and solve uniform motion problems. Set up and solve uniform motion problems.
WHAT DO YOU UNDERSTAND BY WHAT DO YOU UNDERSTAND BY THIS ? THIS ? It s It s a a Mathematical Mathematical Magic Magic Ok, So Today we will learn about some more mathematical tricks .
What do you observe in both the pictures ? What do you observe in both the pictures ? A= 5 B= 8 X= 12 A= # B= ## X= *# I II There are two circles, In first circle the value of A,B and X are specifi There are two circles, In first circle the value of A,B and X are specifi ed but in second circle it is unknown ed but in second circle it is unknown. 6
A letter or symbol that stands for an A letter or symbol that stands for an unknown number is called unknown number is called Variable Variable. . A letter or symbol that stands A letter or symbol that stands for a known number is for a known number is called called Constant Constant.
A LINER EQUATION IS ALSO CALLED A FIRST DEGR EE EQUATION AS THE HIGHEST POWER OF VARIABLE IS 1. EXAMPLE OF LINER EQUATIONS : x + 4 = - 2 2x + 5 = 10 5 3x = 8
THERE ARE TWO WAYS TO SOLVE A LINEAR EQUATION IN ONE VARIABLE 1) INVERSE OPERATION METHOD 2) TRANSPOSITION METHOD LINEAR EQUATION FORMATION ACTIVITY VIDEO
INVERSE OPERATION METHOD INVERSE OPERATION METHOD INVERSE OPERATION METHOD VIDEO INVERSE OPERATION METHOD VIDEO The linear equations in one variable can be solved The linear equations in one variable can be solved mathematically in a systematic method by the inverse operations. mathematically in a systematic method by the inverse operations. In this method, both sides of the linear equation are In this method, both sides of the linear equation are balanced by the basic mathematical operations balanced by the basic mathematical operations inversely for making variable to appear at one side inversely for making variable to appear at one side and its equivalent quantity to appear at the other Side and its equivalent quantity to appear at the other Side of the equation. of the equation. The inverse operations method is always recommendable and The inverse operations method is always recommendable and better than the other method. better than the other method.
Addition form Addition form If a variable is connected to a number by addition at one side If a variable is connected to a number by addition at one side of the equation, then use opposite operation subtraction in of the equation, then use opposite operation subtraction in both sides of the linear equation for eliminating the number both sides of the linear equation for eliminating the number from one side of the equation completely. from one side of the equation completely. x+9=14 x+9=14 In the left In the left- -hand side expression, the number hand side expression, the number 9 9 is added to variable variable x. It should be eliminated from this expression to x. It should be eliminated from this expression to find the value of find the value of x. So, let us subtract x. So, let us subtract 9 9 from the left side expression . But it makes the right side expression . But it makes the right- -hand side expression imbalanced. Therefore, subtract both sides of the equation imbalanced. Therefore, subtract both sides of the equation by by 9 9 for making the equation systematic and balanced. for making the equation systematic and balanced. x+9 x+9 9=14 9=14 9 9 x+9 x+9 9=5 9=5 is added to from the left- -hand hand side expression hand x=5 x=5
Subtraction form Subtraction form If a variable is connected to a number by subtraction at one side of the equation, then use its inverse operation addition at both sides of the linear equation for eliminating the number completely from one side of the equation. x 5=11 The number 5 is subtracted from variable x. In order to solve x, it must be eliminated from left-hand side expression and it is possible by adding 5 to the expression but the expression in the right-hand side become imbalanced due to this operation. However, It can be balanced by adding 5 to both sides of the equation. x 5+5=11+5 x 5+5=16 x=16
Multiplication form Multiplication form If a variable is connected to a number by multiplication at If a variable is connected to a number by multiplication at one side the linear equation, then try its inverse operation one side the linear equation, then try its inverse operation division at both left and right division at both left and right- -hand sides of the equation for solving the variable. solving the variable. hand sides of the equation for 4x=24 4x=24 In this example, the number In this example, the number 4 4 is multiplied to variable For finding the value of For finding the value of x, it s essential to eliminate the x, it s essential to eliminate the number number 4 4 from the expression. It is usually done by the from the expression. It is usually done by the division but it imbalances the right division but it imbalances the right- -hand side of the equation. So, divide both sides the equation by the equation. So, divide both sides the equation by the coefficient of the variable. coefficient of the variable. 4x/4=24/4 4x/4=24/4 is multiplied to variable x. x. hand side of the 4x/4=24/4 4x/4=24/4 x=6 x=6
Division form Division form If a variable is connected to a number by division in one If a variable is connected to a number by division in one side of the linear equation, then apply its inverse operation side of the linear equation, then apply its inverse operation multiplication in both the sides of the mathematical multiplication in both the sides of the mathematical equation for evaluating the variable. equation for evaluating the variable. x/8=2 x/8=2 The number The number 8 8 divides the variable divides the variable x x in this example. To solve for the value of solve for the value of x, it is x, it is necessary number number 8 8 from the expression. It can be done by the from the expression. It can be done by the multiplication and it multiplication and it im imbalances the right balances the right- -hand side expression. Therefore, it is essential to multiply 8 to both expression. Therefore, it is essential to multiply 8 to both sides the equation sides the equation . . 8 8 x/8=8 x/8=8 2 2 8x/8=16 8x/8=16 8x/8=16 8x/8=16 x=16 x=16 in this example. To necessary to eliminate the to eliminate the hand side
TRANSPOSITION METHOD TRANSPOSITION METHOD TRANSPOSE METHOD PDF By transposing a term from one side to ano ther side, we mean changing its sign and c arrying it to the other side. In transpositio n, the plus sign of the term changes into mi nus sign on the other side and vice versa. The transposition method involves the follo wing steps: Obtain the linear equation. TRANSPOSE METHOD VIDEO
Identify the variable and constants. Simplify the L.H.S. and R.H.S. to their simplest forms by removing bracket. Transpose all terms containing variable on L.H.S. an d constant term on R.H.S. Note that the sign of the terms will change in shifting them from L.H.S. to R.H.S. and vice-versa. Simplify L.H.S and R.H.S. in the simplest form so that each side contains just one term. Solve the equation obtained in Step V by dividing both sides by the coefficient of the variable o n L.H.S.
EXAMPLE OF TRANSPOSITION METHOD
Example 1: Translate: Four less than twice some number is 16. Solution: First, choose a variable for the unknown number and identify the key words and phrases. Let x represent the unknown indicated by some number. 2x-4=16 Remember that subtraction is not commutative. For this reason, take care when setting up differences. In this example, 4 2x=16 is an incorrect translation. Answer: 2x 4=16
Example 2: The difference between two integers is 2. The larger integer is 6 less than twice the smaller. Find the integers. Solution: Use the relationship between the two integers in the second sentence, The larger integer is 6 less than twice the smaller, to identify the unknowns in terms of one variable. Let x represent the smaller integer Let 2x-6 represent the larger integer Since the difference is positive, subtract the smaller integer from the larger. (2x-6)-x=2 Solve. 2x-6-x=2 X-6=2 X-6+6=2+6 X=8 Use 2x 6 to find the larger integer. 2x-6=2(8)-6=16-6=10 Answer: The two integers are 8 and 10. These integers clearly solve the problem.
Example-3The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 4x = 48 x = 48/4 x = 12 Therefore, 7x = 7 12 = 84 3x = 3 12 = 36
ADVICE FOR STUDENTS Simplify the process of solving real-world problems by creating mathematical models that describe the relationship among unknowns se algebra to solve the resulting equations. Guessing and checking for solutions is a poor practice. This technique might sometimes produce correct answers, but is unreliable, especially when problems become more complex. Read the problem several times and search for the key words and phrases. Identify the unknowns and assign variables or expressions to the unknown quantities. Look for relationships that allow you to use only one variable. Set up a mathematical model for the situation and use algebra to solve the equation. Check to see if the solution makes sense and present the solution in sentence form. Do not avoid word problems: solving them can be fun and rewarding . With lots of practice you will find that they really are not so bad after all. Modelling and solving applications is one of the major reasons to study algebra. Do not feel discouraged when the first attempt to solve a word problem does not work. This is part of the process. Try something different and learn from incorrect attempts.
Algebra language Bingo Draw up a 3 X 3 grid and pick 9 of these and fill in your grid X +3 3a - 2 b - 3 4x + 6 3b y-9 2x - 5 g-5 m + n 3(x 2) X - 4 2(a + b) 2k 3x + 6 3 + 5 + 7 4A 2p + 2 Y + 3 mf 6y
