Volume 04 Issue 12-2024
64
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
12
Pages:
64-71
OCLC
–
1368736135
A
BSTRACT
This article discusses the process of working with algebraic expressions and preparing students for
understanding equations during primary school mathematics lessons. It highlights the teaching methods
for introducing students to variable expressions and solving problems involving variables.
K
EYWORDS
Algebraic expressions, variable, expressions with variables, equality, equation, root of an equation.
I
NTRODUCTION
In Uzbekistan, significant reforms have been
implemented in recent years as part of the Action
Strategy and its logical continuation, the
Development Strategy. These reforms aim to
continuously improve the system of lifelong
education, provide quality education, and train
qualified personnel, aligning with the practices of
developed countries.
As part of the 2022
–
2026 Development Strategy
of New Uzbekistan, mechanisms have been
developed to expand access to quality education
for children with special needs. Inclusive
education has been introduced as a social norm in
educational
institutions,
ensuring
equal
opportunities for all children.
Journal
Website:
http://sciencebring.co
m/index.php/ijasr
Copyright:
Original
content from this work
may be used under the
terms of the creative
commons
attributes
4.0 licence.
Research Article
TEACHING TO SOLVE ALGEBRAIC PROBLEMS IN PRIMARY
SCHOOL MATHEMATICS LESSONS
Submission Date:
December 02,
2024,
Accepted Date:
December 07, 2024,
Published Date:
December 12, 2024
Crossref doi:
https://doi.org/10.37547/ijasr-04-12-11
Gofurova Mahfuza Abbosovna
A lecturer at Fergana State University, Doctor of Philosophy (PhD) in Pedagogical Sciences, Uzbekistan
Ergasheva Nigora
A second-year student at Fergana State University, Uzbekistan
Volume 04 Issue 12-2024
65
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
12
Pages:
64-71
OCLC
–
1368736135
In the primary mathematics curriculum, the
gradual preparation of students to understand
the concept of variables is envisaged (although
the term itself is not used in primary grades). This
process begins in Grade 1, where students solve
"box problems" (e.g., 5+
□
=75 +
□
= 7 or
□
+
□
=4
□
+
□
= 4). In Grade 3, students are introduced to
simple equations where letters represent
unknown numbers. Additionally, students
encounter letters in expressions where they act as
variables capable of taking on different numerical
values.
By the end of primary school, students should:
1.
Understand the meaning of basic algebraic
expressions.
2.
Calculate the value of expressions by
substituting specific values for variables.
3.
Solve inequalities, such as determining for
which values of aa the inequality 3+a>73 + a > 7
holds true.
Introducing elements of symbolic notation helps
students grasp the concept of variables and
prepares them to use algebraic methods for
problem-solving. In Grades 3
–
4, special attention
is given to solving problems with letters,
constructing
and
comparing
algebraic
expressions, and solving equations step-by-step.
Teaching Equations:
The process of solving equations involves
revisiting earlier material to demonstrate the
relationship between components and results in
addition and subtraction. Visual aids are often
used to reinforce these relationships. For
example:
•
Students solve equations like x+48=90x +
48 = 90 and x−27=33x
- 27 = 33.
•
They learn to represent unknown
numbers using letters (e.g., aa, bb, cc, etc.).
Teaching Multiplication and Division:
In topics like "Multiplication and Division,"
students first explore the relationship between
components and results in multiplication,
followed by division.
Example:
•
Place 2 circles on a desk 4 times.
•
"How many circles are there in total?"
(Answer: 8).
•
"What do the numbers in this example
represent?" (Multiplier, multiplicand, product).
•
Students
then
formulate
division
examples based on the arrangement: 8÷2=48 \div
2 = 4 and 8÷4=28 \div 4 = 2. They compare these
to the multiplication example and generalize the
rule: "If we divide the product by one factor, the
other factor is obtained."
This approach helps students gradually
understand the relationships between operations
and use these relationships to solve equations
effectively. For example, when solving certain
equations, students analyze the relationships
between unknown numbers.
Volume 04 Issue 12-2024
66
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
12
Pages:
64-71
OCLC
–
1368736135
By introducing these methods in primary school,
students build a strong foundation for
understanding algebraic concepts, which will be
further developed in higher grades.
1
X+280=530
X=530-280
X=250
The addend is unknown. To find the unknown
addend, subtract the known addend from the sum.
2
у
- 340 = 260
у
= 340 + 260
X= 600
The minuend is unknown. To find the unknown
minuend, add the subtrahend to the difference.
3
350
- Z -
190
Z = 350 - 190
Z =
160
The subtrahend is unknown. To find the unknown
subtrahend, subtract the difference from the
minuend.
4
70*a = 560
a = 560:70
a =
8
The multiplier is unknown. To find the unknown
multiplier, divide the product by the known
multiplicand.
5
b :
230=4
b=230*4
b=920
The dividend is unknown. To find the unknown
dividend, multiply the divisor by the quotient.
6
900:
с=
50
с=900:50
с
= 18
The divisor is unknown. To find the unknown
divisor, divide the dividend by the quotient.
Solving Example 1:
Given the equation 27+x=2727 + x = 2727+x=27,
the unknown number to be found is represented
by the letter xxx.
To solve the equation, determine which value of
xxx makes the equality true.
The number is 000, because adding 000 to
272727 results in 272727.
Volume 04 Issue 12-2024
67
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
12
Pages:
64-71
OCLC
–
1368736135
The solution of the equation is written as follows:
27+x=2727 + x = 2727+x=27
Solution:
x=27−27x = 27
-
27x=27−27 x=0x = 0x=0
Verification:
27+0=2727 + 0 = 2727+0=27 27=27(Equality is
true).27 = 27 \quad \text{(Equality is
true).}27=27(Equality is true).
Solving Example 2 (written):
a) To find the unknown addend, subtract the
known addend from the sum:
Unknown
addend=Sum−Known
addend\text{Unknown addend} = \text{Sum} -
\text{Known
addend}Unknown
addend=Sum−Known addend
7+8=15
15-7=8
15-8=
b) Introducing the Method to Find the Unknown
Addend
Example 3 (written): Solving equations and
verifying the result.
35+x =70
14+x =24
25+x=50
x=70-35
x=24-14
x
=
50-25
x=35
x= 10
x=25
Tek. 35 + 35 = 70
Tek. 14+10 = 24
Tek.
25 + 25 = 50
70 = 70
24 = 24
50 = 50
Using equations to solve problems simplifies
many tasks. Solving such problems typically
consists of two steps:
Formulating an equation based on the
problem's conditions.
Solving the resulting equation.
Volume 04 Issue 12-2024
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International Journal of Advance Scientific Research
(ISSN
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2750-1396)
VOLUME
04
ISSUE
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Pages:
64-71
OCLC
–
1368736135
Let us solve the following problem using this
method:
Problem:
A steamboat carrying tourists departs from a
riverside station downstream and must return
after 5 hours. The river's current speed is 3 km/h,
and the steamboat's speed in still water is 18
km/h. If the tourists rest on the shore for 3 hours
before returning, how far did the steamboat
travel from the station?
Solution:
1.
Let the distance traveled downstream be xxx
kilometers.
o
The
steamboat
travels
this
distance
downstream at a speed of 18+3=2118 + 3 =
2118+3=21 km/h, taking
𝑥
21
hours.
o
The steamboat returns upstream at a speed of
18−3=1518
-
3 = 1518−3=15 km/h, taking
𝑥
15
hours.
o
The tourists rest on the shore for 3 hours.
Thus, the total time of the trip is:
(
𝑥
21
+
𝑥
15
+ 3
) hours.
According to the problem, the total trip
time is 5 hours.
Equation:
𝑥
21
+
𝑥
15
+ 3 = 5
2) Now solving the resulting equation:
𝑥
21
+
𝑥
15
= 2
By multiplying both sides of the equation by
105 (the least common multiple of 21 and 15),
we obtain:
5x+7x=210,12x=2105x + 7x = 210,
\quad 12x = 2105x+7x=210,12x=210
From this, x=17.5x = 17.5x=17.5.
Thus, the steamboat traveled 17.5 km from the
station.
In the first stage of solving the problem
(formulating the equation), it was necessary to
understand that the speed of the steamboat and
the river's current are added when moving
downstream and subtracted when moving
upstream. The time taken is equal to the distance
divided by the speed.
In the second stage (solving the equation), it was
required to apply the properties of equations
studied in previous sections.
The correctness of the solution can be verified
using the problem's conditions. By considering
the found result as known, any other given value
Volume 04 Issue 12-2024
69
International Journal of Advance Scientific Research
(ISSN
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2750-1396)
VOLUME
04
ISSUE
12
Pages:
64-71
OCLC
–
1368736135
can be calculated. For example, the correctness of
the solution can be verified as follows:
The tourists traveled downstream from the
station for 17.5÷21=5/617.5 \div 21 =
5/617.5÷21=5/6 hours. They returned upstream
for
17.5÷15=11/617.5
\div
15
=
11/617.5÷15=11/6 hours. Adding these times
and the 3-hour rest period gives a total of 555
hours, as stated in the problem conditions.
By providing tasks with similar content in
primary school mathematics lessons, students
can be taught to formulate equations and develop
equation-solving skills systematically.
In addition to numerical equalities and
inequalities, tasks involving inequalities with
variables can also be given.
Solution:
There
are
x
workers
in
the
first
workshop.
There
are
x∙5
workers
in
the
second
workshop.
There are x+(x+5)=2x +5 workers in the third workshop.
According to the condition:
X+(x∙5)+x+(x∙5)=624
12∙x=624
x=624:12
Volume 04 Issue 12-2024
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International Journal of Advance Scientific Research
(ISSN
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2750-1396)
VOLUME
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Pages:
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OCLC
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1368736135
x=52
Answer:
•
In the first workshop, there are 52 workers.
•
In the second workshop, there are 52
⋅
5=26052 \cdot 5 = 26052
⋅
5=260 workers.
•
In the third workshop, there are 260+52=312260 + 52 = 312260+52=312 workers.
C
ONCLUSION
Working with algebraic material in primary
school mathematics is a crucial process. It
involves exploring concepts such as equality and
inequality, identities and their transformations,
roots of equations, and equivalent equations
through examples. These activities help develop
students' calculation skills.
The primary school age is an active phase in the
development of logical thinking. During this
period, children acquire the foundations for
analysis, synthesis, generalization, restriction,
classification, comparison, abstraction, and other
logical operations. These skills form the basis for
successfully mastering the general education
curriculum.
Regularly incorporating tasks and exercises
aimed at developing logical thinking in
mathematics lessons enables young learners to
approach even the simplest laws of daily life with
greater confidence and actively apply their
mathematical knowledge in practice.
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Volume 04 Issue 12-2024
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International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
12
Pages:
64-71
OCLC
–
1368736135
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