Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Try These (Text Book Page No. 51)

Question 1.
Identify the variable and constants among the following terms.
a, 11 – 3x, xy, -89, -m, -n, 5, 5ab, -5 3y, 8pqr, 18, -9t, -1, -8
Solution:
Variable : a, -3x, xy, -m, -n, 5ab, 3y, -9t, 8pqr
Constants : 11, -89, 5, -5, 18, -1, -8

Question 2.
Complete the following table.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 80
Solution:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 2

Try this (Text book Page No. 53)

Question 1.
Can we use the operations multiplication and division to combine terms?
Solution:
No, We can use addition and subtraction to combine terms.
If we use multiplication or division to combine then it become a single term.
Eg : xy, \(\frac{x}{y}\) are monomials.

Try This (Text book Page No. 54)

Question 1.
Complete the following table by forming expressions using the terms given. One is done for you.
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 85
Solution:
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 3

Try this (Text book Page No. 56)

Question 1.
Identify the like terms among the following and group them.
7xy, 19x, 1, 5y, x, 3yx, 15, -13y, 6x, 12xy, -5, 16y, -9x, 15xy, 23, 45y, -8y, 23x, -y, 11
Solution:
7xy, 3yx, 12xy, 15xy, are like terms
19x, x, 6x,-9x, 23x, are like terms
5y, -13y, 16y, 45y, -8y, -y, are like terms
1, 15, -5, 23, 11, are like terms.

Try This (Text book Page No. 57)

Question 1.
Try to find the value of the following expressions if p = 5 and q = 6.
(i) p + q
(ii) q – p
(iii) 2p + 2 > q
(iv) pq – p – q
(v) 5pq – 1
Solution:
(i) Given p = 5; q = 6
p + q = 5 + 6 = 11
(ii) q – p = 6 – 5 = 1
(iii) 2p + 2 > q = 2(5) + 3(6) = 10 + 18 = 28
(iv) pq – p – q = (5) (6) – 5 – 6 = 30 – 5 – 6 = 25 – 6 = 19

Exercise 3.2

Try These (Text book Page No. 59)

Question 1.
Add the terms
(i) 3p, 14p
(ii) m, 12m, 21m
(iii) 11abc, 5abc
(iv) 12y, -y
(v) 4x, 2x, -7x.
Solution:
(i) 3p + 14p = 17p
(ii) m + 12m + 21m = (1 + 12 + 21 )m
= 34 m
(iii) 11abc + 5abc = (11 + 5) abc
= 16 abc
(iv) 12y + (-y) = (12 + (-1))y
= (12 – 1 )y
= 11y
(v) 4x + 2x + (-7x) = (4 + 2+(-7))x
= (6 + (-7))x
= -1x

Ty this (Text Book Page No. 60)

Question 1.
3x; + (y – x) = 3x + y – x, but 3x – (y – x) ≠ 3x – y – x. why ?
Solution:
In the first case
LHS = 3x + (y – x) = 3x + y – x = 3x – x + y = (3 – 1)x + y
= 2x + y
RHS = 3x + y – x = 2x + y
LHS = RHS ⇒ 3x + (y – x) = 3x + y – x
But in the second case
LHS = 3x – (y – x) = 3x – y + x
= (3 + 1)x – y = 4x – y
RHS = 3x – y – x = 3x – x – y
LHS ≠ RHS
∴ 3x – (y – x) ≠ 3x – y – x

Try this (Page No. 1)

Question 1.
What will you get if twice a number is subtracted from thrice the same number?
Solution:
Let the unknown number be x.
Twice the number = 2x.
Thrice the number = 3x.
Twice the number is subtracted from thrice the number = 3x – 2x = (3 – 2)x = x

Exercise 3.3

Try These (Text book Page No. 65)

Question 1.
Try to construct algebraic equations for the following verbal statements.

Question 1.
One third of a number plus 6 to 10.
Solution:
\(\frac{1}{3}\) + 6 = 10

Question 2.
The sum of five times of x and 3 is 28
Solution:
5 (x + 3) = 28

Question 3.
Taking away 8 from y gives 11
Solution:
y – 8 = 11

Question 4.
Perimeter of a square with side a is 16 cm.
Solution:
4 × a = 16

Question 5.
Venkat’s mother’s age is 7 years more than 3 times venkat’s age. His mother’s age is 43 years.
Solution:
3x + 7 = 43, where x is venkat’s age.

Try this (Text book Page No. 65)

Question 1.
Why should we subtract 5 and not some other number ? why don’t we add 5 on both sides? Discuss.
Solution:
Given x + 5 = 12
(i) Our aim is to find the value of x. Which means we have to eliminate the other values from LHS. Since 5 is given with x it should be subtracted.
(ii) If we add 5 on both sides we cannot eliminate the numbers from LHS and we get x + 10.

Try this (Text book Page No. 66)

Question 1.
If the dogs, cats and parrots represents unknown find them. Substitute each of the values so obtained in the equations and verify the answers.
Solution:
(i) 1 dog + 1 dog + 1 dog = 24
3 dog = 24
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 95
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 62

(ii) 1 dog + 1 cat + 1 cat = 14
1 dog + 2cat = 14
8 + 2cat = 14
2cat = 14 – 8
2 cat = 6
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions 63ditional Questions 63″ width=”107″ height=”87″ />

(iii) 1 dog + 1 cat – 1 parrot = 9
8 + 3 – 1 parrot = 9
8 + 3 – 9 = 1 parrot
11 – 9 = 1 parrot
2 = 1 parrot
1 parrot = 2

(iv) 1 dog + 1 cat + 1 parrot = ?
8 + 3 + 2 = 13
Verification:
(i) 8 + 8 + 8 = 24
(ii) 8 + 3 + 3 = 14
(iii) 8+ 3 – 2 = 9
(iv) 8 + 3 + 2 = 13

Try These (Text book Page No. 68)

Question 1.
Kandhan and kaviya are friends. Both of them are having some pen. Kandhan: If you give me one pen then, we will have equal number of pens. Will you? Kaviya: But, if you give me one of your pens, then mine will become twice as yours. Will you?
Construct algebraic equations for this situation, can you guess and find the actual number of pens, they have?
Solution:
Let the number of pens initially Kandhan and Kaviya had be x and y respectively.

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