HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

AP Board Class 6 Maths Chapter 7 Introduction to Algebra Unit Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Book Answers

 AP Board Class 6 Maths Chapter 7 Introduction to Algebra Unit Exercise Textbook Solutions PDF: Download Andhra Pradesh Board STD 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Book Answers

Andhra Pradesh State Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Books Solutions

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AP Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Question 1.
The cost of one fan is Rs. 1500. Then what is the cost of ‘n’ fans?

Given cost of one fan = Rs. 1500
Number of fans = n
Cost of n fans = cost of one fan × no. of fans = 1500 × n
∴ Cost of n fans = 1500n

Question 2.
Srinu has number of pencils. Raheem has 4 times the pencils as of Srinu. How many pencils does Rahim has? Write an expression.
Let number of pencils Srinu has = x
Number of pencils Raheem has = 4 times of Srinu
= 4 × x
∴ Number of pencils Raheem has = 4x

Question 3.
Parvathi has 5 more books than Sofia. How many books are with Parvathi? Write an expression choosing any variable for number of books.
Let number of books Sofia has = y
Given Parvathi has 5 more books than Sofia
Number of books Parvathi has = 5 books more than Sofia
= y + 5
∴ Number of books Parvathi has = y + 5

Question 4.
Which of the following are equations?
i) 10 – 4p = 2
ii) 10 + 8x < – 22
iii) x + 5 = 8
iv) m + 6 = 2
v) 22x – 5 = 8
vi) 4k + 5 > – 100
vii) 4p + 7 = 23
viii) y < – 4
i) 10 – 4p = 2
We know that, a mathematical statement involving equality symbol is called an equation.
10 – 4p = 2 has equality symbol.
So, it is an equation.

ii) 10 + 8x < – 22
We know that, a mathematical statement involving equality symbol is called an equation.
10 + 8x < – 22 has no equality symbol.
So, it is not an equation [so it is an inequation]

iii) x + 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
x + 5 = 8 has equality symbol.
So, it is an equation.

iv) m + 6 = 2 We know that, a mathematical statement involving equality symbol is called an equation.
m + 6 = 2 has equality symbol.
So, it is an equation.

v) 22x – 5 = 8 We know that, a mathematical statement involving equality symbol is called an equation.
22x – 5 = 8 has equality symbol.
So, it is an equation.

vi) 4k + 5 > – 100
We know that, a mathematical statement involving equality symbol is called an equation.
4k + 5 > -100 has no equality symbol.
So, it is not an equation.
It is an inequation.

vii) 4p + 7 = 23
We know that, a mathematical statement involving equality symbol is ailed an equation.
4p + 7 = 23 has equality symbol.
So, it is an equation.

viii) y < – 4
We know that, a mathematical statement involving equality symbol is called an equation.
y < – 4 has no equality symbol.
So, it is not an equation.
It is an inequation.

Question 5.
Write L.H.S and R.H.S of the following equations:
i) 7x + 8 = 22
ii) 9y – 3 = 6
iii) 3k – 10 = 2
iv) 3p – 4q = -19
i) 7x + 8 = 22
Given equation is 7x + 8 = 22
LHS = 7x + 8
RHS = 22

ii) 9y – 3 = 6
Given equation is 9y – 3 = 6
LHS = 9y – 3
RHS = 6

iii) 3k – 10 = 2
Given equation is 3k – 10 = 2
LHS = 3k – 10
RHS = 2

iv) 3p – 4q = -19
Given equation is 3p – 4q = -19
LHS = 3p – 4q
RHS = -19

Question 6.
Solve the following equations by trial and error method.
i) x – 3 = 5
ii) y + 6 = 15
iii) y = -1
iv) 2k – 1 = 3
i) x – 3 = 5
Given equation is x – 3 = 5
If x = 1, then the value of x – 3 = 1 – 3 = -2 ≠ 5
If x = 2, then the value of x – 3 = 2 – 3 = -l ≠ 5
If x = 3, then the value of x – 3 = 3 – 3 = 0 ≠ 5
If x = 4, then the value of x – 3 = 4 – 3 = l ≠ 5
If x = 5, then the value of x – 3 = 5 – 3 = 2 ≠ 5
If x = 6, then the value of x – 3 = 6 – 3 = 3 ≠ 5
If x = 7, then the value of x – 3 = 7 – 3 = 4 ≠ 5
If x = 8, then the value of x – 3 = 8 – 3 = 5 = 5
From the above when x = 8, the both LHS and RHS are equal.
∴ Solution of the equation x – 3 = 5 is x = 8

ii) y + 6 = 15
Given equation is y + 6 = 15
If y = 1, then the value of y + 6 = 1 + 6 = 7 ≠ 15
If y = 2, then the value of y + 6 = 2 + 6 = 8 ≠ 15
If y = 3, then the value of y +6 = 3 + 6 = 9 ≠ 15
If y = 4, then the value of y + 6 = 4 + 6 = 10 ≠ 15
If y = 5, then the value of y + 6 = 5 + 6 = 11 ≠ 15
If y = 6, then the value of y + 6 = 6 + 6 = 12 ≠ 15
If y = 7, then the value hf y + 6 = 7 + 6 = 13 ≠ 15
If y = 8, then the value of y + 6 = 8 + 6 = 14 ≠ 15
If y = 9, then the value of y + 6 = 9 + 6 = 15 = 15
From the above when y = 9, the both LHS and RHS are equal.
∴ Solution of the equation y + 6 = 15 is y = 9

iii) 𝑚2 = -1
Given equation is 𝑚2 = -1
If m = 1, then the value of 𝑚2 = 12 ≠ -1
If m = 2, then the value of 𝑚2 = 22 = 1 ≠ -1
If m = 3, then the value of 𝑚2 = 32 ≠ -1
Here, we are not getting negative values.
If we take (substitute) m as a negative number we will get negative value.
If m = -1, then the value of 𝑚2 = 12 ≠ -1
If m = -2, then the value of 𝑚2 = 22 = -1 = -1
From the above when m = -2, the both LHS and RHS are equal.
∴ Solution of the equation 𝑚2 = -1 is m = -2.

iv) 2k – 1 = 3
Given equation is 2k – 1 = 3
If k = 1, then the value of 2k – 1 = 2(1) – 1 = 2 – 1 = 1 ≠ 3
If k = 2, then the value of 2k – 1 = 2(2) – 1 = 4 – 1 = 3 = 3
From the above when k – 2, the both LHS and RHS are equal.
Solution of the equation 2k – 1 = 3 is k = 2.

Andhra Pradesh Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbooks for Exam Preparations

Andhra Pradesh Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbook Solutions can be of great help in your Andhra Pradesh Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise exam preparation. The AP Board STD 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Books State Board syllabus with maximum efficiency.

FAQs Regarding Andhra Pradesh Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbook Solutions

How to get AP Board Class 6th Maths Chapter 7 Introduction to Algebra Unit Exercise Textbook Answers??

Students can download the Andhra Pradesh Board Class 6 Maths Chapter 7 Introduction to Algebra Unit Exercise Answers PDF from the links provided above.

Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

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