AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Ex 7.3

Question 1.
Identify which of the following are equations.
i) x – 3 = 7
ii) l + 5 > 9
iii) p – 4 < 10
iv) 5 + m = -6
v) 2s – 2 = 12
vi) 3x + 5
Answer:
i) x – 3 = 7
We know that, a mathematical statement involving equality symbol is called an equation.
x – 3 = 7 has equality symbol. So, it is an equation.

ii) l + 5 > 9
We know that, a mathematical statement involving equality symbol is called an equation.
l + 5 > 9 has no equality symbol.
So, it is not an equation. [It is an inequation]

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3

iii) p – 4 < 10
We know that, a mathematical statement involving equality symbol is called an equation.
p – 4 < 10 has no equality symbol.
So, it is not an equation. [It is an inequation]

iv) 5 + m = – 6
We know that, a mathematical statement involving equality symbol is called an equation.
5 + m = – 6 has equality symbol.
So, it is an equation.

v) 2s – 2 = 12
We know that, a mathematical statement involving equality symbol is called an equation.
2s – 2 = 12 has equality symbol.
So, it is an equation.

vi) 3x + 5
It is only an expression. It’s not an equation.

Question 2.
Write LHS and RHS of the following equations.
i) x – 5 = 6
ii) 4y = 12
iii) 2z + 3 = 7
Answer:
i) x – 5 = 6
Given equation is x – 5 = 6
LHS = x – 5
RHS = 6

ii) 4y = 12
Given equation is 4y = 12
LHS = 4y
RHS = 12

iii) 2z + 3 = 7
Given equation is 2z + 3 = 7
LHS = 2z + 3
RHS = 7

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3

Question 3.
Solve the following equation by Trial & Error Method.
i) x + 3 = 5
ii) y – 2 = 7
iii) a + 4 = 9
Answer:
i) x + 3 = 5
Given equation is x + 3 = 5
If x = 1, then the value of x + 3 = 1 + 3 = 4 ≠ 5
If x = 2, then the value of x + 3 = 2 + 3 = 5 = 5
From the above when x = 2, then both LHS and RHS are equal.
∴ Solution of the equation x + 3 = 5 is x = 2

ii) y – 2 = 7
Given equation is y – 2 = 7
If y = 1, then the value of y – 2 = 1 – 2 = -1 ≠ 7
If y = 2, then the value of y – 2 = 2 – 2 = 0 ≠ 7
If y = 3, then the value of y – 2 = 3 – 2 = 1 ≠ 7
If y = 4, then the value of y – 2 = 4 – 2 = 2 ≠ 7
If y = 5, then the value of y – 2 = 5 – 2 = 3 ≠ 7
If y = 6, then the value of y – 2 = 6 – 2 = 4 ≠ 7
If y = 7, then the value of y – 2 = 7 – 2 = 5 ≠ 7
If y = 8, then the value of y – 2 = 8 – 2 = 6 ≠ 7
If y = 9, then the value of y – 2 = 9 – 2 = 7 = 7
From the above when y = 9, the both LHS and RHS are equal.
Solution of the equation y – 2 = 7 is y = 9

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.3

ii) a + 4 = 9
Given equation is a + 4 = 9
If a = 1, then the value of a + 4 = 1 + 4 = 5 ≠ 9
If a = 2, then the value of a + 4 = 2 + 4 = 6 ≠ 9
If a = 3, then the value of a + 4 = 3 + 4 = 7 ≠ 9
If a = 4, then the value of a + 4 = 4 + 4 = 8 ≠ 9
If a = 5, then the value of a + 4 = 5 + 4 = 9 = 9
From the above when a = 5, the both LHS and RHS are equal.
∴ Solution of the equation a + 4 = 9 is a = 5