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]

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

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

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