AP Board 7th Class Maths Solutions Chapter 9 Algebraic Expressions Ex 9.3

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.3

Question 1.
Write standard form and additive inverse of the following expressions.
(i) – 6a
Answer:
Additive inverse of – 6a = – (- 6a) = 6a

(ii) 2 + 7c²
Answer:
Standard form of 2 + 7c² = 7c² + 2
Additive inverse of 7c² + 2
= – (7c² + 2)
= – 7c² – 2

(iii) 6x² + 4x – 5
Answer:
Given expression is in standard form.
Additive inverse of 6x² + 4x – 5
= – (6x² + 4x – 5)
= – 6x² – 4x + 5

(iv) 3c + 7a – 9b
Answer:
Standard form of 3c + 7a – 9b = 7a – 9b + 3c
Additive inverse of 7a – 9b + 3c
= – (7a – 9b + 3c)
= – 7a + 9b – 3c

Question 2.
Write the following expressions in standard form:
(i) 6x + x² – 5
Answer:
Standard form of 6x + x² – 5
= x² + 6x – 5

(ii) 3 – 4a² – 5a
Answer:
Standard form of 3 – 4a² – 5a
= – 4a² – 5a + 3

(iii) – m + 6 + 3m²
Answer:
Standard form of – m + 6 + 3m²
= 3m² – m + 6

(iv) c³ + 1 + c + 2c²
Answer:
Standard form of
c³ + 1 + c + 2c² = c³ + 2c² + c + 1

(v) 9 – p²
Answer:
Standard form of 9 – p² = – p² + 9

Question 3.
Add the following algebraic expressions using both horizontal and vertical methods. Did you get the same answer with both the methods? Verify.
(i) 2x² – 6x +3; 4x² + 9x + 5
Answer:

(ii) a² + 6ab + 8; – 3a² – ab – 2
Answer:

(iii) – p² + 2p – 10; 4 – 5p – 2p²
Answer:

Question 4.
Subtract the second expression from the first expression:
(i) 2x + y , x – y
Answer:
Let A = 2x + y and B = x – y
A – B = A + (- B) Additive inverse of B is
-B = – (x – y) = – x + y
∴ A – B = A + (- B)
= 2x + y + (- x + y)
= 2x + y – x + y
= 2x – x + y + y
= (2 – 1)x + (1 + 1)y
∴ A – B = x + 2y

(ii) a + 2b + c, – a – b – 3c
Answer:
Let A = a + 2b + c and B = – a – b – 3c
A – B = A + (- B)
Additive inverse of B is
– B = – (- a – b – 3c)
= a + b + 3c
∴ A – B = A + (- B)
= a + 2b + c + (a + b + 3c)
= a + 2b + c + a + b + 3c
= (a + a) + (2b + lb) + (c + 3c)
∴ A – B = 2a + 3b + 4c

(iii) 2l² – 3lm + 5m², 3l² – 4lm + 6m²
Answer:
Let A = 2l² – 3lm + 5m² and
B = 3l² – 4lm + 6m²
A – B = A + (- B)
Additive inverse of B is
– B = – (3t² – 4lm + 6m²)
= – 3l² + 4lm – 6m²
∴ A – B = A + (- B)
= (2l² – 3lm + 5m²) + (- 3l² + 4lm – 6m²)
= 2l² – 3lm + 5m² – 3l² + 4lm – 6m²
= 2l² – 3l² – 3lm + 4lm + 5m² – 6m²
= (2 – 3)l² + (- 3 + 4)lm + (5 – 6)m²
= (- 1) l² + 1 lm + (- 1)m²
∴ A – B = – l² + lm – m²

(iv) 7 – x – 3x², 2x² – 5x – 3
Answer:
Let A = 7 – x – 3x² and B = 2x² – 5x – 3
Write the given expressions in standard form.
∴ A = – 3x² – x + 7 and B = 2x² – 5x – 3
A – B = A + (- B)
Additive inverse of B is
– B = – (2x² – 5x – 3)
= – 2x² + 5x + 3
∴ A – B = A + (- B)
= (- 3x² – x + 7) + (- 2x² + 5x + 3)
= – 3x² – x + 7 – 2x² + 5x + 3
= (- 3x² – 2x²) + (- x + 5x) + (7 + 3)
= (- 3 – 2)x² + (- 1 + 5)x + 10
∴ A – B = – 5x² + 4x + 10

(v) 6m³ + 4m² + 7m – 3, 2m³ + 4
Answer:
Let A = 6m³ + 4m² + 7m – 3 and
B = 2 m² + 4
A – B = A + (- B)
Additive inverse of B is
– B = – (2m³ + 4)
= – 2m³ – 4
∴ A – B = A + (- B)
= (6m³ + 4m² + 7m – 3) + (- 2m³ – 4)
= 6m³ + 4m² + 7m – 3 – 2m³ – 4
= (6m³ – 2m³) + 4m² + 7m + (- 3 – 4)
= (6 – 2)m³ + 4m² + 7m + (- 7)
∴ A – B = 4m³ + 4m² + 7m – 7

Question 5.
Find the perimeter of the beside rect¬angle whose length is 6x + y and breadth is 3x – 2y.

Answer:
Given length of rectangle l = 6x + y
breadth b = 3x – 2y
Perimeter of Rectangle = 2 (l + b)
= 2[(6x + y) + (3x – 2y)]
= 2[6x + y + 3x – 2y]
= 2[(6 + 3)x + (1 – 2)y]
= 2[9x + (- 1) y]
= 2[9x – 1y]
= 2 × (9x) – 2 × (1y)
∴ Perimeter of rectangle = (18x – 2y) units.

Question 6.
Find the perimeter of triangle whose sides are a + 3b, a – b and 2a – b.

Answer:
Let the sides of triangle are
x = a + 3b, y = a – b and z = 2a – b
Perimeter of triangle = x + y + z
= (a + 3b) + (a – b) + (2a – b)
= a + 3b + a – b + 2a – b
= (a + a + 2a) + (3b – b – b)
= (1 + 1 + 2)a + (3 – 1 – 1)b
Perimeter of triangle.
= (4a + b) units.

Question 7.
Subtract the sum of x² – 5xy + 2y² and y² – 2xy – 3x² from the sum of 6x² – 8xy – y² and 2xy – 2y² – x².
Answer:
Given expressions are
x² – 5xy + 2y² and y² – 2xy – 3x² and 6x² – 8xy – y² and 2xy – 2y² – x²
Write the given expressions in the standard form
x² – 5xy + 2y² and – 3x² – 2xy + y² and 6x² – 8xy – y² and – x² + 2xy – 2y²
Let the Sum
A = (x² – 5xy + 2y²) + (- 3x² – 2xy + y²)
= x² – 5xy + 2y² – 3x² – 2xy + y²
= x² – 3x² – 5xy – 2xy + 2y² + y²
= (1 – 3)x² + (- 5 – 2)xy + (2 + 1)y²
A = – 2x² – 7xy + 3y²

Let the Sum
B = (6x² – 8xy – y²) + (- x² + 2xy – 2y²)
= 6x² – 8xy – y² – x² + 2xy – 2y²
= (6 – 1 )x² + (- 8 + ²)xy + (- 1 – 2)y²
B = 5x² – 6xy – 3y²
B – A = B + (- A)
Additive inverse of A is
– A = – (A)
= – (- 2x² – 7xy + 3y²)
∴ – A = 2x² + 7xy – 3y²
B – A = B + (- A)
= (5x² – 6xy – 3y²) + (2x² + 7xy – 3y²)
= (5 + 2)x² + (- 6 + 7)xy + (- 3 – 3)y²
∴ B – A = 7x² + xy – 6y²

Question 8.
What should be added to 1 + ²p – 3P² to get p² – p – 1 ?
Answer:
Given expressions are
1 + 2p – 3p² and p² – p – 1
Write the given expressions in the standard form.
– 3p² + ²p + 1 and p² – p – 1
Let A should be added to B to get C.
i.e. A + B = C
∴ A = C – B
Let B = – 3p² + 2p + 1 and
C = p² – p – 1
A = C + (- B )
Additive inverse B is
– B = – (- 3p² + 2p + 1) .
– B = 3p² – 2p – 1 .
A = (p² – p – 1) + (3p² – 2p – 1)
= p² – p – 1 + 3p² – 2p – 1
= (1 + 3)p² + (- 1 – 2)p + (- 1 – 1)
∴ A = 4p² – 3p – ²
∴ 4p² – 3p – ² is added to 1 + 2p – 3p² to get p² – p – 1.

Question 9.
What should be taken away from 3a² – 4b² + 5ab + 20 to get – a² – b² + 6ab + 3 ?
Answer:
Given expressions are
3a² – 4b² + 5ab + 20 and – a² – b² + 6ab +3
Let A be taken away from B to get C. that is A = B – C = B + (- C)
Let B = 3a² – 4b² + 5ab + 20 and C = – a² – b² + 6ab + 3 Additive inverse of C is
(- C) = – (- a² – b² + 6ab + 3)
= a² + b² – 6ab – 3
A = B + (- C)
= (3a² – 4b² + 5ab + 20) + (a² + b² – 6ab – 3)
= 3a² – 4b² + 5ab + 20 + a² + b² – 6ab – 3
= 3a² + a² – 4b² + b² + 5ab – 6ab + 20 – 3
= (3 + 1)a² + (- 4 + 1)b² + (5 – 6) ab + (20 – 3)
A = 4a² – 3b² – 1 ab + 17
So, 4a² – 3b² – 1 ab + 17 is taken away from 3a² – 4b² + 5ab + 20 to get – a² – b² + 6ab + 3.

Question 10.
If A = 4x² + y² – 6xy;
B = 3y² + 12x² + 8xy;
C = 6x² + 8y² + 6xy then,
find(i) A + B + C (ii) (A – B) – C
Answer:
Given A = 4x² + y² – 6xy
B = 3y² + 12x² + 8xy
C = 6x² + 8y² + 6xy
Write the given expressions in standard form.
A = 4x² – 6xy + y²
B = 12x² + 8xy + 3y²
C = 6x² + 6xy + 8y²

(i) A + B + C = (4x² – 6xy + y²) + (12x² + 8xy + 3y²) + (6x² + 6xy + 8y²)
= 4x² – 6xy + y² + 12x² + 8xy + 3y² + 6x² + 6xy + 8y²
= (4x² + 12x² + 6x²) + (- 6xy + 8xy + 6xy) + (y² + 3y² + 8y²)
= (4 + 12 + 6) x² + (- 6 + 8 + 6) xy + (1 + 3 + 8)y²
∴ A + B + C = 22x² + 8xy + 12y²

(ii) (A – B) – C
A + (- B) + (- C)
Additive inverse of B is
– B = – (12x² + 8xy + 3y²)
∴ – B = – 12x² – 8xy – 3y²
Additive inverse of C is
– C = -(6x² + 6xy + 8y²)
∴ – C = – 6x² – 6xy – 8y²
A + (- B) + (- C)
= (4x² – 6xy + y²) + (- 12x² – 8xy – 3y²) + (- 6x² – 6xy – 8y²)
= 4x² – 6xy + y² – 12x² – 8xy – 3y² – 6x² – 6xy – 8y²
= (4x² – 12x² – 6x²) + (- 6xy – 8xy – 6xy) + (y² – 3y² – 8y²)
= (4 – 12 – 6)x² + (- 6 – 8 – 6)xy + (1 – 3 – 8)y²
∴ (A – B) – C = – 14x² – 20xy – 10y²