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 + 7c^{2}

Answer:

Standard form of 2 + 7c^{2} = 7c^{2} + 2

Additive inverse of 7c^{2} + 2

= – (7c^{2} + 2)

= – 7c^{2} – 2

(iii) 6x^{2} + 4x – 5

Answer:

Given expression is in standard form.

Additive inverse of 6x^{2} + 4x – 5

= – (6x^{2} + 4x – 5)

= – 6x^{2} – 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 + x^{2} – 5

Answer:

Standard form of 6x + x^{2} – 5

= x^{2} + 6x – 5

(ii) 3 – 4a^{2} – 5a

Answer:

Standard form of 3 – 4a^{2} – 5a

= – 4a^{2} – 5a + 3

(iii) – m + 6 + 3m^{2}

Answer:

Standard form of – m + 6 + 3m^{2}

= 3m^{2} – m + 6

(iv) c^{3} + 1 + c + 2c^{2}

Answer:

Standard form of

c^{3} + 1 + c + 2c^{2} = c^{3} + 2c^{2} + c + 1

(v) 9 – p^{2}

Answer:

Standard form of 9 – p^{2} = – p^{2} + 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) 2x^{2} – 6x +3; 4x^{2} + 9x + 5

Answer:

(ii) a^{2} + 6ab + 8; – 3a^{2} – ab – 2

Answer:

(iii) – p^{2} + 2p – 10; 4 – 5p – 2p^{2}

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) 2l^{2} – 3lm + 5m^{2}, 3l^{2} – 4lm + 6m^{2}

Answer:

Let A = 2l^{2} – 3lm + 5m^{2} and

B = 3l^{2} – 4lm + 6m^{2}

A – B = A + (- B)

Additive inverse of B is

– B = – (3t^{2} – 4lm + 6m^{2})

= – 3l^{2} + 4lm – 6m^{2}

∴ A – B = A + (- B)

= (2l^{2} – 3lm + 5m^{2}) + (- 3l^{2} + 4lm – 6m^{2})

= 2l^{2} – 3lm + 5m^{2} – 3l^{2} + 4lm – 6m^{2}

= 2l^{2} – 3l^{2} – 3lm + 4lm + 5m^{2} – 6m^{2}

= (2 – 3)l^{2} + (- 3 + 4)lm + (5 – 6)m^{2}

= (- 1) l^{2} + 1 lm + (- 1)m^{2}

∴ A – B = – l^{2} + lm – m^{2}

(iv) 7 – x – 3x^{2}, 2x^{2} – 5x – 3

Answer:

Let A = 7 – x – 3x^{2} and B = 2x^{2} – 5x – 3

Write the given expressions in standard form.

∴ A = – 3x^{2} – x + 7 and B = 2x^{2} – 5x – 3

A – B = A + (- B)

Additive inverse of B is

– B = – (2x^{2} – 5x – 3)

= – 2x^{2} + 5x + 3

∴ A – B = A + (- B)

= (- 3x^{2} – x + 7) + (- 2x^{2} + 5x + 3)

= – 3x^{2} – x + 7 – 2x^{2} + 5x + 3

= (- 3x^{2} – 2x^{2}) + (- x + 5x) + (7 + 3)

= (- 3 – 2)x^{2} + (- 1 + 5)x + 10

∴ A – B = – 5x^{2} + 4x + 10

(v) 6m^{3} + 4m^{2} + 7m – 3, 2m^{3} + 4

Answer:

Let A = 6m^{3} + 4m^{2} + 7m – 3 and

B = 2 m^{2} + 4

A – B = A + (- B)

Additive inverse of B is

– B = – (2m^{3} + 4)

= – 2m^{3} – 4

∴ A – B = A + (- B)

= (6m^{3} + 4m^{2} + 7m – 3) + (- 2m^{3} – 4)

= 6m^{3} + 4m^{2} + 7m – 3 – 2m^{3} – 4

= (6m^{3} – 2m^{3}) + 4m^{2} + 7m + (- 3 – 4)

= (6 – 2)m^{3} + 4m^{2} + 7m + (- 7)

∴ A – B = 4m^{3} + 4m^{2} + 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 x^{2} – 5xy + 2y^{2} and y^{2} – 2xy – 3x^{2} from the sum of 6x^{2} – 8xy – y^{2} and 2xy – 2y^{2} – x^{2}.

Answer:

Given expressions are

x^{2} – 5xy + 2y^{2} and y^{2} – 2xy – 3x^{2} and 6x^{2} – 8xy – y^{2} and 2xy – 2y^{2} – x^{2}

Write the given expressions in the standard form

x^{2} – 5xy + 2y^{2} and – 3x^{2} – 2xy + y^{2} and 6x^{2} – 8xy – y^{2} and – x^{2} + 2xy – 2y^{2}

Let the Sum

A = (x^{2} – 5xy + 2y^{2}) + (- 3x^{2} – 2xy + y^{2})

= x^{2} – 5xy + 2y^{2} – 3x^{2} – 2xy + y^{2}

= x^{2} – 3x^{2} – 5xy – 2xy + 2y^{2} + y^{2}

= (1 – 3)x^{2} + (- 5 – 2)xy + (2 + 1)y^{2}

A = – 2x^{2} – 7xy + 3y^{2}

Let the Sum

B = (6x^{2} – 8xy – y^{2}) + (- x^{2} + 2xy – 2y^{2})

= 6x^{2} – 8xy – y^{2} – x^{2} + 2xy – 2y^{2}

= (6 – 1 )x^{2} + (- 8 + ^{2})xy + (- 1 – 2)y^{2}

B = 5x^{2} – 6xy – 3y^{2}

B – A = B + (- A)

Additive inverse of A is

– A = – (A)

= – (- 2x^{2} – 7xy + 3y^{2})

∴ – A = 2x^{2} + 7xy – 3y^{2}

B – A = B + (- A)

= (5x^{2} – 6xy – 3y^{2}) + (2x^{2} + 7xy – 3y^{2})

= (5 + 2)x^{2} + (- 6 + 7)xy + (- 3 – 3)y^{2}

∴ B – A = 7x^{2} + xy – 6y^{2}

Question 8.

What should be added to 1 + ^{2}p – 3P^{2} to get p^{2} – p – 1 ?

Answer:

Given expressions are

1 + 2p – 3p^{2} and p^{2} – p – 1

Write the given expressions in the standard form.

– 3p^{2} + ^{2}p + 1 and p^{2} – p – 1

Let A should be added to B to get C.

i.e. A + B = C

∴ A = C – B

Let B = – 3p^{2} + 2p + 1 and

C = p^{2} – p – 1

A = C + (- B )

Additive inverse B is

– B = – (- 3p^{2} + 2p + 1) .

– B = 3p^{2} – 2p – 1 .

A = (p^{2} – p – 1) + (3p^{2} – 2p – 1)

= p^{2} – p – 1 + 3p^{2} – 2p – 1

= (1 + 3)p^{2} + (- 1 – 2)p + (- 1 – 1)

∴ A = 4p^{2} – 3p – ^{2}

∴ 4p^{2} – 3p – ^{2} is added to 1 + 2p – 3p^{2} to get p^{2} – p – 1.

Question 9.

What should be taken away from 3a^{2} – 4b^{2} + 5ab + 20 to get – a^{2} – b^{2} + 6ab + 3 ?

Answer:

Given expressions are

3a^{2} – 4b^{2} + 5ab + 20 and – a^{2} – b^{2} + 6ab +3

Let A be taken away from B to get C. that is A = B – C = B + (- C)

Let B = 3a^{2} – 4b^{2} + 5ab + 20 and C = – a^{2} – b^{2} + 6ab + 3 Additive inverse of C is

(- C) = – (- a^{2} – b^{2} + 6ab + 3)

= a^{2} + b^{2} – 6ab – 3

A = B + (- C)

= (3a^{2} – 4b^{2} + 5ab + 20) + (a^{2} + b^{2} – 6ab – 3)

= 3a^{2} – 4b^{2} + 5ab + 20 + a^{2} + b^{2} – 6ab – 3

= 3a^{2} + a^{2} – 4b^{2} + b^{2} + 5ab – 6ab + 20 – 3

= (3 + 1)a^{2} + (- 4 + 1)b^{2} + (5 – 6) ab + (20 – 3)

A = 4a^{2} – 3b^{2} – 1 ab + 17

So, 4a^{2} – 3b^{2} – 1 ab + 17 is taken away from 3a^{2} – 4b^{2} + 5ab + 20 to get – a^{2} – b^{2} + 6ab + 3.

Question 10.

If A = 4x^{2} + y^{2} – 6xy;

B = 3y^{2} + 12x^{2} + 8xy;

C = 6x^{2} + 8y^{2} + 6xy then,

find(i) A + B + C (ii) (A – B) – C

Answer:

Given A = 4x^{2} + y^{2} – 6xy

B = 3y^{2} + 12x^{2} + 8xy

C = 6x^{2} + 8y^{2} + 6xy

Write the given expressions in standard form.

A = 4x^{2} – 6xy + y^{2}

B = 12x^{2} + 8xy + 3y^{2}

C = 6x^{2} + 6xy + 8y^{2}

(i) A + B + C = (4x^{2} – 6xy + y^{2}) + (12x^{2} + 8xy + 3y^{2}) + (6x^{2} + 6xy + 8y^{2})

= 4x^{2} – 6xy + y^{2} + 12x^{2} + 8xy + 3y^{2} + 6x^{2} + 6xy + 8y^{2}

= (4x^{2} + 12x^{2} + 6x^{2}) + (- 6xy + 8xy + 6xy) + (y^{2} + 3y^{2} + 8y^{2})

= (4 + 12 + 6) x^{2} + (- 6 + 8 + 6) xy + (1 + 3 + 8)y^{2}

∴ A + B + C = 22x^{2} + 8xy + 12y^{2}

(ii) (A – B) – C

A + (- B) + (- C)

Additive inverse of B is

– B = – (12x^{2} + 8xy + 3y^{2})

∴ – B = – 12x^{2} – 8xy – 3y^{2}

Additive inverse of C is

– C = -(6x^{2} + 6xy + 8y^{2})

∴ – C = – 6x^{2} – 6xy – 8y^{2}

A + (- B) + (- C)

= (4x^{2} – 6xy + y^{2}) + (- 12x^{2} – 8xy – 3y^{2}) + (- 6x^{2} – 6xy – 8y^{2})

= 4x^{2} – 6xy + y^{2} – 12x^{2} – 8xy – 3y^{2} – 6x^{2} – 6xy – 8y^{2}

= (4x^{2} – 12x^{2} – 6x^{2}) + (- 6xy – 8xy – 6xy) + (y^{2} – 3y^{2} – 8y^{2})

= (4 – 12 – 6)x^{2} + (- 6 – 8 – 6)xy + (1 – 3 – 8)y^{2}

∴ (A – B) – C = – 14x^{2} – 20xy – 10y^{2}