AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 4 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions Exercise 4

Question 1.

Add the following algebraic expressions using both horizontal and vertical methods. Did

you get the same answer with both methods.

(i) x^{2} – 2xy + 3y^{2}; 5y^{2} + 3xy – 6x^{2}

(ii) 4a^{2} + 5b^{2} + 6ab 3ab ; 6a^{2} – 2b^{2} ; 4b^{2} – 5ab

(iii) 2x + 9y – 7z ; 3y + z + 3x ;2x – 4y – z

(iv) 2x^{2} – 6x + 3 ; – 3x^{2} – x – 4 ; 1 + 2x – 3x^{2}

Solution:

(i) x^{2} – 2xy + 3y^{2}; 5y^{2} + 3xy – 6x^{2}

Addition by Horizontal method

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

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

= – 5x^{2} + xy + 8y^{2} .

ii) 4a^{2} + 5b^{2} + 6ab;3ab; 6a^{2} – 2b^{2}; 4a^{2} – 5ab

Horizontal Method

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

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

= 14a^{2} + 3b^{2} + 4ab

iii) 2x^{2} + 9y – 7z ; 3y + z + 3x; 2x4y – z

Horizontal Method

= (2x + – 7z) + (3y + z + 3x) + (2x – 4y – z)

= (2x + 3x + 2x) + (9y + 3y_4y) + ( – 7z + z – z)

= 7x + 8y – 7z

iv) 2x^{2} – 6x – 3; – 3x^{2} – x – 4; 1 + 2x – 3x^{2}

Horizontal Method

(2x^{2} – 6x + 3) + ( – 3x^{2} – x – 4) + (1 + 2x – 3x^{2})

(2x^{2} – 3x^{2} – 3x^{2}) + (- 6x – x + 2x) + (3 – 4 + 1)

= – 4x^{2} – 5x + 0

= – 4x^{2} – 5x

In all the above sums

(i), (ii), (iii), (iv) we got same answer in both horizontal and vertical methods.

Question 2.

Simpli1’: 2x^{2} + 5x – 1 + 8x + x^{2} + 7 – 6x + 3 – 3x^{2}

Solution:

Given algebraic expression is

2x^{2} + 5x – 1 + 8x + x^{2} + 7 – 6x + 3 – 3x^{2}

= (2x^{2} + x^{2} – 3x^{2}) + (5x + 8x – 6x) . ( – 1 + 7 + 3)

= 0 + 7x + 9 = 7x + 9

Question 3.

Find the perimeter of the following rectangle?

Solution:

The perimeter of a rectangle = 2 (length + breadth)

= 2 (6x + y + 3x – 2y)

= 2 (9x – y)

∴ P = (18x – 2y) units

Question 4.

Find the perimeter of a triangle whose sides are 2a + 3b, b – a, 4a – 2b.

Solution:

Perimeter of the triangle ABC = \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CA}}\)

= (b – a) + (3a + 2b) + (4a – 2b)

=( – a + 3a + 4a) +(b + 2b – 2b)

= (6a + b) units

Question 5.

Subtract the second expression from the first expression

(i) 2a + b, a – b

(ii) x + 2y + z , – x – y – 3z

(iii) 3a^{2} – 8ab – 2b^{2}, 3a^{2} – 4ab+6b^{2}

(iv) 4pq – 6p^{2} – 2q^{2}, 9p^{2}

(v) 7 – 2x – 3x^{2}, 2x^{2} – 5 x – 3

(vi) 5x^{2} – 3xy – 7y^{2} , 3x^{2} – xy – 2y^{2}

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

Solution:

(i) 2a + b, a – b = 2a + b – a + b

= (2a – a) + (b + b)

= a + 2b

ii) (x + 2y + z) – ( – x – y – 3z) = x + + z + x + y + 3z

=(x + x) + (2y + y) +(z + 3z)

= 2x + 3y + 4z

iii) (3a^{2} – 8ab – 2b^{2}) – (3a^{2} – 4ab + 6b^{2}) = 3a^{2} – 8ab – 2b^{2} – 3a^{2} +4ab – 6b^{2}

= (3a^{2} – 3a^{2}) + ( – 8ab + 4ab) + (2b^{2} – 6b^{2})

= 0 – 4ab – 8b^{2}

= – 4ab – 8b^{2}

iv) (4pq – 6p^{2} – 2q^{2}) – (9p^{2}) = 4pq – 6p^{2} – 2q^{2} – 9p^{2}

= 4pq – 15p^{2} – 2q^{2}

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

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

= – 5x^{2} + 3x + 10

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

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

= 2x^{2} – 2xy – 5y^{2}

viii) (6m^{3} + 4m^{2} + 7m – 3) – (3m^{3} + 4)= 6m^{3} + 4m^{2} + 7m – 3 – 3m3 – 4

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

= 3m^{3} + 4m^{2} + 7m – 7

Question 6.

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}.

Solution:

The sum of x^{2} – 5xy + 2y^{2} and y^{2} – 2xy – 3x^{2} = (x^{2} – 5xy + 2y^{2}) (y – 2xy – 3x^{2})

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

= – 2x^{2} – 7xy + 3y^{2} ………………….(1)

The sum of 6x^{2} – 8xy – y^{2} and

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

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

= 5x^{2} – 6xy – 3y^{2} ………………. (2)

From (1) and (2)

(2) – (1)= (5x^{2} – 6xy – 3y^{2}) – ( – 2x^{2} – 7xy + 3y^{2})

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

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

= 7x^{2} + xy – 6y^{2}

Question 7.

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

Solution:

Let the added algebraic expression may be ‘A say

(1 + 2x – 3x^{2}) + A = x^{2} – x – 1

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

= x^{2} – x – 1 – 1 – 2x + 3x^{2}

= (x^{2} + 3x^{2}) + ( – x – 2x) +( – 1 – 1)

∴ A = 4x^{2} – 3x – 2

∴ The required added expression (A) = 4x^{2} – 3x – 2

Question 8.

What should be taken away from 3x^{2} – 4y^{2} + 5xy +20 to get – x^{2} – y^{2} + 6xy + 20.

Solution:

Let the subtracted algebraic expression may be B’ say

3x^{2} – 4y^{2} + 5xy – B = – x^{2} – y^{2} + 6xy + 20

B = (3x^{2} – 4y^{2} + 5xy) – (- x^{2} – y^{2} + 6xy + 20)

= 3x^{2} – 4y^{2} + 5xy + x^{2} +y^{2} – 6xy – 20

= (3x^{2} + x^{2}) ( – 4y^{2} + y^{2}) + (5xy – 6xy) – 20

∴ B = 4x^{2} – 3y^{2} – xy – 20

∴ The required subtracted expression is B = 4x^{2} – 3y^{2} – xy – 20

Question 9.

The sum of 3 expressions is 8 + 13a + 7a^{2}. Two of them are 2a^{2} + 3a + 2 and 3a2 – 4a + 1. Find the third expression.

Solution:

Given that f he sum of 3 expressions is 8 + 13a + 7a^{2} ……………..(1)

Two of them are 2a^{2} + 3a + 2 and 3a^{2} – 4a + 1.

∴ The sum of above two expressions (2a^{2} + 3a + 2) + (3a^{2} – 4a + 1)

= 2a^{2} + 3a + 2 + 3a^{2} – 4a + 1.

= (2a^{2} + 3a^{2}) + (3a – 4a) + (2 + 1)

= 5a^{2} – a + 3

∴ The required 3rd expression (1) – (2)

(1) – (2) =(7a^{2} + 13a + 8) – (5a^{2} – a + 3) = 7a^{2} + 13a + 8 – – 5a^{2} – a – 3

=(7a^{2} – 5a^{2}) +(13a + a)+ (8 – 3)

= 2a^{2} + 14a + 5

Question 10.

If A = 4x^{2} + y^{2} – 6xy;

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

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

Find (i) A + B + C (ii) (A – B) – C

Solution:

(i) im 7

(ii) (A – B) – C

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

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

A – B = – 8x^{2} – 2y^{2} – 14xy

∴ (A – B) – C = ( – 8x^{2} – 2y^{2} – 14xy) – (6x^{2} + 8y^{2} + 6xy)

= ( – 8x^{2} – 6x^{2}) + -2y^{2} – 8y^{2}) + ( – 14xy – 6xy)

= – 14x^{2} – 10y^{2} – 20xy

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

iii) 2A + B

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

∴ 2A + B = (8x^{2} + 2y^{2} – 12xy) + (3y^{2} + 12x^{2} + 8xy)

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

∴ 2A + B = 20x^{2} + 5y^{2} – 4xy

iv) A – 3B

38 = 3(3y^{2} + 12x^{2} + 8xy) = 9y^{2} + 36x^{2} + 24xy

∴ A – 3B = (4x^{2} + y^{2} – 6xy) – (9y^{2} + 36x^{2} + 24xy)

= (4x^{2} – 36x^{2}) (y^{2} – 9y^{2}) ( 6xy – 24 xy)

= 32x^{2} – 8y^{2} – 30xy

∴ A – 3B = – 32x^{2} – 8y^{2} – 30xy (or)

= – [32x^{2} + 8y + 30xy]