AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions Ex 4

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) x² – 2xy + 3y²; 5y² + 3xy – 6x²
(ii) 4a² + 5b² + 6ab 3ab ; 6a² – 2b² ; 4b² – 5ab
(iii) 2x + 9y – 7z ; 3y + z + 3x ;2x – 4y – z
(iv) 2x² – 6x + 3 ; – 3x² – x – 4 ; 1 + 2x – 3x²
Solution:
(i) x² – 2xy + 3y²; 5y² + 3xy – 6x²
Addition by Horizontal method
(x² – 2xy + 3y²) + ( – 6x² + 3xy + 5y²)
= (x² – 6x²) + ( – 2xy + 3xy) + (3y² + 5y²)
= – 5x² + xy + 8y² .

ii) 4a² + 5b² + 6ab;3ab; 6a² – 2b²; 4a² – 5ab
Horizontal Method
(4a² + 5b² + 6ab) + 3ab + (6a² – 2b²) (4a² – 5ab)
= (4a² + 6a² + 4a²) + (5b² – 2b²) + (6ab + 3ab – 5ab)
= 14a² + 3b² + 4ab

iii) 2x² + 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) 2x² – 6x – 3; – 3x² – x – 4; 1 + 2x – 3x²
Horizontal Method
(2x² – 6x + 3) + ( – 3x² – x – 4) + (1 + 2x – 3x²)
(2x² – 3x² – 3x²) + (- 6x – x + 2x) + (3 – 4 + 1)
= – 4x² – 5x + 0
= – 4x² – 5x

In all the above sums
(i), (ii), (iii), (iv) we got same answer in both horizontal and vertical methods.

Question 2.
Simpli1’: 2x² + 5x – 1 + 8x + x² + 7 – 6x + 3 – 3x²
Solution:
Given algebraic expression is
2x² + 5x – 1 + 8x + x² + 7 – 6x + 3 – 3x²
= (2x² + x² – 3x²) + (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) 3a² – 8ab – 2b², 3a² – 4ab+6b²
(iv) 4pq – 6p² – 2q², 9p²
(v) 7 – 2x – 3x², 2x² – 5 x – 3
(vi) 5x² – 3xy – 7y² , 3x² – xy – 2y²
(vii) 6m³ + 4m² + 7m – 3 , 3m³ + 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) (3a² – 8ab – 2b²) – (3a² – 4ab + 6b²) = 3a² – 8ab – 2b² – 3a² +4ab – 6b²
= (3a² – 3a²) + ( – 8ab + 4ab) + (2b² – 6b²)
= 0 – 4ab – 8b²
= – 4ab – 8b²

iv) (4pq – 6p² – 2q²) – (9p²) = 4pq – 6p² – 2q² – 9p²
= 4pq – 15p² – 2q²

v) (7 – 2x – 3x²) – (2x² – 5x – 3)=7 – 2x – 3x² – 2x² + 5x + 3
= ( – 3x² – 2x²) + ( – 2x + 5x) + (7 + 3)
= – 5x² + 3x + 10

vi) (5x² – 3xy – 7y²) – (3x² – xy – 2y²) = 5x²– 3xy – 7y² – 3x² + xy + 2y²
= (5x² – 3x²) + ( – 3xy + xy) + ( – 7y² + 2y²)
= 2x² – 2xy – 5y²

viii) (6m³ + 4m² + 7m – 3) – (3m³ + 4)= 6m³ + 4m² + 7m – 3 – 3m3 – 4
= (6m³ – 3m³) + 4m² + 7m + (- 3 – 4)
= 3m³ + 4m² + 7m – 7

Question 6.
Subtract the sum of x² – 5xy + 2y² and y² – 2xy – 3x² from the sum of 6x² – 8xy – y² and 2xy – 2y² – x².
Solution:
The sum of x² – 5xy + 2y² and y² – 2xy – 3x² = (x² – 5xy + 2y²) (y – 2xy – 3x²)
= (x² – 3x²) + ( – 5xy – 2xy) + (2y² + y²)
= – 2x² – 7xy + 3y² ………………….(1)

The sum of 6x² – 8xy – y² and
2xy – 2y² – x² = (6x² – 8xy – y²) + (2xy – 2y² – x²)
= (6x² – x²) + ( – 8xy + 2xy) + (-y² – 2y²)
= 5x² – 6xy – 3y² ………………. (2)

From (1) and (2)
(2) – (1)= (5x² – 6xy – 3y²) – ( – 2x² – 7xy + 3y²)
= 5x² – 6xy – 3y² + 2x² + 7xy – 3y²
= (5x² + 2x²) + ( – 6xy + 7xy) + ( – 3y² – 3y²)
= 7x² + xy – 6y²

Question 7.
What should be added to 1 + 2x – 3x² to get x² – x – 1?
Solution:
Let the added algebraic expression may be ‘A say
(1 + 2x – 3x²) + A = x² – x – 1
=A =(x² – x – 1) – (1 + 2x – 3x²)
= x² – x – 1 – 1 – 2x + 3x²
= (x² + 3x²) + ( – x – 2x) +( – 1 – 1)
∴ A = 4x² – 3x – 2
∴ The required added expression (A) = 4x² – 3x – 2

Question 8.
What should be taken away from 3x² – 4y² + 5xy +20 to get – x² – y² + 6xy + 20.
Solution:
Let the subtracted algebraic expression may be B’ say
3x² – 4y² + 5xy – B = – x² – y² + 6xy + 20
B = (3x² – 4y² + 5xy) – (- x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + x² +y² – 6xy – 20
= (3x² + x²) ( – 4y² + y²) + (5xy – 6xy) – 20
∴ B = 4x² – 3y² – xy – 20
∴ The required subtracted expression is B = 4x² – 3y² – xy – 20

Question 9.
The sum of 3 expressions is 8 + 13a + 7a². Two of them are 2a² + 3a + 2 and 3a2 – 4a + 1. Find the third expression.
Solution:
Given that f he sum of 3 expressions is 8 + 13a + 7a² ……………..(1)
Two of them are 2a² + 3a + 2 and 3a² – 4a + 1.
∴ The sum of above two expressions (2a² + 3a + 2) + (3a² – 4a + 1)
= 2a² + 3a + 2 + 3a² – 4a + 1.
= (2a² + 3a²) + (3a – 4a) + (2 + 1)
= 5a² – a + 3
∴ The required 3rd expression (1) – (2)
(1) – (2) =(7a² + 13a + 8) – (5a² – a + 3) = 7a² + 13a + 8 – – 5a² – a – 3
=(7a² – 5a²) +(13a + a)+ (8 – 3)
= 2a² + 14a + 5

Question 10.
If A = 4x² + y² – 6xy;
B = 3y²+ 12x² + 8xy;
C = 6x² + 8y² + 6xy
Find (i) A + B + C (ii) (A – B) – C
Solution:
(i) im 7
(ii) (A – B) – C
(A – B) = (4x² + y² – 6xy) – (3y² + 12x² + 8xy)
= (4x² – 12x²) + (y² – 3y²) + ( – 6xy – 8xy)
A – B = – 8x² – 2y² – 14xy
∴ (A – B) – C = ( – 8x² – 2y² – 14xy) – (6x² + 8y² + 6xy)
= ( – 8x² – 6x²) + -2y² – 8y²) + ( – 14xy – 6xy)
= – 14x² – 10y² – 20xy
∴ (A – B) – C = – (14x² + 10y² + 20xy)

iii) 2A + B
2A = 2(4x² + y² – 6xy) = 8x² + 2y² – 12xy
∴ 2A + B = (8x² + 2y² – 12xy) + (3y² + 12x² + 8xy)
= (8x² + 12x²) + (2y² + 3y²) + ( – 12xy + 8xy)
∴ 2A + B = 20x² + 5y² – 4xy

iv) A – 3B
38 = 3(3y² + 12x² + 8xy) = 9y² + 36x² + 24xy
∴ A – 3B = (4x² + y² – 6xy) – (9y² + 36x² + 24xy)
= (4x² – 36x²) (y² – 9y²) ( 6xy – 24 xy)
= 32x² – 8y² – 30xy
∴ A – 3B = – 32x² – 8y² – 30xy (or)
= – [32x² + 8y + 30xy]