# RD Sharma Solutions For Class 7 Maths Chapter - 7 Algebraic Expressions

In order to excel in exams, students can view and download the PDF of RD Sharma Solutions for Class 7 Maths Chapter 7 Algebraic Expressions. RD Sharma Solutions are useful for students as it helps them in scoring high marks in the examination. These solutions are prepared by subject matter experts at BYJUâ€™S, describing the complete method of solving problems.

Chapter 7, Algebraic Expressions, contains four exercises. RD Sharma Solutions for Class 7 given here includes the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

• Definition and meaning of algebraic expressions
• Types of algebraic expressions
• Definitions of factors and coefficients
• Like and unlike terms
• Finding the value of algebraic expressions
• Operations on algebraic expressions
• Addition of positive like terms
• Addition of negative like terms
• Addition of positive and negative like terms
• Addition of algebraic expressions with like and unlike terms
• Subtraction of algebraic expressions
• The use of grouping symbols in writing algebraic expressions
• Removal of brackets

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Exercise 7.1 Page No: 7.7

1. Identify the monomials, binomials, trinomials and quadrinomials from the following expressions:

(i) a2

(ii) a2Â âˆ’ b2

(iii) x3Â + y3Â + z3

(iv) x3Â + y3Â + z3Â + 3xyz

(v) 7 + 5

(vi) a b c + 1

(vii) 3x â€“ 2 + 5

(viii) 2x â€“ 3y + 4

(ix) x y + y z + z x

(x) ax3Â + bx2Â + cx + d

Solution:

(i) Given a2

a2Â is a monomial expression because it contains only one term

(ii) Given a2Â âˆ’ b2

a2Â âˆ’ b2Â is a binomial expression because it contains two terms

(iii) Given x3Â + y3Â + z3

x3Â + y3Â + z3 is a trinomial because it contains three terms

(iv) Given x3Â + y3Â + z3Â + 3xyz

x3Â + y3Â + z3Â + 3xyz is a quadrinomial expression because it contains four terms

(v) Given 7 + 5

7 + 5 is a monomial expression because it contains only one term

(vi) Given a b c + 1

a b c + 1 is a binomial expression because it contains two terms

(vii) Given 3x â€“ 2 + 5

3x â€“ 2 + 5 is a binomial expression because it contains two terms

(viii) Given 2x â€“ 3y + 4

2x â€“ 3y + 4 is a trinomial because it contains three terms

(ix) Given x y + y z + z x

x y + y z + z x is a trinomial because it contains three terms

(x) Given ax3Â + bx2Â + cx + d

ax3Â + bx2Â + cx + d is a quadrinomial expression because it contains four terms

2. Write all the terms of each of the following algebraic expressions:

(i) 3x

(ii) 2x â€“ 3

(iii) 2x2Â âˆ’ 7

(iv) 2x2Â + y2Â âˆ’ 3xy + 4

Solution:

(i) Given 3x

3x is the only term of the given algebraic expression.

(ii) Given 2x â€“ 3

2x and -3 are the terms of the given algebraic expression.

(iii) Given 2x2Â âˆ’ 7

2x2Â and âˆ’7 are the terms of the given algebraic expression.

(iv) Given 2x2Â + y2Â âˆ’ 3xy + 4

2x2, y2, âˆ’3xy and 4 are the terms of the given algebraic expression.

3. Identify the terms and also mention the numerical coefficients of those terms:

(i) 4xy, -5x2y, -3yx, 2xy2

(ii) 7a2bc,-3ca2b,-(5/2) abc2, 3/2abc2,(-4/3)cba2

Solution:

(i) Like terms 4xy, -3yx and Numerical coefficients 4, -3

(ii) Like terms (7a2bc, âˆ’3ca2b) and (âˆ’4/3cba2)Â and their Numerical coefficients 7, -3,

(-4/3)

Like terms are (âˆ’5/2abc2)Â and (3/2 abc2)Â and numerical coefficients areÂ (âˆ’5/2) and (3/2)

4. Identify the like terms in the following algebraic expressions:

(i) a2Â + b2Â -2a2Â + c2Â + 4a

(ii) 3x + 4xy âˆ’ 2yz + 52zy

(iii) abc + ab2c + 2acb2Â + 3c2ab + b2ac âˆ’ 2a2bc + 3cab2

Solution:

(i) Given a2Â + b2Â -2a2Â + c2Â + 4a

The like terms in the given algebraic expressions are a2Â and âˆ’2a2.

(ii) Given 3x + 4xy âˆ’ 2yz + 52zy

The like terms in the given algebraic expressions are -2yz and 52zy.

(iii) Given abc + ab2c + 2acb2Â + 3c2ab + b2ac âˆ’ 2a2bc + 3cab2

The like terms in the given algebraic expressions are ab2c, 2acb2, b2ac and 3cab2.

5. Write the coefficient of x in the following:

(i) â€“12x

(ii) â€“7xy

(iii) xyz

(iv) â€“7ax

Solution:

(i) Given -12x

The numerical coefficient of x is -12.

(ii) Given -7xy

The numerical coefficient of x is -7y.

(iii) Given xyz

The numerical coefficient of x is yz.

(iv) Given -7ax

The numerical coefficient of x is -7a.

6. Write the coefficient of x2 in the following:

(i) âˆ’3x2

(ii) 5x2yz

(iii) 5/7x2z

(iv) (-3/2) ax2Â + yx

Solution:

(i) Given âˆ’3x2

The numerical coefficient of x2Â is -3.

(ii) Given 5x2yz

The numerical coefficient of x2Â is 5yz.

(iii) Given 5/7x2z

The numerical coefficient of x2Â is 5/7z.

(iv) Given (-3/2) ax2Â + yx

The numerical coefficient of x2Â is â€“ (3/2) a.

7. Write the coefficient of:

(i) y in â€“3y

(ii) a in 2ab

(iii) z in â€“7xyz

(iv) p in â€“3pqr

(v) y2Â in 9xy2z

(vi) x3Â in x3Â +1

(vii) x2Â in âˆ’ x2

Solution:

(i) Given â€“3y

The coefficient of y is -3.

(ii) Given 2ab

The coefficient of a is 2b.

(iii) Given -7xyz

The coefficient of z is -7xy.

(iv) Given -3pqr

The coefficient of p is -3qr.

(v) Given 9xy2z

The coefficient of y2Â is 9xz.

(vi) Given x3Â +1

The coefficient of x3Â is 1.

(vii) Given âˆ’ x2

The coefficient of x2Â is -1.

8. Write the numerical coefficient of each in the following:

(i) xy

(ii) -6yz

(iii) 7abc

(iv) -2x3y2z

Solution:

(i) Given xy

The numerical coefficient in the term xy is 1.

(ii) Given -6yz

The numerical coefficient in the term – 6yz is – 6.

(iii) Given 7abc

The numerical coefficient in the term 7abc is 7.

(iv) Given -2x3y2z

The numerical coefficient in the term âˆ’2x3y2z is -2.

9. Write the numerical coefficient of each term in the following algebraic expressions:

(i) 4x2y â€“ (3/2)xy + 5/2 xy2

(ii) (â€“5/3)x2y + (7/4)xyz + 3

Solution:

(i) Given 4x2y â€“ (3/2) xy + 5/2 xy2

Numerical coefficient of following algebraic expressions are given below

 Term Coefficient 4x2y 4 â€“ (3/2) xy -(3/2) 5/2 xy2 (5/2)

(ii) Given (â€“5/3)x2y + (7/4)xyz + 3

Numerical coefficient of following algebraic expressions are given below

 Term Coefficient (â€“5/3)x2y (-5/3) (7/4)xyz (7/4) 3 3

10. Write the constant term of each of the following algebraic expressions:

(i) x2y âˆ’ xy2Â + 7xy âˆ’ 3

(ii) a3Â âˆ’ 3a2Â + 7a + 5

Solution:

(i) Given x2y âˆ’ xy2Â + 7xy âˆ’ 3

The constant term in the given algebraic expressions is -3.

(ii) Given a3Â âˆ’ 3a2Â + 7a + 5

The constant term in the given algebraic expressions is 5.

11. Evaluate each of the following expressions for x = -2, y = -1, z = 3:

(i) (x/y) + (y/z) + (z/x)

(ii) x2 + y2 + z2 â€“ xy â€“ yz â€“ zx

Solution:

(i) Given x = -2, y = -1, z = 3

Consider (x/y) + (y/z) + (z/x)

On substituting the given values we get,

= (-2/-1) + (-1/3) + (3/-2)

The LCM of 3 and 2 is 6

= (12 â€“ 2 â€“ 9)/6

= (1/6)

(ii) Given x = -2, y = -1, z = 3

Consider x2 + y2 + z2 â€“ xy â€“ yz â€“ zx

On substituting the given values we get,

= (-2)2 + (-1)2 + 32 â€“ (-2) (-1) â€“ (-1) (3) â€“ (3) (-2)

= 4 + 1 + 9 â€“ 2 + 3 + 6

= 23 â€“ 2

= 21

12. Evaluate each of the following algebraic expressions for x = 1, y = -1, z = 2, a = -2, b = 1, c = -2:

(i) ax + by + cz

(ii) ax2Â + by2Â â€“ cz

(iii) axy + byz + cxy

Solution:

(i) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2

Consider ax + by + cz

On substituting the given values

= (-2) (1) + (1) (-1) + (-2) (2)

= â€“2 â€“ 1 â€“ 4

= â€“7

(ii) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2

Consider ax2Â + by2Â â€“ cz

On substituting the given values

= (-2) Ã— 12Â + 1 Ã— (-1)2Â â€“ (-2) Ã— 2

= -2 + 1 â€“ (-4)

= -1 + 4

= 3

(iii) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2

Consider axy + byz + cxy

= (-2) Ã— 1 Ã— -1 + 1 Ã— -1 Ã— 2 + (-2) Ã— 1 Ã— (-1)

= 2 + (-2) + 2

= 4 â€“ 2

= 2

Exercise 7.2 Page No: 7.13

(i)Â 3x and 7x

(ii) -5xy and 9xy

Solution:

(i) Given 3x and 7x

3x + 7x = (3 + 7) x

= 10x

(ii) Given -5xy and 9xy

-5xy + 9xy = (-5 + 9) xy

= 4xy

2. Simplify each of the following:

(i) 7x3y +9yx3

(ii) 12a2b + 3ba2

Solution:

(i) Given 7x3y +9yx3

7x3y + 9yx3Â = (7 + 9) x3y

= 16x3y

(ii) Given

12a2b + 3ba2Â = (12 + 3) a2b

= 15a2b

(i) 7abc, -5abc, 9abc, -8abc

(ii) 2x2y, – 4x2y, 6x2y, -5x2y

Solution:

(i) Given 7abc, -5abc, 9abc, -8abc

Consider 7abc + (-5abc) + (9abc) + (-8abc)

= 7abc â€“ 5abc + 9abc â€“ 8abc

= (7 â€“ 5 + 9 â€“ 8) abc [by taking abc common]

= (16 â€“ 13) abc

= 3abc

(ii) Given 2x2y, – 4x2y, 6x2y, -5x2y

2x2y +(-4x2y) + (6x2y) + (-5x2y)

= 2x2y – 4x2y + 6x2y – 5x2y

= (2- 4 + 6 – 5) x2y [by taking x2 y common]

= (8 – 9) x2y

= -x2y

(i) x3Â -2x2y + 3xy2– y3, 2x3– 5xy2Â + 3x2y – 4y3

(ii) a4Â – 2a3b + 3ab3Â + 4a2b2Â + 3b4, – 2a4Â – 5ab3Â + 7a3b – 6a2b2Â + b4

Solution:

(i) Given x3Â -2x2y + 3xy2– y3, 2x3– 5xy2Â + 3x2y – 4y3

Collecting positive and negative like terms together, we get

= x3Â +2x3Â – 2x2y + 3x2y + 3xy2Â – 5xy2Â – y3– 4y3

= 3x3Â + x2y – 2xy2Â – 5y3

(ii) Given a4Â – 2a3b + 3ab3Â + 4a2b2Â + 3b4, – 2a4Â – 5ab3Â + 7a3b – 6a2b2Â + b4

= a4Â – 2a3b + 3ab3Â + 4a2b2Â + 3b4Â – 2a4Â – 5ab3Â + 7a3b – 6a2b2Â + b4

Collecting positive and negative like terms together, we get

= a4Â – 2a4– 2a3b + 7a3b + 3ab3Â – 5ab3Â + 4a2b2Â – 6a2b2Â + 3b4Â + b4

= – a4Â + 5a3b – 2ab3Â – 2a2b2Â + 4b4

(i) 8a â€“ 6ab + 5b, â€“6a â€“ ab â€“ 8b and â€“4a + 2ab + 3b

(ii) 5x3Â + 7 + 6x – 5x2, 2x2Â â€“ 8 – 9x, 4x – 2x2Â + 3 x 3, 3 x 3 – 9x – x2Â and x – x2Â – x3Â â€“ 4

Solution:

(i) Given 8a â€“ 6ab + 5b, â€“6a â€“ ab â€“ 8b and â€“4a + 2ab + 3b

= (8a â€“ 6ab + 5b) + (â€“6a â€“ ab â€“ 8b) + (â€“4a + 2ab + 3b)

Collecting positive and negative like terms together, we get

= 8a â€“ 6a â€“ 4a â€“ 6ab â€“ ab + 2ab + 5b â€“ 8b + 3b

= 8a â€“ 10a â€“ 7ab + 2ab + 8b â€“ 8b

= â€“2a â€“ 5ab

(ii) Given 5x3Â + 7 + 6x – 5x2, 2x2Â â€“ 8 – 9x, 4x – 2x2Â + 3 x 3, 3 x 3 – 9x – x2Â and x – x2Â – x3Â â€“ 4

= (5 x 3 + 7+ 6x – 5x2) + (2 x 2 â€“ 8 – 9x) + (4x – 2x2Â + 3 x 3) + (3 x 3 – 9x-x2) + (x – x2Â – x3Â – 4)

Collecting positive and negative like terms together, we get

5x3Â + 3x3Â + 3x3Â – x3Â – 5x2Â + 2x2Â – 2x2– x2Â – x2Â + 6x – 9x + 4x – 9x + x + 7 â€“ 8 – 4

= 10x3Â – 7x2Â – 7x â€“ 5

(i) x â€“ 3y â€“ 2z

5x + 7y â€“ 8z

3x â€“ 2y + 5z

(ii) 4ab â€“ 5bc + 7ca

â€“3ab + 2bc â€“ 3ca

5ab â€“ 3bc + 4ca

Solution:

(i) Given x â€“ 3y â€“ 2z, 5x + 7y â€“ 8z and 3x â€“ 2y + 5z

= (x â€“ 3y â€“ 2z) + (5x + 7y â€“ 8z) + (3x â€“ 2y + 5z)

Collecting positive and negative like terms together, we get

= x + 5x + 3x â€“ 3y + 7y â€“ 2y â€“ 2z â€“ 8z + 5z

= 9x â€“ 5y + 7y â€“ 10z + 5z

= 9x + 2y â€“ 5z

(ii) Given 4ab â€“ 5bc + 7ca, â€“3ab + 2bc â€“ 3ca and 5ab â€“ 3bc + 4ca

= (4ab â€“ 5bc + 7ca) + (â€“3ab + 2bc â€“ 3ca) + (5ab â€“ 3bc + 4ca)

Collecting positive and negative like terms together, we get

= 4ab â€“ 3ab + 5ab â€“ 5bc + 2bc â€“ 3bc + 7ca â€“ 3ca + 4ca

= 9ab â€“ 3ab â€“ 8bc + 2bc + 11ca â€“ 3ca

= 6ab â€“ 6bc + 8ca

7. Add 2x2Â – 3x + 1 to the sum of 3x2Â – 2x and 3x + 7.

Solution:

Given 2x2Â – 3x + 1, 3x2Â – 2x and 3x + 7

sum of 3x2Â – 2x and 3x + 7

= (3x2Â – 2x) + (3x +7)

=3x2Â – 2x + 3x + 7

= (3x2Â + x + 7)

Now, required expression = 2x2Â – 3x + 1+ (3x2Â + x + 7)

= 2x2Â + 3x2Â – 3x + x + 1 + 7

= 5x2Â – 2x + 8

8. Add x2Â + 2xy + y2Â to the sum of x2Â – 3y2and 2x2Â – y2Â + 9.

Solution:

Given x2Â + 2xy + y2,Â x2Â – 3y2and 2x2Â – y2Â + 9.

First we have to find the sum of x2Â – 3y2Â and 2x2Â – y2Â + 9

= (x2Â – 3y2) + (2x2Â – y2Â + 9)

= x2Â + 2x2Â – 3y2Â – y2+ 9

= 3x2Â – 4y2Â + 9

Now, required expression = (x2Â + 2xy + y2) + (3x2Â – 4y2Â + 9)

= x2Â + 3x2Â + 2xy + y2Â – 4y2Â + 9

= 4x2Â + 2xyÂ  – 3y2+ 9

9. Add a3+ b3Â – 3 to the sum of 2a3Â – 3b3Â – 3ab + 7 and -a3Â + b3Â + 3ab – 9.

Solution:

Given a3+ b3Â â€“ 3, 2a3Â – 3b3Â – 3ab + 7 and -a3Â + b3Â + 3ab – 9.

First, we need to find the sum of 2a3Â – 3b3– 3ab + 7 and – a3Â + b3Â + 3ab – 9.

= (2a3Â – 3b3– 3ab + 7) + (- a3Â + b3Â + 3ab – 9)

Collecting positive and negative like terms together, we get

= 2a3Â – a3– 3b3+ b3Â – 3ab + 3ab + 7 – 9

= a3Â – 2b3Â – 2

Now, the required expression = (a3Â + b3Â – 3) + (a3Â – 2b3Â – 2).

= a3+ a3+ b3– 2b3Â – 3 – 2

= 2a3Â – b3Â – 5

10. Subtract:

(i) 7a2b from 3a2b

(ii) 4xy from -3xy

Solution:

(i) Given 7a2b from 3a2b

= 3a2b -7a2b

= (3 -7) a2b

= – 4a2b

(ii) Given 4xy from -3xy

= â€“3xy â€“ 4xy

= â€“7xy

11. Subtract:

(i) – 4x from 3y

(ii) – 2x from â€“ 5y

Solution:

(i) Given – 4x from 3y

= (3y) â€“ (â€“4x)

= 3y + 4x

(ii) Given – 2x from â€“ 5y

= (-5y) â€“ (â€“2x)

= â€“5y + 2x

12. Subtract:

(i) 6x3Â âˆ’7x2Â + 5x âˆ’ 3 from 4 âˆ’ 5x + 6x2Â âˆ’ 8x3

(ii) âˆ’ x2Â âˆ’3z from 5x2Â â€“ y + z + 7

(iii) x3Â + 2x2y + 6xy2Â âˆ’ y3Â from y3âˆ’3xy2âˆ’4x2y

Solution:

(i) Given 6x3Â âˆ’7x2Â + 5x âˆ’ 3 and 4 âˆ’ 5x + 6x2Â âˆ’ 8x3

= (4 – 5x + 6x2Â – 8x3) – (6x3Â – 7x2Â + 5x – 3)

= 4 – 5x + 6x2Â – 8x3Â – 6x3Â + 7x2Â – 5x + 3

= – 8x3– 6x3Â + 7x2Â + 6x2– 5x – 5x + 3 + 4

= – 14x3Â + 13x2Â – 10x +7

(ii) Given âˆ’ x2Â âˆ’3z and 5x2Â â€“ y + z + 7

= (5x2Â – y + z + 7) – (- x2Â – 3z)

= 5x2Â – y + z + 7 + x2Â + 3z

= 5x2+ x2Â – y + z + 3z + 7

= 6x2Â – y + 4z + 7

(iii) Given x3Â + 2x2y + 6xy2Â âˆ’ y3Â and y3âˆ’3xy2âˆ’4x2y

= (y3Â – 3xy2Â – 4x2y) – (x3Â + 2x2y + 6xy2Â – y3)

= y3Â – 3xy2Â – 4x2y – x3Â – 2x2y – 6xy2Â + y3

= y3Â + y3– 3xy2– 6xy2– 4x2y – 2x2y – x3

= 2y3– 9xy2Â – 6x2y – x3

13. From

(i) p3 â€“ 4 + 3p2, take away 5p2Â âˆ’ 3p3Â + p âˆ’ 6

(ii) 7 + x âˆ’ x2, take away 9 + x + 3x2Â + 7x3

(iii) 1âˆ’ 5y2, take away y3Â + 7y2Â + y + 1

(iv) x3Â âˆ’ 5x2Â + 3x + 1, take away 6x2Â âˆ’ 4x3Â + 5 + 3x

Solution:

(i) Given p3 â€“ 4 + 3p2, take away 5p2Â âˆ’ 3p3Â + p âˆ’ 6

= (p3Â – 4 + 3p2) – (5p2Â – 3p3Â + p – 6)

= p3Â – 4 + 3p2Â – 5p2Â + 3p3Â – p + 6

= p3Â + 3p3Â + 3p2Â – 5p2– p – 4+ 6

= 4p3Â – 2p2Â – p + 2

(ii) Given 7 + x âˆ’ x2, take away 9 + x + 3x2Â + 7x3

= (7 + x – x2) – (9 + x + 3x2Â + 7x3)

= 7 + x – x2Â – 9 – x – 3x2Â – 7x3

= – 7x3– x2Â – 3x2Â + 7 – 9

= – 7x3Â – 4x2Â – 2

(iii) Given 1âˆ’ 5y2, take away y3Â + 7y2Â + y + 1

= (1 – 5y2) – (y3+ 7y2Â + y + 1)

= 1 – 5y2Â – y3Â – 7y2Â – y – 1

= – y3– 5y2Â – 7y2Â – y

= – y3– 12y2Â – y

(iv) Given x3Â âˆ’ 5x2Â + 3x + 1, take away 6x2Â âˆ’ 4x3Â + 5 + 3x

= (x3Â – 5x2Â + 3x + 1) – (6x2Â – 4x3Â + 5 +3x)

= x3Â – 5x2Â + 3x + 1 – 6x2Â + 4x3Â – 5 – 3x

= x3+ 4x3Â – 5x2Â – 6x2Â + 1 – 5

= 5x3Â – 11x2Â â€“ 4

14. From the sum of 3x2Â âˆ’ 5x + 2 and âˆ’ 5x2Â âˆ’ 8x + 9 subtract 4x2Â âˆ’ 7x + 9.

Solution:

First we have to add 3x2Â âˆ’ 5x + 2 and âˆ’ 5x2Â âˆ’ 8x + 9 then from the result we have to subtract 4x2Â âˆ’ 7x + 9.

= {(3x2Â – 5x + 2) + (- 5x2Â – 8x + 9)} – (4x2Â – 7x + 9)

= {3x2Â – 5x + 2 – 5x2Â – 8x + 9} –Â (4x2Â – 7x + 9)

= {3x2Â – 5x2Â – 5x – 8x + 2 + 9} –Â (4x2Â – 7x + 9)

= {- 2x2Â – 13x +11} – (4x2Â – 7x + 9)

= – 2x2Â – 13x + 11 – 4x2Â + 7x â€“ 9

= – 2x2Â – 4x2Â – 13x + 7x + 11 – 9

= – 6x2Â – 6x + 2

15. Subtract the sum of 13x â€“ 4y + 7z and â€“ 6z + 6x + 3y from the sum of 6x â€“ 4y â€“ 4z andÂ Â  2x + 4y â€“ 7.

Solution:

First we have to find the sum of 13x â€“ 4y + 7z and â€“ 6z + 6x + 3y

Therefore, sum of (13x â€“ 4y + 7z) and (â€“6z + 6x + 3y)

= (13x â€“ 4y + 7z) + (â€“6z + 6x + 3y)

= (13x â€“ 4y + 7z â€“ 6z + 6x + 3y)

= (13x + 6x â€“ 4y + 3y + 7z â€“ 6z)

= (19x â€“ y + z)

Now we have to find the sum of (6x â€“ 4y â€“ 4z) and (2x + 4y â€“ 7)

= (6x â€“ 4y â€“ 4z) + (2x + 4y â€“ 7)

= (6x â€“ 4y â€“ 4z + 2x + 4y â€“ 7)

= (6x + 2x â€“ 4z â€“ 7)

= (8x â€“ 4z â€“ 7)

Now, required expression = (8x â€“ 4z â€“ 7) â€“ (19x â€“ y + z)

= 8x â€“ 4z â€“ 7 â€“ 19x + y â€“ z

= 8x â€“ 19x + y â€“ 4z â€“ z â€“ 7

= â€“11x + y â€“ 5z â€“ 7

16. From the sum of x2Â + 3y2Â âˆ’ 6xy, 2x2Â âˆ’ y2Â + 8xy, y2Â + 8 and x2Â âˆ’ 3xy subtract âˆ’3x2Â + 4y2Â â€“ xy + x â€“ y + 3.

Solution:

First we have to find the sum of (x2Â + 3y2Â – 6xy), (2x2Â – y2Â + 8xy), (y2Â + 8) and (x2Â – 3xy)

={(x2Â + 3y2Â – 6xy) + (2x2Â – y2Â + 8xy) + ( y2Â + 8) + (x2Â – 3xy)}

={x2Â + 3y2Â – 6xy + 2x2Â – y2Â + 8xy + y2Â + 8 + x2Â – 3xy}

= {x2+ 2x2+ x2Â + 3y2– y2Â + y2– 6xy + 8xy – 3xy + 8}

= 4x2Â + 3y2Â – xy + 8

Now, from the result subtract the âˆ’3x2Â + 4y2Â â€“ xy + x â€“ y + 3.

Therefore, required expression = (4x2Â + 3y2Â – xy + 8) – (- 3x2Â + 4y2Â – xy + x – y + 3)

= 4x2Â + 3y2Â – xy + 8 + 3x2Â – 4y2Â + xy – x + y – 3

= 4x2Â + 3x2+ 3y2– 4y2– x + y – 3 + 8

= 7x2Â – y2– x + y + 5

17. What should be added to xy â€“ 3yz + 4zx to get 4xy â€“ 3zx + 4yz + 7?

Solution:

By subtracting xy â€“ 3yz + 4zx from 4xy â€“ 3zx + 4yz + 7, we get the required expression.

Therefore, required expression = (4xy â€“ 3zx + 4yz + 7) â€“ (xy â€“ 3yz + 4zx)

= 4xy â€“ 3zx + 4yz + 7 â€“ xy + 3yz â€“ 4zx

= 4xy â€“ xy â€“ 3zx â€“ 4zx + 4yz + 3yz + 7

= 3xy â€“ 7zx + 7yz + 7

18. What should be subtracted from x2Â â€“ xy + y2Â â€“ x + y + 3 to obtain âˆ’x2Â + 3y2Â âˆ’ 4xy + 1?

Solution:

Let ‘E’ be the required expression. Then, we have

x2Â – xy + y2– x + y + 3 – E = – x2Â + 3y2Â – 4xy + 1

Therefore, E = (x2Â – xy + y2– x + y + 3) – (- x2Â + 3y2Â – 4xy + 1)

= x2Â – xy + y2– x + y + 3 + x2Â – 3y2Â + 4xy – 1

Collecting positive and negative like terms together, we get

= x2Â + x2– xy + 4xy + y2– 3y2Â – x + y + 3 – 1

= 2x2+ 3xy- 2y2– x + y + 2

19. How much is x â€“ 2y + 3z greater than 3x + 5y â€“ 7?

Solution:

By subtracting x â€“ 2y + 3z from 3x + 5y â€“ 7 we can get the required expression,

Required expression = (x â€“ 2y + 3z) â€“ (3x + 5y â€“ 7)

= x â€“ 2y + 3z â€“ 3x â€“ 5y + 7

Collecting positive and negative like terms together, we get

= x â€“ 3x â€“ 2y + 5y + 3z + 7

= â€“2x â€“ 7y + 3z + 7

20. How much is x2Â âˆ’ 2xy + 3y2Â less than 2x2Â âˆ’ 3y2Â + xy?

Solution:

By subtracting the x2Â âˆ’ 2xy + 3y2Â  from 2x2Â âˆ’ 3y2Â + xy we can get the required expression,

Required expression = (2x2Â – 3y2Â + xy) – (x2Â – 2xy + 3y2)

= 2x2Â – 3y2Â + xy – x2Â + 2xy – 3y2

Collecting positive and negative like terms together, we get

= 2x2– x2Â – 3y2Â – 3y2Â + xy + 2xy

= x2Â – 6y2Â + 3xy

21. How much does a2Â âˆ’ 3ab + 2b2Â exceed 2a2Â âˆ’ 7ab + 9b2?

Solution:

By subtracting 2a2Â âˆ’ 7ab + 9b2 from a2Â âˆ’ 3ab + 2b2Â we get the required expression

Required expression = (a2– 3ab + 2b2) – (2a2Â – 7ab + 9b2)

= a2– 3ab + 2b2Â – 2a2Â + 7ab – 9b2

Collecting positive and negative like terms together, we get

=Â a2Â – 2a2 – 3ab + 7ab + 2b2Â – 9b2

= – a2Â + 4ab – 7b2

22. What must be added to 12x3Â âˆ’ 4x2Â + 3x âˆ’ 7 to make the sum x3Â + 2x2Â âˆ’ 3x + 2?

Solution:

Let ‘E’ be the required expression. Thus, we have

12x3Â – 4x2Â + 3x – 7 + E = x3Â + 2x2Â – 3x + 2

Therefore, E = (x3Â + 2x2Â – 3x + 2) – (12x3Â – 4x2Â + 3x – 7)

=Â  x3Â + 2x2Â – 3x + 2 – 12x3Â + 4x2Â – 3x + 7

Collecting positive and negative like terms together, we get

= x3– 12x3+ 2x2Â + 4x2Â – 3x – 3x + 2 + 7

= – 11x3Â + 6x2Â – 6x + 9

23. If P = 7x2Â + 5xy âˆ’ 9y2, Q = 4y2Â âˆ’ 3x2Â âˆ’ 6xy and R = âˆ’4x2Â + xy + 5y2, show that P + Q + R = 0.

Solution:

Given P = 7x2Â + 5xy âˆ’ 9y2, Q = 4y2Â âˆ’ 3x2Â âˆ’ 6xy and R = âˆ’4x2Â + xy + 5y2

Now we have to prove P + Q + R = 0,

Consider P + Q + R = (7x2Â + 5xy – 9y2) + (4y2Â – 3x2Â – 6xy) + (- 4x2Â + xy + 5y2)

= 7x2Â + 5xy – 9y2Â + 4y2Â – 3x2Â – 6xy – 4x2Â + xy + 5y2

Collecting positive and negative like terms together, we get

= 7x2– 3x2Â – 4x2Â + 5xy – 6xy + xy – 9y2Â + 4y2Â + 5y2

= 7x2– 7x2Â + 6xy – 6xyÂ Â – 9y2Â + 9y2

= 0

24. If P = a2Â âˆ’ b2Â + 2ab, Q = a2Â + 4b2Â âˆ’ 6ab, R = b2Â + b, S = a2Â âˆ’ 4ab and T = âˆ’2a2Â + b2Â â€“ ab + a. Find P + Q + R + S â€“ T.

Solution:

Given P = a2Â âˆ’ b2Â + 2ab, Q = a2Â + 4b2Â âˆ’ 6ab, R = b2Â + b, S = a2Â âˆ’ 4ab and T = âˆ’2a2Â + b2Â â€“ ab + a.

Now we have to find P + Q + R + S â€“ T

Substituting all values we get

Consider P + Q + R + S – T = {(a2Â – b2Â + 2ab) + (a2Â + 4b2Â – 6ab) + (b2Â + b) + (a2Â – 4ab)} – (-2a2Â + b2Â – ab + a)

= {a2Â – b2Â + 2ab + a2Â + 4b2Â – 6ab + b2Â + b + a2Â – 4ab}- (- 2a2Â + b2Â – ab + a)

= {3a2Â + 4b2Â – 8ab + b } – (-2a2Â + b2Â – ab + a)

= 3a2+ 4b2Â – 8ab + b + 2a2Â – b2Â + ab – a

Collecting positive and negative like terms together, we get

3a2+ 2a2Â + 4b2Â – b2Â – 8ab + ab – a + b

= 5a2Â + 3b2– 7ab – a + b

Exercise 7.3 Page No: 7.16

1. Place the last two terms of the following expressions in parentheses preceded by a minus sign:

(i) x + y â€“ 3z + yÂ Â Â Â

(ii) 3x â€“ 2y â€“ 5z â€“ 4

(iii) 3a â€“ 2b + 4c â€“ 5

(iv) 7a + 3b + 2c + 4

(v) 2a2Â – b2Â – 3ab + 6

(vi) a2Â + b2Â – c2Â + ab – 3ac

Solution:

(i) Given x + y â€“ 3z + y

x + y â€“ 3z + y = x + y â€“ (3z â€“ y)

(ii) Given 3x â€“ 2y â€“ 5z â€“ 4

3x â€“ 2y â€“ 5z â€“ 4 = 3x â€“ 2y â€“ (5z + 4)

(iii) Given 3a â€“ 2b + 4c â€“ 5

3a â€“ 2b + 4c â€“ 5 = 3a â€“ 2b â€“ (â€“4c + 5)

(iv) Given 7a + 3b + 2c + 4

7a + 3b + 2c + 4 = 7a + 3b â€“ (â€“2c â€“ 4)

(v) Given 2a2Â – b2Â – 3ab + 6

2a2Â – b2Â – 3ab + 6 = 2a2Â – b2Â – (3ab – 6)

(vi) Given a2Â + b2Â – c2Â + ab – 3ac

a2Â + b2Â – c2Â + ab – 3ac = a2Â + b2Â – c2Â – (- ab + 3ac)

2. Write each of the following statements by using appropriate grouping symbols:

(i) The sum of a â€“ b and 3a â€“ 2b + 5 is subtracted from 4a + 2b â€“ 7.

(ii) Three times the sum of 2x + y â€“ [5 â€“ (x â€“ 3y)] and 7x â€“ 4y + 3 is subtracted from 3x â€“ 4y + 7

(iii) The subtraction of x2Â – y2Â + 4xy from 2x2Â + y2Â – 3xy is added to 9x2Â – 3y2– xy.

Solution:

(i) Given the sum of a â€“ b and 3a â€“ 2b + 5 = [(a â€“ b) + (3a â€“ 2b + 5)].

This is subtracted from 4a + 2b â€“ 7.

Thus, the required expression is (4a + 2b â€“ 7) â€“ [(a â€“ b) + (3a â€“ 2b + 5)]

(ii) Given three times the sum of 2x + y â€“ {5 â€“ (x â€“ 3y)} and 7x â€“ 4y + 3 = 3[(2x + y â€“ {5 â€“ (x â€“ 3y)}) + (7x â€“ 4y + 3)]

This is subtracted from 3x â€“ 4y + 7.

Thus, the required expression is (3x â€“ 4y + 7) â€“ 3[(2x + y â€“ {5 â€“ (x â€“ 3y)}) + (7x â€“ 4y + 3)]

(iii) Given the product of subtraction of x2– y2Â + 4xy from 2x2Â + y2Â – 3xy is given by {(2x2Â + y2Â – 3xy) â€“ (x2-y2Â + 4xy)}

When the above equation is added to 9x2Â – 3y2Â – xy, we get

{(2x2Â + y2Â – 3xy) â€“ (x2Â – y2Â + 4xy)} + (9x2Â – 3y2– xy))

This is the required expression.

Exercise 7.4 Page No: 7.20

Simplify each of the following algebraic expressions by removing grouping symbols.

1. 2x + (5x â€“ 3y)

Solution:

Given 2x + (5x â€“ 3y)

Since the â€˜+â€™ sign precedes the parentheses, we have to retain the sign of each term in the parentheses when we remove them.

= 2x + 5x â€“ 3y

On simplifying, we get

= 7x â€“ 3y

2. 3x – (y – 2x)

Solution:

Given 3x â€“ (y â€“ 2x)

Since the â€˜â€“â€™ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= 3x â€“ y + 2x

On simplifying, we get

= 5x â€“ y

3. 5a â€“ (3b â€“ 2a + 4c)

Solution:

Given 5a â€“ (3b â€“ 2a + 4c)

Since the â€˜-â€˜sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them.

= 5a â€“ 3b + 2a â€“ 4c

On simplifying, we get

= 7a â€“ 3b â€“ 4c

4. -2(x2Â – y2Â + xy) – 3(x2Â +y2Â – xy)

Solution:

Given – 2(x2Â – y2Â + xy) – 3(x2Â +y2Â – xy)

Since the â€˜â€“â€™ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= -2x2Â + 2y2Â – 2xy – 3x2Â – 3y2Â + 3xy

On rearranging,

= -2x2Â – 3x2Â + 2y2Â – 3y2Â – 2xy + 3xy

On simplifying, we get

= -5x2Â – y2Â + xy

5. 3x + 2y â€“ {x â€“ (2y â€“ 3)}

Solution:

Given 3x + 2y â€“ {x â€“ (2y â€“ 3)}

First, we have to remove the parentheses. Then, we have to remove the braces.

Then we get,

= 3x + 2y â€“ {x â€“ 2y + 3}

= 3x + 2y â€“ x + 2y â€“ 3

On simplifying, we get

= 2x + 4y â€“ 3

6. 5a â€“ {3a â€“ (2 â€“ a) + 4}

Solution:

Given 5a â€“ {3a â€“ (2 â€“ a) + 4}

First, we have to remove the parentheses. Then, we have to remove the braces.

Then we get,

= 5a â€“ {3a â€“ 2 + a + 4}

= 5a â€“ 3a + 2 â€“ a â€“ 4

On simplifying, we get

= 5a â€“ 4a â€“ 2

= a â€“ 2

7. a â€“ [b â€“ {a â€“ (b â€“ 1) + 3a}]

Solution:

Given a â€“ [b â€“ {a â€“ (b â€“ 1) + 3a}]

First we have to remove the parentheses, then the curly brackets, and then the square brackets.

Then we get,

= a â€“ [b â€“ {a â€“ (b â€“ 1) + 3a}]

= a â€“ [b â€“ {a â€“ b + 1 + 3a}]

= a â€“ [b â€“ {4a â€“ b + 1}]

= a â€“ [b â€“ 4a + b â€“ 1]

= a â€“ [2b â€“ 4a â€“ 1]

On simplifying, we get

= a â€“ 2b + 4a + 1

= 5a â€“ 2b + 1

8. Â a â€“ [2b â€“ {3a â€“ (2b â€“ 3c)}]

Solution:

Given a â€“ [2b â€“ {3a â€“ (2b â€“ 3c)}]

First we have to remove the parentheses, then the braces, and then the square brackets.

Then we get,

= a â€“ [2b â€“ {3a â€“ (2b â€“ 3c)}]

= a â€“ [2b â€“ {3a â€“ 2b + 3c}]

= a â€“ [2b â€“ 3a + 2b â€“ 3c]

= a â€“ [4b â€“ 3a â€“ 3c]

On simplifying we get,

= a â€“ 4b + 3a + 3c

= 4a â€“ 4b + 3c

9. -x + [5y â€“ {2x â€“ (3y â€“ 5x)}]

Solution:

Given -x + [5y â€“ {2x â€“ (3y â€“ 5x)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= â€“ x + [5y â€“ {2x â€“ (3y â€“ 5x)}]

= â€“ x + [5y â€“ {2x â€“ 3y + 5x)]

= â€“ x + [5y â€“ {7x â€“ 3y}]

= â€“ x + [5y â€“ 7x + 3y]

= â€“ x + [8y â€“ 7x]

On simplifying we get

= â€“ x + 8y â€“ 7x

= â€“ 8x + 8y

10. 2a â€“ [4b â€“ {4a â€“ 3(2a â€“ b)}]

Solution:

Given 2a â€“ [4b â€“ {4a â€“ 3(2a â€“ b)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= 2a â€“ [4b â€“ {4a â€“ 3(2a â€“ b)}]

= 2a â€“ [4b â€“ {4a â€“ 6a + 3b}]

= 2a â€“ [4b â€“ {- 2a + 3b}]

= 2a â€“ [4b + 2a â€“ 3b]

= 2a â€“ [b + 2a]

On simplifying, we get

= 2a â€“ b â€“ 2a

= â€“ b

11. -a â€“ [a + {a + b â€“ 2a â€“ (a â€“ 2b)} – b]

Solution:

Given -a â€“ [a + {a + b â€“ 2a â€“ (a â€“ 2b)} – b]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= â€“ a â€“ [a + {a + b â€“ 2a â€“ (a â€“ 2b)} â€“ b]

= â€“ a â€“ [a + {a + b â€“ 2a â€“ a + 2b} â€“ b]

= â€“ a â€“ [a + {- 2a + 3b} â€“ b]

= â€“ a â€“ [a â€“ 2a + 3b â€“ b]

= â€“ a â€“ [- a + 2b]

On simplifying, we get

= â€“ a + a â€“ 2b

= â€“ 2b

12. 2x â€“ 3y â€“ [3x â€“ 2y -{x â€“ z â€“ (x â€“ 2y)}]

Solution:

Given 2x â€“ 3y â€“ [3x â€“ 2y -{x â€“ z â€“ (x â€“ 2y)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= 2x â€“ 3y â€“ [3x â€“ 2y â€“ {x â€“ z â€“ (x â€“ 2y)})

= 2x â€“ 3y â€“ [3x â€“ 2y â€“ {x â€“ z â€“ x + 2y}]

= 2x â€“ 3y â€“ [3x â€“ 2y â€“ {- z + 2y}]

= 2x â€“ 3y â€“ [3x â€“ 2y + z â€“ 2y]

= 2x â€“ 3y â€“ [3x â€“ 4y + z]

On simplifying, we get

= 2x â€“ 3y â€“ 3x + 4y â€“ z

= – x + y â€“ z

13. 5 + [x â€“ {2y â€“ (6x + y â€“ 4) + 2x} â€“ {x â€“ (y â€“ 2)}]

Solution:

Given 5 + [x â€“ {2y â€“ (6x + y â€“ 4) + 2x} â€“ {x â€“ (y â€“ 2)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= 5 + [x â€“ {2y â€“ (6x + y â€“ 4) + 2x} â€“ {x â€“ (y â€“ 2)}]

= 5 + [x â€“ {2y â€“ 6x â€“ y + 4 + 2x} â€“ {x â€“ y + 2}]

= 5 + [x â€“ {y â€“ 4x + 4} â€“ {x â€“ y + 2}]

= 5 + [x â€“ y + 4x â€“ 4 â€“ x + y â€“ 2]

= 5 + [4x â€“ 6]

= 5 + 4x â€“ 6

= 4x â€“ 1

14. x2Â – [3x + [2x – (x2Â – 1)] + 2]

Solution:

Given x2Â – [3x + [2x – (x2Â – 1)] + 2]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= x2Â – [3x + [2x – (x2Â – 1)] + 2]

= x2Â – [3x + [2x – x2Â + 1] + 2]

= x2Â – [3x + 2x – x2Â + 1 + 2]

= x2Â – [5x – x2Â + 3]

On simplifying we get

= x2Â – 5x + x2Â – 3

= 2x2Â – 5x â€“ 3

15. 20 – [5xy + 3[x2Â – (xy – y) – (x – y)]]

Solution:

Given 20 – [5xy + 3[x2Â – (xy – y) – (x – y)]]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= 20 – [5xy + 3[x2Â – (xy – y) – (x – y)]]

= 20 – [5xy + 3[x2Â â€“ xy + y â€“ x + y]]

= 20 – [5xy + 3[x2Â â€“ xy + 2y – x]]

= 20 – [5xy + 3x2Â – 3xy + 6y – 3x]

= 20 – [2xy + 3x2Â + 6y – 3x]

On simplifying we get

= 20 – 2xy – 3x2Â – 6y + 3x

= – 3x2Â – 2xy – 6y + 3x + 20

16. 85 â€“ [12x â€“ 7(8x â€“ 3) â€“ 2{10x â€“ 5(2 â€“ 4x)}]

Solution:

Given 85 â€“ [12x â€“ 7(8x â€“ 3) â€“ 2{10x â€“ 5(2 â€“ 4x)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= 85 â€“ [12x â€“ 7(8x â€“ 3) â€“ 2{10x â€“ 5(2 â€“ 4x)}]

= 85 â€“ [12x â€“ 56x + 21 â€“ 2{10x â€“ 10 + 20x}]

= 85 â€“ [12x â€“ 56x + 21 â€“ 2{30x â€“ 10}]

= 85 â€“ [12x â€“ 56x + 21 â€“ 60x + 20]

= 85 â€“ [12x â€“ 116x + 41]

= 85 â€“ [- 104x + 41]

On simplifying, we get

= 85 + 104x â€“ 41

= 44 + 104x

17. xy [yz â€“ zx â€“ {yx â€“ (3y â€“ xz) â€“ (xy â€“ zy)}]

Solution:

Given xy [yz â€“ zx â€“ {yx â€“ (3y â€“ xz) â€“ (xy â€“ zy)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= xy â€“ [yz â€“ zx â€“ {yx â€“ (3y â€“ xz) â€“ (xy â€“ zy)}]

= xy â€“ [yz â€“ zx â€“ {yx â€“ 3y + xz â€“ xy + zy}]

= xy â€“ [yz â€“ zx â€“ {- 3y + xz + zy}]

= xy â€“ [yz â€“ zx + 3y â€“ xz â€“ zy]

= xy â€“ [- zx + 3y â€“ xz]

On simplifying, we get

= xy â€“ [- 2zx + 3y]

= xy + 2xz â€“ 3y