RD Sharma Solutions For Class 7 Maths Chapter 7 Algebraic Expressions Free Online
Exercise 7.1 Page No: 7.7
1. Identify the monomials, binomials, trinomials and quadrinomials from the following expressions:
(i) a2
(ii) a2 − b2
(iii) x3 + y3 + z3
(iv) x3 + y3 + z3 + 3xyz
(v) 7 + 5
(vi) a b c + 1
(vii) 3x – 2 + 5
(viii) 2x – 3y + 4
(ix) x y + y z + z x
(x) ax3 + bx2 + cx + d
Solution:
(i) Given a2
a2 is a monomial expression because it contains only one term
(ii) Given a2 − b2
a2 − b2 is a binomial expression because it contains two terms
(iii) Given x3 + y3 + z3
x3 + y3 + z3 is a trinomial because it contains three terms
(iv) Given x3 + y3 + z3 + 3xyz
x3 + y3 + z3 + 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 – 2 + 5
3x – 2 + 5 is a binomial expression because it contains two terms
(viii) Given 2x – 3y + 4
2x – 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 ax3 + bx2 + cx + d
ax3 + bx2 + 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 – 3
(iii) 2x2 − 7
(iv) 2x2 + y2 − 3xy + 4
Solution:
(i) Given 3x
3x is the only term of the given algebraic expression.
(ii) Given 2x – 3
2x and -3 are the terms of the given algebraic expression.
(iii) Given 2x2 − 7
2x2 and −7 are the terms of the given algebraic expression.
(iv) Given 2x2 + y2 − 3xy + 4
2x2, y2, −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, −3ca2b) and (−4/3cba2) and their Numerical coefficients – 7, -3,
(-4/3)
Like terms are (−5/2abc2) and (3/2 abc2) and numerical coefficients are (−5/2) and (3/2)
4. Identify the like terms in the following algebraic expressions:
(i) a2 + b2 -2a2 + c2 + 4a
(ii) 3x + 4xy − 2yz + 52zy
(iii) abc + ab2c + 2acb2 + 3c2ab + b2ac − 2a2bc + 3cab2
Solution:
(i) Given a2 + b2 -2a2 + c2 + 4a
The like terms in the given algebraic expressions are a2 and −2a2.
(ii) Given 3x + 4xy − 2yz + 52zy
The like terms in the given algebraic expressions are -2yz and 5/2zy.
(iii) Given abc + ab2c + 2acb2 + 3c2ab + b2ac − 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) –12x
(ii) –7xy
(iii) xyz
(iv) –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) −3x2
(ii) 5x2yz
(iii) 5/7x2z
(iv) (-3/2) ax2 + yx
Solution:
(i) Given −3x2
The numerical coefficient of x2 is -3.
(ii) Given 5x2yz
The numerical coefficient of x2 is 5yz.
(iii) Given5/7x2z
The numerical coefficient of x2 is 57z.
(iv) Given (-3/2) ax2 + yx
The numerical coefficient of x2 is – (3/2) a.
7. Write the coefficient of:
(i) y in –3y
(ii) a in 2ab
(iii) z in –7xyz
(iv) p in –3pqr
(v) y2 in 9xy2z
(vi) x3 in x3 +1
(vii) x2 in − x2
Solution:
(i) Given –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 y2 is 9xz.
(vi) Given x3 +1
The coefficient of x3 is 1.
(vii) Given − x2
The coefficient of x2 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 −2x3y2z is -2.
9. Write the numerical coefficient of each term in the following algebraic expressions:
(i) 4x2y – (3/2)xy + 5/2 xy2
(ii) (–5/3)x2y + (7/4)xyz + 3
Solution:
(i) Given 4x2y – (3/2) xy + 5/2 xy2
Numerical coefficient of following algebraic expressions are given below
Term
|
Coefficient
|
4x2y
|
4
|
– (3/2) xy
|
-(3/2)
|
5/2 xy2
|
(5/2)
|
(ii) Given (–5/3)x2y + (7/4)xyz + 3
Numerical coefficient of following algebraic expressions are given below
Term
|
Coefficient
|
(–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 − xy2 + 7xy − 3
(ii) a3 − 3a2 + 7a + 5
Solution:
(i) Given x2y − xy2 + 7xy − 3
The constant term in the given algebraic expressions is -3.
(ii) Given a3 − 3a2 + 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 – xy – yz – 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 – 2 – 9)/6
= (1/6)
(ii) Given x = -2, y = -1, z = 3
Consider x2 + y2 + z2 – xy – yz – zx
On substituting the given values we get,
= (-2)2 + (-1)2 + 32 – (-2) (-1) – (-1) (3) – (3) (-2)
= 4 + 1 + 9 – 2 + 3 + 6
= 23 – 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) ax2 + by2 – 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)
= –2 – 1 – 4
= –7
(ii) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2
Consider ax2 + by2 – cz
On substituting the given values
= (-2) × 12 + 1 × (-1)2 – (-2) × 2
= 4 + 1 – (-4)
= 5 + 4
= 9
(iii) Given x = 1, y = -1, z = 2, a = -2, b = 1, c = -2
Consider axy + byz + cxy
= (-2) × 1 × -1 + 1 × -1 × 2 + (-2) × 1 × (-1)
= 2 + (-2) + 2
= 4 – 2
= 2
Exercise 7.2 Page No: 7.13
1. Add the following:
(i) 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 + 9yx3 = (7 + 9) x3y
= 16x3y
(ii) Given
12a2b + 3ba2 = (12 + 3) a2b
= 15a2b
3. Add the following:
(i) 7abc, -5abc, 9abc, -8abc
(ii) 2x2y, – 4x2y, 6x2y, -5x2y
Solution:
(i) Given 7abc, -5abc, 9abc, -8abc
Consider 7abc + (-5abc) + (9abc) + (-8abc)
= 7abc – 5abc + 9abc – 8abc
= (7 – 5 + 9 – 8) abc [by taking abc common]
= (16 – 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
4. Add the following expressions:
(i) x3 -2x2y + 3xy2– y3, 2x3– 5xy2 + 3x2y – 4y3
(ii) a4 – 2a3b + 3ab3 + 4a2b2 + 3b4, – 2a4 – 5ab3 + 7a3b – 6a2b2 + b4
Solution:
(i) Given x3 -2x2y + 3xy2– y3, 2x3– 5xy2 + 3x2y – 4y3
Collecting positive and negative like terms together, we get
= x3 +2x3 – 2x2y + 3x2y + 3xy2 – 5xy2 – y3– 4y3
= 3x3 + x2y – 2xy2 – 5y3
(ii) Given a4 – 2a3b + 3ab3 + 4a2b2 + 3b4, – 2a4 – 5ab3 + 7a3b – 6a2b2 + b4
= a4 – 2a3b + 3ab3 + 4a2b2 + 3b4 – 2a4 – 5ab3 + 7a3b – 6a2b2 + b4
Collecting positive and negative like terms together, we get
= a4 – 2a4– 2a3b + 7a3b + 3ab3 – 5ab3 + 4a2b2 – 6a2b2 + 3b4 + b4
= – a4 + 5a3b – 2ab3 – 2a2b2 + 4b4
5. Add the following expressions:
(i) 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b
(ii) 5x3 + 7 + 6x – 5x2, 2x2 – 8 – 9x, 4x – 2x2 + 3 x 3, 3 x 3 – 9x – x2 and x – x2 – x3 – 4
Solution:
(i) Given 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b
= (8a – 6ab + 5b) + (–6a – ab – 8b) + (–4a + 2ab + 3b)
Collecting positive and negative like terms together, we get
= 8a – 6a – 4a – 6ab – ab + 2ab + 5b – 8b + 3b
= 8a – 10a – 7ab + 2ab + 8b – 8b
= –2a – 5ab
(ii) Given 5x3 + 7 + 6x – 5x2, 2x2 – 8 – 9x, 4x – 2x2 + 3 x 3, 3 x 3 – 9x – x2 and x – x2 – x3 – 4
= (5 x 3 + 7+ 6x – 5x2) + (2 x 2 – 8 – 9x) + (4x – 2x2 + 3 x 3) + (3 x 3 – 9x-x2) + (x – x2 – x3 – 4)
Collecting positive and negative like terms together, we get
5x3 + 3x3 + 3x3 – x3 – 5x2 + 2x2 – 2x2– x2 – x2 + 6x – 9x + 4x – 9x + x + 7 – 8 – 4
= 10x3 – 7x2 – 7x – 5
6. Add the following:
(i) x – 3y – 2z
5x + 7y – 8z
3x – 2y + 5z
(ii) 4ab – 5bc + 7ca
–3ab + 2bc – 3ca
5ab – 3bc + 4ca
Solution:
(i) Given x – 3y – 2z, 5x + 7y – 8z and 3x – 2y + 5z
= (x – 3y – 2z) + (5x + 7y – 8z) + (3x – 2y + 5z)
Collecting positive and negative like terms together, we get
= x + 5x + 3x – 3y + 7y – 2y – 2z – 8z + 5z
= 9x – 5y + 7y – 10z + 5z
= 9x + 2y – 5z
(ii) Given 4ab – 5bc + 7ca, –3ab + 2bc – 3ca and 5ab – 3bc + 4ca
= (4ab – 5bc + 7ca) + (–3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)
Collecting positive and negative like terms together, we get
= 4ab – 3ab + 5ab – 5bc + 2bc – 3bc + 7ca – 3ca + 4ca
= 9ab – 3ab – 8bc + 2bc + 11ca – 3ca
= 6ab – 6bc + 8ca
7. Add 2x2 – 3x + 1 to the sum of 3x2 – 2x and 3x + 7.
Solution:
Given 2x2 – 3x + 1, 3x2 – 2x and 3x + 7
sum of 3x2 – 2x and 3x + 7
= (3x2 – 2x) + (3x +7)
=3x2 – 2x + 3x + 7
= (3x2 + x + 7)
Now, required expression = 2x2 – 3x + 1+ (3x2 + x + 7)
= 2x2 + 3x2 – 3x + x + 1 + 7
= 5x2 – 2x + 8
8. Add x2 + 2xy + y2 to the sum of x2 – 3y2and 2x2 – y2 + 9.
Solution:
Given x2 + 2xy + y2, x2 – 3y2and 2x2 – y2 + 9.
First we have to find the sum of x2 – 3y2 and 2x2 – y2 + 9
= (x2 – 3y2) + (2x2 – y2 + 9)
= x2 + 2x2 – 3y2 – y2+ 9
= 3x2 – 4y2 + 9
Now, required expression = (x2 + 2xy + y2) + (3x2 – 4y2 + 9)
= x2 + 3x2 + 2xy + y2 – 4y2 + 9
= 4x2 + 2xy – 3y2+ 9
= (x2 – 3y2) + (2x2 – y2 + 9)
= x2 + 2x2 – 3y2 – y2+ 9
= 3x2 – 4y2 + 9
Now, required expression = (x2 + 2xy + y2) + (3x2 – 4y2 + 9)
= x2 + 3x2 + 2xy + y2 – 4y2 + 9
= 4x2 + 2xy – 3y2+ 9
9. Add a3+ b3 – 3 to the sum of 2a3 – 3b3 – 3ab + 7 and -a3 + b3 + 3ab – 9.
Solution:
Given a3+ b3 – 3, 2a3 – 3b3 – 3ab + 7 and -a3 + b3 + 3ab – 9.
First, we need to find the sum of 2a3 – 3b3– 3ab + 7 and – a3 + b3 + 3ab – 9.
= (2a3 – 3b3– 3ab + 7) + (- a3 + b3 + 3ab – 9)
Collecting positive and negative like terms together, we get
= 2a3 – a3– 3b3+ b3 – 3ab + 3ab + 7 – 9
= a3 – 2b3 – 2
Now, the required expression = (a3 + b3 – 3) + (a3 – 2b3 – 2).
= a3+ a3+ b3– 2b3 – 3 – 2
= 2a3 – b3 – 5
= (2a3 – 3b3– 3ab + 7) + (- a3 + b3 + 3ab – 9)
Collecting positive and negative like terms together, we get
= 2a3 – a3– 3b3+ b3 – 3ab + 3ab + 7 – 9
= a3 – 2b3 – 2
Now, the required expression = (a3 + b3 – 3) + (a3 – 2b3 – 2).
= a3+ a3+ b3– 2b3 – 3 – 2
= 2a3 – b3 – 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
= –3xy – 4xy
= –7xy
11. Subtract:
(i) – 4x from 3y
(ii) – 2x from – 5y
Solution:
(i) Given – 4x from 3y
= (3y) – (–4x)
= 3y + 4x
(ii) Given – 2x from – 5y
= (-5y) – (–2x)
= –5y + 2x
12. Subtract:
(i) 6x3 −7x2 + 5x − 3 from 4 − 5x + 6x2 − 8x3
(ii) − x2 −3z from 5x2 – y + z + 7
(iii) x3 + 2x2y + 6xy2 − y3 from y3−3xy2−4x2y
Solution:
(i) Given 6x3 −7x2 + 5x − 3 and 4 − 5x + 6x2 − 8x3
= (4 – 5x + 6x2 – 8x3) – (6x3 – 7x2 + 5x – 3)
= 4 – 5x + 6x2 – 8x3 – 6x3 + 7x2 – 5x + 3
= – 8x3– 6x3 + 7x2 + 6x2– 5x – 5x + 3 + 4
= 4 – 5x + 6x2 – 8x3 – 6x3 + 7x2 – 5x + 3
= – 8x3– 6x3 + 7x2 + 6x2– 5x – 5x + 3 + 4
= – 14x3 + 13x2 – 10x +7
(ii) Given − x2 −3z and 5x2 – y + z + 7
= (5x2 – y + z + 7) – (- x2 – 3z)
= 5x2 – y + z + 7 + x2 + 3z
= 5x2+ x2 – y + z + 3z + 7
= 6x2 – y + 4z + 7
= 5x2 – y + z + 7 + x2 + 3z
= 5x2+ x2 – y + z + 3z + 7
= 6x2 – y + 4z + 7
(iii) Given x3 + 2x2y + 6xy2 − y3 and y3−3xy2−4x2y
= (y3 – 3xy2 – 4x2y) – (x3 + 2x2y + 6xy2 – y3)
= y3 – 3xy2 – 4x2y – x3 – 2x2y – 6xy2 + y3
= y3 + y3– 3xy2– 6xy2– 4x2y – 2x2y – x3
= 2y3– 9xy2 – 6x2y – x3
= y3 – 3xy2 – 4x2y – x3 – 2x2y – 6xy2 + y3
= y3 + y3– 3xy2– 6xy2– 4x2y – 2x2y – x3
= 2y3– 9xy2 – 6x2y – x3
13. From
(i) p3 – 4 + 3p2, take away 5p2 − 3p3 + p − 6
(ii) 7 + x − x2, take away 9 + x + 3x2 + 7x3
(iii) 1− 5y2, take away y3 + 7y2 + y + 1
(iv) x3 − 5x2 + 3x + 1, take away 6x2 − 4x3 + 5 + 3x
Solution:
(i) Given p3 – 4 + 3p2, take away 5p2 − 3p3 + p − 6
= (p3 – 4 + 3p2) – (5p2 – 3p3 + p – 6)
= p3 – 4 + 3p2 – 5p2 + 3p3 – p + 6
= p3 + 3p3 + 3p2 – 5p2– p – 4+ 6
= 4p3 – 2p2 – p + 2
= p3 – 4 + 3p2 – 5p2 + 3p3 – p + 6
= p3 + 3p3 + 3p2 – 5p2– p – 4+ 6
= 4p3 – 2p2 – p + 2
(ii) Given 7 + x − x2, take away 9 + x + 3x2 + 7x3
= (7 + x – x2) – (9 + x + 3x2 + 7x3)
= 7 + x – x2 – 9 – x – 3x2 – 7x3
= – 7x3– x2 – 3x2 + 7 – 9
= – 7x3 – 4x2 – 2
= 7 + x – x2 – 9 – x – 3x2 – 7x3
= – 7x3– x2 – 3x2 + 7 – 9
= – 7x3 – 4x2 – 2
(iii) Given 1− 5y2, take away y3 + 7y2 + y + 1
= (1 – 5y2) – (y3+ 7y2 + y + 1)
= 1 – 5y2 – y3 – 7y2 – y – 1
= – y3– 5y2 – 7y2 – y
= – y3– 12y2 – y
= 1 – 5y2 – y3 – 7y2 – y – 1
= – y3– 5y2 – 7y2 – y
= – y3– 12y2 – y
(iv) Given x3 − 5x2 + 3x + 1, take away 6x2 − 4x3 + 5 + 3x
= (x3 – 5x2 + 3x + 1) – (6x2 – 4x3 + 5 +3x)
= x3 – 5x2 + 3x + 1 – 6x2 + 4x3 – 5 – 3x
= x3+ 4x3 – 5x2 – 6x2 + 1 – 5
= 5x3 – 11x2 – 4
= x3 – 5x2 + 3x + 1 – 6x2 + 4x3 – 5 – 3x
= x3+ 4x3 – 5x2 – 6x2 + 1 – 5
= 5x3 – 11x2 – 4
14. From the sum of 3x2 − 5x + 2 and − 5x2 − 8x + 9 subtract 4x2 − 7x + 9.
Solution:
First we have to add 3x2 − 5x + 2 and − 5x2 − 8x + 9 then from the result we have to subtract 4x2 − 7x + 9.
= {(3x2 – 5x + 2) + (- 5x2 – 8x + 9)} – (4x2 – 7x + 9)
= {3x2 – 5x + 2 – 5x2 – 8x + 9} – (4x2 – 7x + 9)
= {3x2 – 5x2 – 5x – 8x + 2 + 9} – (4x2 – 7x + 9)
= {3x2 – 5x2 – 5x – 8x + 2 + 9} – (4x2 – 7x + 9)
= {- 2x2 – 13x +11} – (4x2 – 7x + 9)
= – 2x2 – 13x + 11 – 4x2 + 7x – 9
= – 2x2 – 13x + 11 – 4x2 + 7x – 9
= – 2x2 – 4x2 – 13x + 7x + 11 – 9
= – 6x2 – 6x + 2
= – 6x2 – 6x + 2
15. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7.
Solution:
First we have to find the sum of 13x – 4y + 7z and – 6z + 6x + 3y
Therefore, sum of (13x – 4y + 7z) and (–6z + 6x + 3y)
= (13x – 4y + 7z) + (–6z + 6x + 3y)
= (13x – 4y + 7z – 6z + 6x + 3y)
= (13x + 6x – 4y + 3y + 7z – 6z)
= (19x – y + z)
Now we have to find the sum of (6x – 4y – 4z) and (2x + 4y – 7)
= (6x – 4y – 4z) + (2x + 4y – 7)
= (6x – 4y – 4z + 2x + 4y – 7)
= (6x + 2x – 4z – 7)
= (8x – 4z – 7)
Now, required expression = (8x – 4z – 7) – (19x – y + z)
= 8x – 4z – 7 – 19x + y – z
= 8x – 19x + y – 4z – z – 7
= –11x + y – 5z – 7
16. From the sum of x2 + 3y2 − 6xy, 2x2 − y2 + 8xy, y2 + 8 and x2 − 3xy subtract −3x2 + 4y2 – xy + x – y + 3.
Solution:
First we have to find the sum of (x2 + 3y2 – 6xy), (2x2 – y2 + 8xy), (y2 + 8) and (x2 – 3xy)
={(x2 + 3y2 – 6xy) + (2x2 – y2 + 8xy) + ( y2 + 8) + (x2 – 3xy)}
={x2 + 3y2 – 6xy + 2x2 – y2 + 8xy + y2 + 8 + x2 – 3xy}
= {x2+ 2x2+ x2 + 3y2– y2 + y2– 6xy + 8xy – 3xy + 8}
= 4x2 + 3y2 – xy + 8
Now, from the result subtract the −3x2 + 4y2 – xy + x – y + 3.
={(x2 + 3y2 – 6xy) + (2x2 – y2 + 8xy) + ( y2 + 8) + (x2 – 3xy)}
={x2 + 3y2 – 6xy + 2x2 – y2 + 8xy + y2 + 8 + x2 – 3xy}
= {x2+ 2x2+ x2 + 3y2– y2 + y2– 6xy + 8xy – 3xy + 8}
= 4x2 + 3y2 – xy + 8
Now, from the result subtract the −3x2 + 4y2 – xy + x – y + 3.
Therefore, required expression = (4x2 + 3y2 – xy + 8) – (- 3x2 + 4y2 – xy + x – y + 3)
= 4x2 + 3y2 – xy + 8 + 3x2 – 4y2 + xy – x + y – 3
= 4x2 + 3x2+ 3y2– 4y2– x + y – 3 + 8
= 7x2 – y2– x + y + 5
= 4x2 + 3y2 – xy + 8 + 3x2 – 4y2 + xy – x + y – 3
= 4x2 + 3x2+ 3y2– 4y2– x + y – 3 + 8
= 7x2 – y2– x + y + 5
17. What should be added to xy – 3yz + 4zx to get 4xy – 3zx + 4yz + 7?
Solution:
By subtracting xy – 3yz + 4zx from 4xy – 3zx + 4yz + 7, we get the required expression.
Therefore, required expression = (4xy – 3zx + 4yz + 7) – (xy – 3yz + 4zx)
= 4xy – 3zx + 4yz + 7 – xy + 3yz – 4zx
= 4xy – xy – 3zx – 4zx + 4yz + 3yz + 7
= 3xy – 7zx + 7yz + 7
18. What should be subtracted from x2 – xy + y2 – x + y + 3 to obtain −x2 + 3y2 − 4xy + 1?
Solution:
Let ‘E’ be the required expression. Then, we have
x2 – xy + y2– x + y + 3 – E = – x2 + 3y2 – 4xy + 1
Therefore, E = (x2 – xy + y2– x + y + 3) – (- x2 + 3y2 – 4xy + 1)
= x2 – xy + y2– x + y + 3 + x2 – 3y2 + 4xy – 1
Collecting positive and negative like terms together, we get
= x2 + x2– xy + 4xy + y2– 3y2 – x + y + 3 – 1
= 2x2+ 3xy- 2y2– x + y + 2
x2 – xy + y2– x + y + 3 – E = – x2 + 3y2 – 4xy + 1
Therefore, E = (x2 – xy + y2– x + y + 3) – (- x2 + 3y2 – 4xy + 1)
= x2 – xy + y2– x + y + 3 + x2 – 3y2 + 4xy – 1
Collecting positive and negative like terms together, we get
= x2 + x2– xy + 4xy + y2– 3y2 – x + y + 3 – 1
= 2x2+ 3xy- 2y2– x + y + 2
19. How much is x – 2y + 3z greater than 3x + 5y – 7?
Solution:
By subtracting x – 2y + 3z from 3x + 5y – 7 we can get the required expression,
Required expression = (x – 2y + 3z) – (3x + 5y – 7)
= x – 2y + 3z – 3x – 5y + 7
Collecting positive and negative like terms together, we get
= x – 3x – 2y + 5y + 3z + 7
= –2x – 7y + 3z + 7
20. How much is x2 − 2xy + 3y2 less than 2x2 − 3y2 + xy?
Solution:
By subtracting the x2 − 2xy + 3y2 from 2x2 − 3y2 + xy we can get the required expression,
Required expression = (2x2 – 3y2 + xy) – (x2 – 2xy + 3y2)
= 2x2 – 3y2 + xy – x2 + 2xy – 3y2
Collecting positive and negative like terms together, we get
= 2x2– x2 – 3y2 – 3y2 + xy + 2xy
= x2 – 6y2 + 3xy
= 2x2 – 3y2 + xy – x2 + 2xy – 3y2
Collecting positive and negative like terms together, we get
= 2x2– x2 – 3y2 – 3y2 + xy + 2xy
= x2 – 6y2 + 3xy
21. How much does a2 − 3ab + 2b2 exceed 2a2 − 7ab + 9b2?
Solution:
By subtracting 2a2 − 7ab + 9b2 from a2 − 3ab + 2b2 we get the required expression
Required expression = (a2– 3ab + 2b2) – (2a2 – 7ab + 9b2)
= a2– 3ab + 2b2 – 2a2 + 7ab – 9b2
Collecting positive and negative like terms together, we get
= a2 – 2a2 – 3ab + 7ab + 2b2 – 9b2
= – a2 + 4ab – 7b2
= a2– 3ab + 2b2 – 2a2 + 7ab – 9b2
Collecting positive and negative like terms together, we get
= a2 – 2a2 – 3ab + 7ab + 2b2 – 9b2
= – a2 + 4ab – 7b2
22. What must be added to 12x3 − 4x2 + 3x − 7 to make the sum x3 + 2x2 − 3x + 2?
Solution:
Let ‘E’ be the required expression. Thus, we have
12x3 – 4x2 + 3x – 7 + E = x3 + 2x2 – 3x + 2
Therefore, E = (x3 + 2x2 – 3x + 2) – (12x3 – 4x2 + 3x – 7)
= x3 + 2x2 – 3x + 2 – 12x3 + 4x2 – 3x + 7
Collecting positive and negative like terms together, we get
= x3– 12x3+ 2x2 + 4x2 – 3x – 3x + 2 + 7
= – 11x3 + 6x2 – 6x + 9
12x3 – 4x2 + 3x – 7 + E = x3 + 2x2 – 3x + 2
Therefore, E = (x3 + 2x2 – 3x + 2) – (12x3 – 4x2 + 3x – 7)
= x3 + 2x2 – 3x + 2 – 12x3 + 4x2 – 3x + 7
Collecting positive and negative like terms together, we get
= x3– 12x3+ 2x2 + 4x2 – 3x – 3x + 2 + 7
= – 11x3 + 6x2 – 6x + 9
23. If P = 7x2 + 5xy − 9y2, Q = 4y2 − 3x2 − 6xy and R = −4x2 + xy + 5y2, show that P + Q + R = 0.
Solution:
Given P = 7x2 + 5xy − 9y2, Q = 4y2 − 3x2 − 6xy and R = −4x2 + xy + 5y2
Now we have to prove P + Q + R = 0,
Now we have to prove P + Q + R = 0,
Consider P + Q + R = (7x2 + 5xy – 9y2) + (4y2 – 3x2 – 6xy) + (- 4x2 + xy + 5y2)
= 7x2 + 5xy – 9y2 + 4y2 – 3x2 – 6xy – 4x2 + xy + 5y2
Collecting positive and negative like terms together, we get
= 7x2– 3x2 – 4x2 + 5xy – 6xy + xy – 9y2 + 4y2 + 5y2
= 7x2– 7x2 + 6xy – 6xy – 9y2 + 9y2
= 0
= 7x2 + 5xy – 9y2 + 4y2 – 3x2 – 6xy – 4x2 + xy + 5y2
Collecting positive and negative like terms together, we get
= 7x2– 3x2 – 4x2 + 5xy – 6xy + xy – 9y2 + 4y2 + 5y2
= 7x2– 7x2 + 6xy – 6xy – 9y2 + 9y2
= 0
24. If P = a2 − b2 + 2ab, Q = a2 + 4b2 − 6ab, R = b2 + b, S = a2 − 4ab and T = −2a2 + b2 – ab + a. Find P + Q + R + S – T.
Solution:
Given P = a2 − b2 + 2ab, Q = a2 + 4b2 − 6ab, R = b2 + b, S = a2 − 4ab and T = −2a2 + b2 – ab + a.
Now we have to find P + Q + R + S – T
Substituting all values we get
Consider P + Q + R + S – T = {(a2 – b2 + 2ab) + (a2 + 4b2 – 6ab) + (b2 + b) + (a2 – 4ab)} – (-2a2 + b2 – ab + a)
= {a2 – b2 + 2ab + a2 + 4b2 – 6ab + b2 + b + a2 – 4ab}- (- 2a2 + b2 – ab + a)
= {3a2 + 4b2 – 8ab + b } – (-2a2 + b2 – ab + a)
= 3a2+ 4b2 – 8ab + b + 2a2 – b2 + ab – a
Collecting positive and negative like terms together, we get
3a2+ 2a2 + 4b2 – b2 – 8ab + ab – a + b
= 5a2 + 3b2– 7ab – a + b
= {3a2 + 4b2 – 8ab + b } – (-2a2 + b2 – ab + a)
= 3a2+ 4b2 – 8ab + b + 2a2 – b2 + ab – a
Collecting positive and negative like terms together, we get
3a2+ 2a2 + 4b2 – b2 – 8ab + ab – a + b
= 5a2 + 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 – 3z + y
(ii) 3x – 2y – 5z – 4
(iii) 3a – 2b + 4c – 5
(iv) 7a + 3b + 2c + 4
(v) 2a2 – b2 – 3ab + 6
(vi) a2 + b2 – c2 + ab – 3ac
Solution:
(i) Given x + y – 3z + y
x + y – 3z + y = x + y – (3z – y)
(ii) Given 3x – 2y – 5z – 4
3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)
(iii) Given 3a – 2b + 4c – 5
3a – 2b + 4c – 5 = 3a – 2b – (–4c + 5)
(iv) Given 7a + 3b + 2c + 4
7a + 3b + 2c + 4 = 7a + 3b – (–2c – 4)
(v) Given 2a2 – b2 – 3ab + 6
2a2 – b2 – 3ab + 6 = 2a2 – b2 – (3ab – 6)
(vi) Given a2 + b2 – c2 + ab – 3ac
a2 + b2 – c2 + ab – 3ac = a2 + b2 – c2 – (- ab + 3ac)
2. Write each of the following statements by using appropriate grouping symbols:
(i) The sum of a – b and 3a – 2b + 5 is subtracted from 4a + 2b – 7.
(ii) Three times the sum of 2x + y – [5 – (x – 3y)] and 7x – 4y + 3 is subtracted from 3x – 4y + 7
(iii) The subtraction of x2 – y2 + 4xy from 2x2 + y2 – 3xy is added to 9x2 – 3y2– xy.
Solution:
(i) Given the sum of a – b and 3a – 2b + 5 = [(a – b) + (3a – 2b + 5)].
This is subtracted from 4a + 2b – 7.
Thus, the required expression is (4a + 2b – 7) – [(a – b) + (3a – 2b + 5)]
(ii) Given three times the sum of 2x + y – {5 – (x – 3y)} and 7x – 4y + 3 = 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]
This is subtracted from 3x – 4y + 7.
Thus, the required expression is (3x – 4y + 7) – 3[(2x + y – {5 – (x – 3y)}) + (7x – 4y + 3)]
(iii) Given the product of subtraction of x2– y2 + 4xy from 2x2 + y2 – 3xy is given by {(2x2 + y2 – 3xy) – (x2-y2 + 4xy)}
When the above equation is added to 9x2 – 3y2 – xy, we get
{(2x2 + y2 – 3xy) – (x2 – y2 + 4xy)} + (9x2 – 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 – 3y)
Solution:
Given 2x + (5x – 3y)
Since the ‘+’ sign precedes the parentheses, we have to retain the sign of each term in the parentheses when we remove them.
= 2x + 5x – 3y
On simplifying, we get
= 7x – 3y
2. 5a – (3b – 2a + 4c)
Solution:
Given 3x – (y – 2x)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have
= 3x – y + 2x
On simplifying, we get
= 5x – y
3. 5a – (3b – 2a + 4c)
Solution:
Given 5a – (3b – 2a + 4c)
Since the ‘-‘sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them.
= 5a – 3b + 2a – 4c
On simplifying, we get
= 7a – 3b – 4c
4. -2(x2 – y2 + xy) – 3(x2 +y2 – xy)
Solution:
Given – 2(x2 – y2 + xy) – 3(x2 +y2 – xy)
Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have
= -2x2 + 2y2 – 2xy – 3x2 – 3y2 + 3xy
On rearranging,
= -2x2 – 3x2 + 2y2 – 3y2 – 2xy + 3xy
On simplifying, we get
= -5x2 – y2 + xy
5. 3x + 2y – {x – (2y – 3)}
Solution:
Given 3x + 2y – {x – (2y – 3)}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 3x + 2y – {x – 2y + 3}
= 3x + 2y – x + 2y – 3
On simplifying, we get
= 2x + 4y – 3
6. 5a – {3a – (2 – a) + 4}
Solution:
Given 5a – {3a – (2 – a) + 4}
First, we have to remove the parentheses. Then, we have to remove the braces.
Then we get,
= 5a – {3a – 2 + a + 4}
= 5a – 3a + 2 – a – 4
On simplifying, we get
= 5a – 4a – 2
= a – 2
7. a – [b – {a – (b – 1) + 3a}]
Solution:
Given a – [b – {a – (b – 1) + 3a}]
First we have to remove the parentheses, then the curly brackets, and then the square brackets.
Then we get,
= a – [b – {a – (b – 1) + 3a}]
= a – [b – {a – b + 1 + 3a}]
= a – [b – {4a – b + 1}]
= a – [b – 4a + b – 1]
= a – [2b – 4a – 1]
On simplifying, we get
= a – 2b + 4a + 1
= 5a – 2b + 1
8. a – [2b – {3a – (2b – 3c)}]
Solution:
Given a – [2b – {3a – (2b – 3c)}]
First we have to remove the parentheses, then the braces, and then the square brackets.
Then we get,
= a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
= a – [2b – 3a + 2b – 3c]
= a – [4b – 3a – 3c]
On simplifying we get,
= a – 4b + 3a + 3c
= 4a – 4b + 3c
9. -x + [5y – {2x – (3y – 5x)}]
Solution:
Given -x + [5y – {2x – (3y – 5x)}]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= – x + [5y – {2x – (3y – 5x)}]
= – x + [5y – {2x – 3y + 5x)]
= – x + [5y – {7x – 3y}]
= – x + [5y – 7x + 3y]
= – x + [8y – 7x]
On simplifying we get
= – x + 8y – 7x
= – 8x + 8y
10. 2a – [4b – {4a – 3(2a – b)}]
Solution:
Given 2a – [4b – {4a – 3(2a – b)}]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= 2a – [4b – {4a – 3(2a – b)}]
= 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b – {- 2a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – [b + 2a]
On simplifying, we get
= 2a – b – 2a
= – b
11. -a – [a + {a + b – 2a – (a – 2b)} – b]
Solution:
Given -a – [a + {a + b – 2a – (a – 2b)} – b]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= – a – [a + {a + b – 2a – (a – 2b)} – b]
= – a – [a + {a + b – 2a – a + 2b} – b]
= – a – [a + {- 2a + 3b} – b]
= – a – [a – 2a + 3b – b]
= – a – [- a + 2b]
On simplifying, we get
= – a + a – 2b
= – 2b
12. 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
Solution:
Given 2x – 3y – [3x – 2y -{x – z – (x – 2y)}]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= 2x – 3y – [3x – 2y – {x – z – (x – 2y)})
= 2x – 3y – [3x – 2y – {x – z – x + 2y}]
= 2x – 3y – [3x – 2y – {- z + 2y}]
= 2x – 3y – [3x – 2y + z – 2y]
= 2x – 3y – [3x – 4y + z]
On simplifying, we get
= 2x – 3y – 3x + 4y – z
= – x + y – z
13. 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
Solution:
Given 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
= 5 + [x – {2y – 6x – y + 4 + 2x} – {x – y + 2}]
= 5 + [x – {y – 4x + 4} – {x – y + 2}]
= 5 + [x – y + 4x – 4 – x + y – 2]
= 5 + [4x – 6]
= 5 + 4x – 6
= 4x – 1
14. x2 – [3x + [2x – (x2 – 1)] + 2]
Solution:
Given x2 – [3x + [2x – (x2 – 1)] + 2]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= x2 – [3x + [2x – (x2 – 1)] + 2]
= x2 – [3x + [2x – x2 + 1] + 2]
= x2 – [3x + 2x – x2 + 1 + 2]
= x2 – [5x – x2 + 3]
On simplifying we get
= x2 – 5x + x2 – 3
= 2x2 – 5x – 3
15. 20 – [5xy + 3[x2 – (xy – y) – (x – y)]]
Solution:
Given 20 – [5xy + 3[x2 – (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[x2 – (xy – y) – (x – y)]]
= 20 – [5xy + 3[x2 – xy + y – x + y]]
= 20 – [5xy + 3[x2 – xy + 2y – x]]
= 20 – [5xy + 3x2 – 3xy + 6y – 3x]
= 20 – [2xy + 3x2 + 6y – 3x]
On simplifying we get
= 20 – 2xy – 3x2 – 6y + 3x
= – 3x2 – 2xy – 6y + 3x + 20
16. 85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]
Solution:
Given 85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= 85 – [12x – 7(8x – 3) – 2{10x – 5(2 – 4x)}]
= 85 – [12x – 56x + 21 – 2{10x – 10 + 20x}]
= 85 – [12x – 56x + 21 – 2{30x – 10}]
= 85 – [12x – 56x + 21 – 60x + 20]
= 85 – [12x – 116x + 41]
= 85 – [- 104x + 41]
On simplifying, we get
= 85 + 104x – 41
= 44 + 104x
17. xy [yz – zx – {yx – (3y – xz) – (xy – zy)}]
Solution:
Given xy [yz – zx – {yx – (3y – xz) – (xy – zy)}]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= xy – [yz – zx – {yx – (3y – xz) – (xy – zy)}]
= xy – [yz – zx – {yx – 3y + xz – xy + zy}]
= xy – [yz – zx – {- 3y + xz + zy}]
= xy – [yz – zx + 3y – xz – zy]
= xy – [- zx + 3y – xz]
On simplifying, we get
= xy – [- 2zx + 3y]
= xy + 2xz – 3y
