In mathematics, an algebraic expression is an expression which is built from constant, integers, variables and algebraic operators. For example “5x^{3} − 6xy + a” is an algebraic expression. Learn the concepts of algebraic expressions by solving the questions from RS Aggarwal for class 6 which covers all the basic and advanced concepts present in the chapter. To help students we have also provided the RS Aggarwal class 6 solutions chapter 8 algebraic expressions.

**1. Write the following using literals, numbers and signs of basic operations:**

**(i) x increased by 12**

(i) x increased by 12 is (1+12).

**(ii) y decreased by 7**

(ii)y decreased by 7 is (y-7).

**(iii) The difference of a and b, when a > b**

(iii) The difference of a and b, when alb is (a-b).

**(iv) The product of x and y added to their sum**

(iv) The product of x and y is my. The sum of x and y is (x+y). So, product of x and y added to their sum is xy+(x+y).

**(v) One-third of x multiplied by the sum of a and b**

(v) One third of x is \(frac{x}{3}\).

The sum of a and b is (a+b).

One-third of x multiplied by the sum of a and b = \(frac{x}{3}\)*(a+b) = \(frac{x(a+b)}{3}\)

**(vi) 5 times x added to 7 times y**

(vi) 5 times x added to 7 times y= (5*x)+(7*y), which is equal to 5x+7y.

**(vii) Sum of x and the quotient of y by 5**

(vii) Sum of x and the quotient of y by 5 is x+y5.

**(viii) x taken away from 4**

(viii) x taken away from 4 is (4-x).

**(ix) 2 less than the quotient of x by y**

(ix) 2 less than the quotient of x by y is \(frac{x}{y}\)-2.

**(x) x multiplied by itself**

(x) x multiplied by itself is x*x=\(x^{2}\).

**(xi) Thrice x increased by y**

(xi) Twice x increased by y is (2*x) + y = 2x+y.

**(xii) Thrice x added to y squared**

(xii) Thrice x added to y squared is (3*x)+(y*y) = 3x+\(y^{2}\).

**(xiii) x minus twice y**

(xiii) x minus twice y is x-(2*y) = x-2y.

**(xiv) x cubed less than y cubed**

(xiv) x cubed less than y cubed is (y*y*y)-(x*x*x)= \(y^{3}\). -\(x^{3}\)..

**(xv) The quotient of x by 8 is multiplied by y**

(xv) The quotient of x by 8 is multiplied by y is \(frac{x}{8}\)*y=\(frac{xy}{8}\).

**2) Ranjit scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?**

Ranjit’s score in English = 80 marks

Ranjit’s score in Hindi = x marks .

Total score in the two subjects = (Ranjit’s score in English + Ranjit’s score in Hindi) .

Total score in the two subjects = (80 + x) marks.

**3) Write the following in the exponential form:**

**(i) b x b x b x … 15 times**

(i)b*b*b*…15times = \(b^{15}\)

**(ii) y x y x y x … 20 times**

(ii) y*y*y*…20times=\(y^{20}\)

**(iii) 14 x a x a x a x a x b x b x b**

(iii) 14 x a x a x a x a x b x b x b = 14x( a x a x a x a ) x ( b x b x b ) =14\(a^{4}\) \(b^{3}\)

**(iv) 6 * x * x * y * y**

(iv) 6 *x * x * y * y= 6x(x*x) x (y*y)=6\(x^{2}\) \(y^{2}\)

**(v) 3 * z * z * z * y * y * x**

(v) 3 * z * z * z * y * y * x= 3*(z * z * z)x(y * y)* x=3\(z^{3}\) \(y^{2}\)*x

**4) Write down the following in the product form:**

**(i) \(x^{2}\)y**

(i) \(x^{2}\) \(y^{4}\) = x * x * y * y * y * y

**(ii) 6\(y^{5}\)**

(ii) 6 \(y^{5}\) =6 * y * y * y * y * y

**(iii) 9x\(y^{2}\)z**

(iii) 9x \(y^{2}\) z = 9 * x * y * y * z

**(iv) 10\(a^{3}\) \(b^{3}\) \(c^{3}\)**

(iv) 10 \(a^{3}\) \(b^{3}\) \(c^{3}\) = 10 * a * a * a * b * b * b * c * c * c

**EXERCISE – 8B**

**1) If a = 2 and b = 3, find the value of**

**(i) a+b **

Substituting a = 2 and b = 3 in the given expression:

2+3 = 5

**(ii) \(a^{2}\) +ab **

Substituting a = 2 and b = 3 in the given expression:

\(2^{2}\) +(2×3)=4+6=10

**(iii) ab-\(a^{2}\) **

Substituting a = 2 and b = 3 in the given expression:

(2×3)- \(2^{2}\) = 6-4 = 2

**(iv) 2a-3b **

Substituting a = 2 and b = 3 in the given expression:

(2×2)-(3×3)=4-9=-5

**(v) 5\(a^{2}\) -2ab **

Substituting a=2 and b=3 in the given expression:

5x\(2^{2}\) – 2 x 2 x 3 = 5 x 4 – 12 = 20 – 12 = 13

**(vi) \(a^{3}\) – \(b^{3}\) **

Substituting a=2 and b=3 in the given expression:

\(2^{3}\) – \(3^{3}\) =2 x 2 x 2 – 3 x 3 x 3 = 13 – 27 = – 19

**2) If x = 1, y = 2 and z = 5, find the value of
(i) 3x-2y+4z**

Substituting x = 1, y = 2 and z = 5 in the given expression:

3(1) – 2(2) + 4(5) = 3 – 4 + 20 = 19

**(ii) \(x^{2}+y^{2}+z^{2}\)**

Substituting x = 1, y = 2 and z = 5 in the given expression:

\(1^{2}\) + \(2^{2}\) + \(5^{2}\)

= (1 *1)+(2*2)+(5*5)

= 1+4+25=30

**(iii) 2\(x^{2}\)-3\(y^{2}\)+\(z^{2}\) **

Substituting x = 1, y = 2 and z = 5 in the given expression:

2 x \(1^{2}\) – 3x\(2^{2}\) + \(5^{2}\)

=2 x (1×1) – 3 x (2×2) + (5×5)

= 2-12+25

=15

**(iv) xy + yz – zx **

Substituting x = 1, y = 2 and z = 5 in the given expression:

(1×2)+(2×5)-(5×1)

=2+10-5

=7

**(v) 2\(x^{2}\)y-5yz+x\(y^{2}\) **

Substituting x = 1, y = 2 and z = 5 in the given expression:

2x\(1^{2}\)x2-5x2x5+1x\(2^{2}\)

= 4 – 50 + 4

= – 42

**(vi) \(x^{3}\)- \(y^{3}\) – \(z^{3}\)**

Substituting x = 1, y = 2 and z = 5 in the given expression:

\(1^{3}\) – \(2^{3}\) – \(5^{3}\)

=(1 * 1 * 1) – (2 *2 * 2) – (5 * 5 * 5)

= 1 – 8 – 125

= -132

**3) If p = -2, q = -1 and r = 3, find the value of**

**(i) \(p^{2}\) + \(q^{2}\) – \(r^{2}\)**

Substituting p = -2, q = -1 and r = 3 in the given expression:

\((-2)^{2}\) + \((-1)^{2}\) – \(3^{2}\)

=\((-2)^{2}\) + \((-1)^{-1}\)- (3×3)

= 4 + 1 – 9

=4.

**(ii) 2\(p^{2}\) – \(q^{2}\) +3\(r^{2}\)**

Substituting p = -2, q = -1 and r = 3 in the given expression:

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

=2 x (-2x-2)-(-1x-1)+3x(3×3)

= 8-1+27

=34.

**(iii) p – q – r **

Substituting p = -2, q = -1 and r = 3 in the given expression:

(-2) – (-1) – (3)

= -2+1-3

=4

**(iv) \(p^{3}\)+ \(q^{3}\) + \(r^{3}\) +3pqr**

Substituting p = -2, q = -1 and r = 3 in the given expression:

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

=( -2 x -2 x -2) + (-1 x -1 x -1) + (3 x 3 x 3) + 3 x (6)

=(-8)+(-1)+(27)+18

=36

**(v) 3\(p^{2}\)q+5p\(q^{2}\)+2pqr **

Substituting p = -2, q = -1 and r = 3 in the given expression:

3 x \((-2)^{2}\) x (-1)+5x(-2)x(-1)2+2x(-2x-1×3)

=3 x (-2 x -2) x (-1) + 5 x (-2) x (-1x-1) + 2 x (-2) x -1 x 3

=12-10+12

=10.

**(vi) \(p^{4}\) + \(q^{4}\) – \(r^{4}\)**

Substituting p = -2, q = -1 and r = 3 in the given expression:

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

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

=16 + 1 – 81

=64

**4) Write the coefficient of**

(i) **x in 13x**

Coefficient of x in 13x is 13.

(ii) **y in -5y**

Coefficient of y in -5y is -5.

(iii) **a in 6ab**

Coefficient of a in 6ab is 6b.

(iv) **z in -7xz**

Coefficient of z in -7xzis -7x.

(v) **p in -2pqr**

Coefficient of p in -2pqr is -2qr.

(vi) **y^{2}\) in 8x\(y^{2}\)z**

Coefficient of \(y^{2}\) in 8x\(y^{2}\) z is 8xz.

(vii) **\(x^{3}\) in \(x^{3}\)**

Coefficient of \(x^{3}\) in \(x^{3}\) is 1.

(viii)** \(x^{2}\) in \((-x)^{2}\)**

Coefficient of \(x^{2}\) in \((-x)^{2}\) is -1.

**5) Write the numerical co-efficient of :**

(i) **ab**

Numerical coefficient of ab is 1.

(ii) **-6bc**

Numerical coefficient of -6bc is -6.

(iii)** 7xyz**

Numerical coefficient of 7xyz is 7.

(iv) **-2\(x^{3}\) \(y^{2}\)z**

Numerical coefficient of -2\(x^{3}\) \(y^{2}\)z is -2.

**6) Write the constant term of :**

A term of expression having no literal factors is called a constant term.

(i)** 3\(x^{2}\) + 5x + 8**

In the expression 3\(x^{2}\) + 5x + 8, the constant term is 8.

(ii) **2\(x^{2}\) – 9**

In the expression 2\(x^{2}\) – 9, the constant term is -9.

(iii) **4\(y^{2}\) – 5y + 35**

In the expression 4\(y^{2}\) – 5y + 35, the constant term is 35.

(iv)**\(z^{3}\) – 2\(z^{2}\) + z- 83**

In the expression \(z^{3}\) – 2\(z^{2}\) + z- 83 , the constant term is -83.

**7) Identify the monomials, binomials and trinomials in the following:**

**(i) -2zyz**

**(ii) 5 + 7\(x^{3}\) \(y^{3}\) \(z^{3}\)**

**(iii) -5\(x^{3}\)**

**(iv) a + b – 2c**

**(v) xy + yz – zx**

**(vi) \(x^{5}\)**

**(vii) a\(x^{3}\) + b\(x^{2}\) + cx + d**

**(viii) -14**

**(ix) 2x + 1**

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.

The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.

The expressions given in (iv) and (v) contain three terms. So, both of them are trinomial.

The expression given in (vii) contains four terms. So, it does not represent any of the given types.

**8) Write all the terms of the algebraic expressions:**

** (i) \(x^{5}\) – 6\(y^{4}\) + 7\(x^{2}\)y – 9**

Expression 4\(x^{5}\) – 6\(y^{4}\) + 7\(x^{2}\)y – 9 has four terms, namely 4\(x^{5}\) – 6\(y^{4}\) + 7\(x^{2}\)y and -9.

(ii) **9\(x^{3}\) – 5\(z^{4}\) + 7\(z^{3}\)y – xyz**

Expression 9\(x^{3}\) – 5\(z^{4}\) + 7\(z^{3}\)y – xyz has four terms, namely 9\(x^{3}\) , 5\(z^{4}\) , 7\(z^{3}\)y and – xyz.

**9) Identify the like terms in the following:**

The terms that have same literals are called like terms.

**(i) \(a^{2}\), \(b^{2}\), -2\(a^{2}\), \(c^{2}\), 4a**

\(a^{2}\) and 2\(a^{2}\) are like terms.

**(ii) 3x, 4xy, -yz, 12zy**

-yz and 12zy are like terms.

**(iii) -2x\(y^{2}\), \(x^{2}\)y, 5\(y^{2}\)x, \(y^{2}\)z**

-2x\(y^{2}\) and 5\(y^{2}\)x are like terms.

**(iv) abc, a\(b^{2}\)c, ac\(b^{2}\), \(c^{2}\)ab, \(b^{2}\)ac, \(a^{2}\)bc, ca\(b^{2}\)**

a\(b^{2}\)c , ac\(b^{2}\) , \(b^{2}\)ac and ca\(b^{2}\) are like terms.