Algebraic Expressions

(i) Thrice a number x = 3x. 6 more than this = 3x + 6.

(ii) 5 times x = 5x. 5x subtracted from 13 = 13 − 5x.

(iii) x squared = x², y squared = y². Both added = x² + y².

(iv) Product of p and q = pq. 3 times the product = 3pq. 7 added to it = 3pq + 7.

(v) Product of x with itself = x². Three times of x = 3x. 3x subtracted from x² = x² − 3x.

(vi) Product of m and n = mn. Sum of m and n = m + n. The sum subtracted from their product = mn − (m + n).

Charge per km = ₹ 9

Charge for x km = ₹ 9x

Fixed charge = ₹ 50

∴ Total invoice = ₹ (9x + 50)

Hence, the algebraic expression of the invoice is ₹ (9x + 50).

(i) The required expression is 5a − 3b + c.

(ii) The required expression is x² − 5x + 6.

(iii) The required expression is x²y + xy − xy².

(i) The terms of 3 − 7x are 3 and −7x.

(ii) The terms of 2 - 5a + $\frac{3}{2}b$ are 2, −5a and $\frac{3}{2}b$.

(iii) The terms of 3x⁵ + 4y³ − 7xy² + 3 are 3x⁵, 4y³, −7xy² and 3.

(i) The expression −4x + 5y has the terms −4x and 5y.

Factors of −4x are −4 and x.

Factors of 5y are 5 and y.

(ii) The expression xy + 2x²y² has the terms xy and 2x²y².

Factors of xy are x and y.

Factors of 2x²y² are 2, x, x, y and y.

(iii) The expression 1.2ab − 2.4b + 3.6a has the terms 1.2ab, −2.4b and 3.6a.

Factors of 1.2ab are 1.2, a and b.

Factors of −2.4b are −2.4 and b.

Factors of 3.6a are 3.6 and a.

(i) The expression 8x + 3y² has two terms 8x and 3y². The factors of 8x are 8 and x, and the factors of 3y² are 3, y and y.

(ii) The expression y − y³ has two terms y and −y³. The factor of y is y, and the factors of −y³ are −1, y, y and y.

(iii) The expression 5xy² + 7x²y has two terms 5xy² and 7x²y. The factors of 5xy² are 5, x, y and y, and the factors of 7x²y are 7, x, x and y.

(iv) The expression −ab + 2b² − 3a² has three terms −ab, 2b² and −3a². The factors of −ab are −1, a and b; the factors of 2b² are 2, b and b; and the factors of −3a² are −3, a and a.

(i) The numerical coefficient of −7x is −7.

(ii) The numerical coefficient of −2x³y² is −2.

(iii) The numerical coefficient of 6abcd² is 6.

(iv) The numerical coefficient of $\frac{2}{3}pq^2$ is $\frac{2}{3}$.

(i) −4bx = (−4b) × x, so the coefficient of x is −4b.

(ii) 5xyz = (5yz) × x, so the coefficient of x is 5yz.

(iii) −x = (−1) × x, so the coefficient of x is −1.

(iv) −3x²y = (−3xy) × x, so the coefficient of x is −3xy.

We have −7xy²z³.

(i) −7xy²z³ = (7x)(−y²z³), so the coefficient of 7x is −y²z³.

(ii) −7xy²z³ = (−xy²)(7z³), so the coefficient of −xy² is 7z³.

(iii) −7xy²z³ = (xyz)(−7yz²), so the coefficient of xyz is −7yz².

(iv) −7xy²z³ = (7yz²)(−xyz), so the coefficient of 7yz² is −xyz.

(i) In 3 − 7x, the non-constant term is −7x.

Numerical coefficient of −7x = −7.

(ii) In 1 + 2x − 3x², the non-constant terms are 2x and −3x².

Numerical coefficient of 2x = 2.

Numerical coefficient of −3x² = −3.

(iii) In 1.2a + 0.8b, the non-constant terms are 1.2a and 0.8b.

Numerical coefficient of 1.2a = 1.2.

Numerical coefficient of 0.8b = 0.8.

(i) In 13y² − 8xy, the term containing x is −8xy.

Coefficient of x in −8xy = −8y.

(ii) In 7x − xy², the terms containing x are 7x and −xy².

Coefficient of x in 7x = 7.

Coefficient of x in −xy² = −y².

(iii) In 5 − 7xyz + 4x²y, the terms containing x are −7xyz and 4x²y.

Coefficient of x in −7xyz = −7yz.

Coefficient of x in 4x²y = 4xy.

(i) In 8 − xy², the term containing y² is −xy².

Coefficient of y² in −xy² = −x.

(ii) In 5y² + 7x − 3xy², the terms containing y² are 5y² and −3xy².

Coefficient of y² in 5y² = 5.

Coefficient of y² in −3xy² = −3x.

(iii) In 2x²y − 15xy² + 7y², the term containing y² is -15xy² and 7y².

Coefficient of y² in -15xy² = -15x

Coefficient of y² in 7y² = 7.

An expression with 1 term is a monomial, 2 terms is a binomial and 3 terms is a trinomial.

(i) 4y − 7z has 2 terms → Binomial.

(ii) −5xy² has 1 term → Monomial.

(iii) x + y − xy has 3 terms → Trinomial.

(iv) ab² − 5b − 3a has 3 terms → Trinomial.

(v) 4p²q − 5pq² has 2 terms → Binomial.

(vi) 2017 has 1 term → Monomial.

(vii) 1 + x + x² has 3 terms → Trinomial.

(viii) 5x² − 7 + 3x + 4 = 5x² + 3x − 3 has 3 terms → Trinomial. [Note that −7 + 4 = −3]

Two terms are like terms if they have the same literal (variable) part.

(i) −7x and $\frac{5}{2}x$ have the same literal part x → Like terms.

(ii) −29x and −29y have different variables → Unlike terms.

(iii) 2xy and 2xyz have different variables → Unlike terms.

(iv) 4m²p and 4mp² have different powers → Unlike terms.

(v) 12xz and 12x²z² have different powers → Unlike terms.

(vi) −5pq and 7qp have the same literal part (pq = qp) → Like terms.

Like terms have the same literal (variable) part.

(i) Only x²y and −2x²y are like terms.

(ii) The like terms are 3a²b, −6a²b and 2abc, 4abc.

(iii) The groups of like terms are:

10pq, −7qp, 78qp (terms in pq);

7p, 2405p (terms in p);

8q, −100q (terms in q);

−p²q², 12q²p² (terms in p²q²);

−23, 41 (constant terms);

−5p², 701p² (terms in p²);

13p²q, qp² (terms in p²q).

The degree of a polynomial in x is the highest power of x present in it.

(i) In x² − 6x⁷ + x⁸, the highest power of x is 8 → Degree 8.

(ii) In 3 − 2x, the highest power of x is 1 → Degree 1.

(iii) −2 = −2x⁰, so the power of x is 0 → Degree 0.

(iv) In 1 − x², the highest power of x is 2 → Degree 2.

In a polynomial in two or more variables, the degree is the greatest sum of the powers of the variables in any term.

(i) In 3x² − 5xy² + 7, the degrees of the terms are 2, (1 + 2) = 3 and 0. The greatest is 3 → Degree 3.

(ii) In xy² − y³ + 3y⁴ − 2, the degrees of the terms are (1 + 2) = 3, 3, 4 and 0. The greatest is 4 → Degree 4.

(iii) In 7 − 2x³ − 5xy³ + 9y⁵, the degrees of the terms are 0, 3, (1 + 3) = 4 and 5. The greatest is 5 → Degree 5.

(i) True. Both 5y and 5 + y depend on the value of y, so they are variables.

(ii) False. 7x is a single term (a monomial) with 7 as its numerical coefficient and x as its literal coefficient.

(iii) False. 5 + xy has 2 terms, so it is a binomial (not a trinomial).

(iv) False. 7a × bc = 7abc is a single term (a monomial).

(v) True. All the powers of x are non-negative integers.

(vi) False. $\frac{3}{x}$ has a negative power of x, so the expression is not a polynomial.

(vii) False. The coefficient of x in −3xy is −3y.

The sum of like terms is a like term whose coefficient is the sum of the coefficients.

(i) 7x + (−3x) = (7 − 3)x = 4x.

(ii) 6x + (−11x) = (6 − 11)x = −5x.

(iii) 5x² + (−9x²) = (5 − 9)x² = −4x².

(iv) 3ab² + (−5ab²) = (3 − 5)ab² = −2ab².

(v) $\frac{1}{2}pq + \left(-\frac{1}{3}pq\right) = \left(\frac{1}{2} - \frac{1}{3}\right)pq = \frac{3 - 2}{6}pq = \frac{1}{6}pq$.

(vi) $5x^3y + \left(-\frac{2}{3}x^3y\right) = \left(5 - \frac{2}{3}\right)x^3y = \frac{15 - 2}{3}x^3y = \frac{13}{3}x^3y$.

(i) 3x + (−5x) + 7x = (3 − 5 + 7)x = 5x.

(ii) 7xy + 2xy + (−8xy) = (7 + 2 − 8)xy = xy.

(iii) −2abc + 3abc + abc = (−2 + 3 + 1)abc = 2abc.

(iv) 3mn + (−5mn) + 8mn + (−4mn) = (3 − 5 + 8 − 4)mn = 2mn.

(v) 2x³ + 3x³ + (−4x³) + (−5x³) = (2 + 3 − 4 − 5)x³ = −4x³.

(i) 21b − 32 + 7b − 20b

= (21 + 7 − 20)b − 32

= 8b − 32.

(ii) 12m² − 9m + 5m − 4m² − 7m + 10

= (12 − 4)m² + (−9 + 5 − 7)m + 10

= 8m² − 11m + 10.

(iii) −z² + 13z² − 5z + 7z³ − 15z

= 7z³ + (−1 + 13)z² + (−5 − 15)z

= 7z³ + 12z² − 20z.

(iv) 5x²y − 5x² + 3yx² − 3y² + x² − y² + 8xy² − 3y²

= (5x²y + 3x²y) + (−5x² + x²) + (−3y² − y² − 3y²) + 8xy²

= 8x²y − 4x² − 7y² + 8xy²

= 8x²y + 8xy² − 4x² − 7y².

(v) p − (p − q) − (q − p) − q

= p − p + q − q + p − q

= (p − p + p) + (q − q − q)

= p − q.

(vi) 3a − 2b − ab − (a − b + ab) + 3ab + b − a

= 3a − 2b − ab − a + b − ab + 3ab + b − a

= (3a − a − a) + (−2b + b + b) + (−ab − ab + 3ab)

= a + ab.

(vii) (3y² + 5y − 4) − (8y − y² − 4)

= 3y² + 5y − 4 − 8y + y² + 4

= (3y² + y²) + (5y − 8y) + (−4 + 4)

= 4y² − 3y.

(i) 5xy + (−7xy) + 3x²

= (5 − 7)xy + 3x²

= 3x² − 2xy.

(ii) 4x²y + (−3xy²) + (−5xy²) + 5x²y

= (4 + 5)x²y + (−3 − 5)xy²

= 9x²y − 8xy².

(iii) (−7mn + 5) + (12mn + 2) + (8mn − 8) + (−2mn − 3)

= (−7 + 12 + 8 − 2)mn + (5 + 2 − 8 − 3)

= 11mn − 4.

(iv) (a + b − 3) + (b − a + 3) + (a − b + 3)

= (a − a + a) + (b + b − b) + (−3 + 3 + 3)

= a + b + 3.

(v) (14x + 10y − 12xy − 13) + (18 − 7x − 10y + 8xy) + 4xy

= (14 − 7)x + (10 − 10)y + (−12 + 8 + 4)xy + (−13 + 18)

= 7x + 5.

(vi) (5m − 7n) + (3n − 4m + 2) + (2m − 3mn − 5)

= (5 − 4 + 2)m + (−7 + 3)n − 3mn + (2 − 5)

= 3m − 4n − 3mn − 3.

(vii) (3x³ − 5x² + 2x + 1) + (3x − 2x² − x³) + (2x² − 7x + 9)

= (3 − 1)x³ + (−5 − 2 + 2)x² + (2 + 3 − 7)x + (1 + 9)

= 2x³ − 5x² − 2x + 10.

(viii) (7a² − 5a + 2) + (3a² − 7) + (2a + 9) + (1 + 2a − 5a²)

= (7 + 3 − 5)a² + (−5 + 2 + 2)a + (2 − 7 + 9 + 1)

= 5a² − a + 5.

(i) 2x² + 3y² − 5xy + 5x² − y² + 6xy − 3x²

= (2 + 5 − 3)x² + (3 − 1)y² + (−5 + 6)xy

= 4x² + 2y² + xy.

(ii) 3xy² − 5x²y + 7xy − 8xy² − 4xy + 6x²y

= (3 − 8)xy² + (−5 + 6)x²y + (7 − 4)xy

= −5xy² + x²y + 3xy.

(iii) 5x⁴ − 7x² + 8x − 1 + 3x³ − 9x² + 7 − 3x⁴ + 11x − 2 + 8x²

= (5 − 3)x⁴ + 3x³ + (−7 − 9 + 8)x² + (8 + 11)x + (−1 + 7 − 2)

= 2x⁴ + 3x³ − 8x² + 19x + 4.

To subtract, we change the sign of each term of the expression to be subtracted and then add.

(i) y² − (−5y²) = y² + 5y² = 6y².

(ii) −2xy − (−7xy) = −2xy + 7xy = 5xy.

(iii) Here a(b − 5) = ab − 5a and b(5 − a) = 5b − ab.

b(5 − a) − a(b − 5) = (5b − ab) − (ab − 5a)

= 5b − ab − ab + 5a

= 5a + 5b − 2ab.

(iv) (4m² − 3mn + 8) − (−m² + 5mn)

= 4m² − 3mn + 8 + m² − 5mn

= 5m² − 8mn + 8.

(v) (3ab − 2a² − 2b²) − (5a² − 7ab + 5b²)

= 3ab − 2a² − 2b² − 5a² + 7ab − 5b²

= (−2 − 5)a² + (3 + 7)ab + (−2 − 5)b²

= 10ab − 7a² − 7b².

(vi) (5p² + 3q² − pq) − (4pq − 5q² − 3p²)

= 5p² + 3q² − pq − 4pq + 5q² + 3p²

= (5 + 3)p² + (3 + 5)q² + (−1 − 4)pq

= 8p² + 8q² − 5pq.

(vii) (7x² − 8xy + 3y² − 5) − (7xy + 5x² − 7y² + 3)

= 7x² − 8xy + 3y² − 5 − 7xy − 5x² + 7y² − 3

= (7 − 5)x² + (−8 − 7)xy + (3 + 7)y² + (−5 − 3)

= 2x² − 15xy + 10y² − 8.

(viii) (x⁴ − 3x³ − 2x² + 3) − (2x⁴ − 7x² + 5x + 3)

= x⁴ − 3x³ − 2x² + 3 − 2x⁴ + 7x² − 5x − 3

= (1 − 2)x⁴ − 3x³ + (−2 + 7)x² − 5x + (3 − 3)

= −x⁴ − 3x³ + 5x² − 5x.

First, find the sum of 10p − r and 5p + 2q.

Sum = (10p − r) + (5p + 2q) = 15p + 2q − r

Now, subtract p − 2q + r from this sum.

(15p + 2q − r) − (p − 2q + r)

= 15p + 2q − r − p + 2q − r

= (15 − 1)p + (2 + 2)q + (−1 − 1)r

= 14p + 4q − 2r.

Sum of 4 + 3x and 5 − 4x + 2x²:

= (4 + 3x) + (5 − 4x + 2x²) = 2x² − x + 9

Sum of 3x² − 5x and −x² + 2x + 5:

= (3x² − 5x) + (−x² + 2x + 5) = 2x² − 3x + 5

Now, subtracting the second sum from the first:

(2x² − x + 9) − (2x² − 3x + 5)

= 2x² − x + 9 − 2x² + 3x − 5

= 2x + 4.

Required expression = (x² + y² + 5xy) − (x² − y² + 2xy)

= x² + y² + 5xy − x² + y² − 2xy

= (1 − 1)x² + (1 + 1)y² + (5 − 2)xy

= 2y² + 3xy.

Required expression = (−7mn + 2m² + 3n²) − (m² + 2mn + n²)

= −7mn + 2m² + 3n² − m² − 2mn − n²

= (2 − 1)m² + (−7 − 2)mn + (3 − 1)n²

= m² + 2n² − 9mn.

Required difference = (y⁴ − 12y² + y + 14) − (17y³ + 34y² − 51y + 68)

= y⁴ − 12y² + y + 14 − 17y³ − 34y² + 51y − 68

= y⁴ − 17y³ + (−12 − 34)y² + (1 + 51)y + (14 − 68)

= y⁴ − 17y³ − 46y² + 52y − 54.

Required difference = (93p² − 55p + 4) − (13p³ − 5p² + 17p − 90)

= 93p² − 55p + 4 − 13p³ + 5p² − 17p + 90

= −13p³ + (93 + 5)p² + (−55 − 17)p + (4 + 90)

= −13p³ + 98p² − 72p + 94.

Required expression = (3x² − 4y² + 5xy + 20) − (−x² − y² + 6xy + 20)

= 3x² − 4y² + 5xy + 20 + x² + y² − 6xy − 20

= (3 + 1)x² + (−4 + 1)y² + (5 − 6)xy + (20 − 20)

= 4x² − 3y² − xy.

Sum of 2y² + 3yz, −y² − yz − z² and yz + 2z²:

= (2y² + 3yz) + (−y² − yz − z²) + (yz + 2z²)

= (2 − 1)y² + (3 − 1 + 1)yz + (−1 + 2)z²

= y² + 3yz + z²

Sum of 3y² − z² and −y² + yz + z²:

= (3y² − z²) + (−y² + yz + z²)

= (3 − 1)y² + yz + (−1 + 1)z²

= 2y² + yz

Now, subtracting the second sum from the first:

(y² + 3yz + z²) − (2y² + yz)

= y² + 3yz + z² − 2y² − yz

= (1 − 2)y² + (3 − 1)yz + z²

= −y² + 2yz + z².

Given, m = 2.

(i) Substituting m = 2 in 3m − 5, we get:

3m − 5 = 3(2) − 5 = 6 − 5

∴ 3m − 5 = 1

(ii) Substituting m = 2 in 9 − 5m, we get:

9 − 5m = 9 − 5(2) = 9 − 10

∴ 9 − 5m = −1

(iii) Substituting m = 2 in 3m² − 2m − 7, we get:

3m² − 2m − 7 = 3(2)² − 2(2) − 7

= 3(4) − 4 − 7

= 12 − 4 − 7

∴ 3m² − 2m − 7 = 1

(iv) Substituting m = 2 in $\frac{5}{2}m - 4$, we get:

$\frac{5}{2}m - 4 = \frac{5}{2}(2) - 4 = 5 - 4$

∴ $\frac{5}{2}$ m − 4 = 1

Given, p = −2.

(i) Substituting p = −2 in 4p + 7, we get:

4p + 7 = 4(−2) + 7 = −8 + 7

∴ 4p + 7 = −1

(ii) Substituting p = −2 in −3p² + 4p + 7, we get:

−3p² + 4p + 7 = −3(−2)² + 4(−2) + 7

= −3(4) − 8 + 7

= −12 − 8 + 7

∴ −3p² + 4p + 7 = −13

(iii) Substituting p = −2 in −2p³ − 3p² + 4p + 7, we get:

−2p³ − 3p² + 4p + 7 = −2(−2)³ − 3(−2)² + 4(−2) + 7

= −2(−8) − 3(4) − 8 + 7

= 16 − 12 − 8 + 7

∴ −2p³ − 3p² + 4p + 7 = 3

Given, a = 2 and b = −2.

(i) Substituting a = 2 and b = −2 in a² + b², we get:

a² + b² = (2)² + (−2)² = 4 + 4

∴ a² + b² = 8

(ii) Substituting a = 2 and b = −2 in a² + ab + b², we get:

a² + ab + b² = (2)² + (2)(−2) + (−2)²

= 4 − 4 + 4

∴ a² + ab + b² = 4

(iii) Substituting a = 2 and b = −2 in a² − b², we get:

a² − b² = (2)² − (−2)² = 4 − 4

∴ a² − b² = 0

Given, a = 0 and b = −1.

(i) Substituting a = 0 and b = −1 in 2a² + b² + 1, we get:

2a² + b² + 1 = 2(0)² + (−1)² + 1

= 0 + 1 + 1

∴ 2a² + b² + 1 = 2

(ii) Substituting a = 0 and b = −1 in a² + ab + 2, we get:

a² + ab + 2 = (0)² + (0)(−1) + 2

= 0 + 0 + 2

∴ a² + ab + 2 = 2

(iii) Substituting a = 0 and b = −1 in 2a²b + 2ab² + ab, we get:

2a²b + 2ab² + ab = 2(0)²(−1) + 2(0)(−1)² + (0)(−1)

= 0 + 0 + 0

∴ 2a²b + 2ab² + ab = 0

Given, p = −10.

Substituting p = −10 in p² − 2p − 100, we get:

p² − 2p − 100 = (−10)² − 2(−10) − 100

= 100 + 20 − 100

∴ p² − 2p − 100 = 20

Given, z = 10.

Substituting z = 10 in z³ − 3(z − 10), we get:

z³ − 3(z − 10) = (10)³ − 3(10 − 10)

= 1000 − 3(0)

= 1000 − 0

∴ z³ − 3(z − 10) = 1000

(i) x + 7 + 4(x − 5)

= x + 7 + 4x − 20

= 5x − 13

When x = 2:

5x − 13 = 5(2) − 13 = 10 − 13

∴ The value is −3.

(ii) 3(x + 2) + 5x − 7

= 3x + 6 + 5x − 7

= 8x − 1

When x = 2:

8x − 1 = 8(2) − 1 = 16 − 1

∴ The value is 15.

(iii) 6x + 5(x − 2)

= 6x + 5x − 10

= 11x − 10

When x = 2:

11x − 10 = 11(2) − 10 = 22 − 10

∴ The value is 12.

(iv) 4(2x − 1) + 3x + 11

= 8x − 4 + 3x + 11

= 11x + 7

When x = 2:

11x + 7 = 11(2) + 7 = 22 + 7

∴ The value is 29.

(i) 2a − 2b − 4 − 5 + a

= (2a + a) − 2b + (−4 − 5)

= 3a − 2b − 9

When a = −1 and b = −2:

3a − 2b − 9 = 3(−1) − 2(−2) − 9

= −3 + 4 − 9

∴ The value is −8.

(ii) 2(a² + ab) + 3 − ab

= 2a² + 2ab + 3 − ab

= 2a² + ab + 3

When a = −1 and b = −2:

2a² + ab + 3 = 2(−1)² + (−1)(−2) + 3

= 2(1) + 2 + 3

= 2 + 2 + 3

∴ The value is 7.

(i) The terms with different algebraic factors are called unlike terms.

(ii) The number of terms in a monomial is one.

(iii) An algebraic expression having two unlike terms is called a binomial.

(iv) Since 3a²b and −7ba² have the same literal part a²b, they are like terms.

(v) Since −6a²b and −6ab² have different powers of a and b, they are unlike terms.

(vi) 3x² − 2xy + 5x² = 8x² − 2xy, which has 2 unlike terms.

(vii) The factors of the term −3p²q² are −3, p, p, q, q.

(viii) Perimeter = 2a + b + (a + b) = 3a + 2b.

(ix) At x = 1, 2x³ − 7x² + 5x − 3 = 2(1) − 7(1) + 5(1) − 3 = 2 − 7 + 5 − 3 = −3.

(x) −7a²bc = (a)(−7abc), so the coefficient of a is −7abc.

(xi) The highest power of x in 3 − 5x² + 7x³ − x⁴ is 4, so the degree is 4.

(xii) In 3x² − 2xy² + 5, the degrees of the terms are 2, (1 + 2) = 3 and 0, so the degree is 3.

(i) False. 5x + 7 − 2x = 3x + 7, which has 2 terms, so it is a binomial.

(ii) False. (7x − 10) − (3x − 5) = 7x − 10 − 3x + 5 = 4x − 5, not 4x − 15.

(iii) False. −3x³y = (3x)(−x²y), so the coefficient of 3x is −x²y.

(iv) False. The constant term in 2x² − 3xy − 7 is −7.

(v) True. $xy(x^2 + y^2) = 3 \times \frac{1}{3}\left(3^2 + \left(\frac{1}{3}\right)^2\right) = 1 \times \left(9 + \frac{1}{9}\right) = 9\frac{1}{9}$.

(vi) False. (3x − y + 5) − (x + y) = 3x − y + 5 − x − y = 2x − 2y + 5, which has 3 terms (a trinomial).

(vii) False. The sum of 2 and p is 2 + p, not 2p.

(viii) False. The sum of x² + x and y² + y is x² + x + y² + y.

(ix) True. In like terms, the variables along with their powers are the same.

(x) False. A polynomial may have more than one term, while a monomial has only one term.

(xi) False. For example, (−x) + (x + 5) = 5, which is a monomial.

(xii) False. For example, (x + 5) − x = 5, which is a monomial.

(xiii) False. Adding a monomial can cancel at most one term of a trinomial, so the result has at least two terms.

(xiv) True. For example, z + (x + y) = x + y + z, which is a trinomial.

Five times the sum of y and 2 = 5(y + 2).

Thrice the square of x = 3x².

Thrice the square subtracted from five times the sum = 5(y + 2) − 3x².

Hence, option 3 is the correct option.

7x − 5(x² + y²) = 7x − 5x² − 5y², which has 3 terms.

Hence, option 3 is the correct option.

−5a³bc = (5a²)(−abc), so the coefficient of 5a² is −abc.

Hence, option 4 is the correct option.

Like terms have the same literal part. Here −5xy and 7yx have the same literal part (xy = yx).

Hence, option 4 is the correct option.

3x(3 − 2y) = 9x − 6xy and 2(xy + x²) = 2xy + 2x².

The like terms are −6xy and 2xy (both contain xy).

Hence, option 2 is the correct option.

1. 3xy² + 5y − x²y has 3 terms → trinomial.

2. 2x²y − 5y − 2x²y = −5y has 1 term → monomial.

3. 3xy² + 5y − xy² = 2xy² + 5y has 2 terms → binomial.

4. xy + yz + zx has 3 terms → trinomial.

Hence, option 3 is the correct option.

3xy² + 2y²z − y²x + y(xz + yz) − 5

= 3xy² + 2y²z − xy² + xyz + y²z − 5

= (3 − 1)xy² + (2 + 1)y²z + xyz − 5

= 2xy² + 3y²z + xyz − 5

This has 4 unlike terms.

Hence, option 2 is the correct option.

x³ + y³ = (2)³ + (−2)³ = 8 + (−8) = 0.

Hence, option 1 is the correct option.

−xy − (−5xy) = −xy + 5xy = (−1 + 5)xy = 4xy.

Hence, option 4 is the correct option.

(5y − 3x − 9) − (7x + 5y − 3)

= 5y − 3x − 9 − 7x − 5y + 3

= (−3 − 7)x + (5 − 5)y + (−9 + 3)

= −10x − 6.

Hence, option 2 is the correct option.

$\frac{5}{3}x^2 + 1 = \frac{5}{3}(-2)^2 + 1 = \frac{5}{3}(4) + 1 = \frac{20}{3} + 1 = \frac{20 + 3}{3} = \frac{23}{3}$.

Hence, option 4 is the correct option.

The degrees of the terms are (3 + 1) = 4, (1 + 4) = 5, 1 and 0. The greatest is 5.

Hence, option 1 is the correct option.

Statement I: 5x² × 3y² + 6z² = 15x²y² + 6z², which has 2 terms, so it is a binomial, not a trinomial. Hence Statement I is false.

Statement II: A trinomial is indeed an algebraic expression having three unlike terms. Hence Statement II is true.

Hence, option 2 is the correct option.

Statement I: (4x² − 3y³ + 6z) − (−2x² + 5y³) = 4x² − 3y³ + 6z + 2x² − 5y³ = 6x² − 8y³ + 6z. Hence Statement I is true.

Statement II: For example, (x² + x + 5) − x = x² + 5, a binomial. So Statement II is true.

Hence, option 3 is the correct option.

Statement I: When x = 2 and y = −2,

$-\frac{1}{2}x + 5 = -\frac{1}{2}(2) + 5 = -1 + 5 = 4$

$\frac{2}{3}y^2 = \frac{2}{3}(-2)^2 = \frac{2}{3}(4) = \frac{8}{3} = 2\frac{2}{3}$

Since $4 \gt 2\frac{2}{3}$, the statement $-\frac{1}{2}x + 5 \gt \frac{2}{3}y^2$ is correct. Hence Statement I is true.

Statement II: The value of an algebraic expression does depend on the values of the variables involved. Hence Statement II is true.

Hence, option 3 is the correct option.

(i) The given expression $\frac{3}{2}x^2y - \frac{1}{2}xy^2$ + 6x²y² has 3 terms.

Since the expression has three terms, it is called a trinomial.

(ii) The terms of the expression are:

$\frac{3}{2}x^2y, -\frac{1}{2}xy^2$ and 6x²y²

(iii) In the term $-\frac{1}{2}xy^2$:

Numerical coefficient = $-\frac{1}{2}$

Literal coefficient = xy²

(iv) In the term $-\frac{1}{2}xy^2$, the coefficient of x is the remaining factor after removing x from the term.

$-\frac{1}{2}xy^2 = -\frac{1}{2} \times x \times y^2$

Hence, the coefficient of x is $-\frac{1}{2}y^2$.

(i) In $\frac{2}{5}x^3 - 7x^2 - \frac{1}{2}x + 3$, the highest power of x is 3.

Hence, the degree of the polynomial is 3.

(ii) In $\frac{2}{3}xy^2 - 5xy + \frac{3}{5}y^2x^2 + 2x$, the degrees of the terms are (1 + 2) = 3, (1 + 1) = 2, (2 + 2) = 4 and 1. The greatest is 4.

Hence, the degree of the polynomial is 4.

(i) 5x × y = 5xy has 1 term → Monomial.

(ii) 3 − 5x has 2 terms → Binomial.

(iii) $\frac{1}{2}(7x - 3y + 5z) = \frac{7}{2}x - \frac{3}{2}y + \frac{5}{2}z$ has 3 terms → Trinomial.

(iv) 3x² − 1.2xy has 2 terms → Binomial.

(v) −3x³y⁴z⁵ has 1 term → Monomial.

(vi) 5x(2x − 3y) + 7x² = 10x² − 15xy + 7x² = 17x² − 15xy has 2 terms → Binomial.

(i) (x² + y² − 2xy) + (−2x² − y² − 2xy) + (3x² + y² + xy)

= (1 − 2 + 3)x² + (1 − 1 + 1)y² + (−2 − 2 + 1)xy

= 2x² + y² − 3xy.

Hence, the sum = 2x² + y² − 3xy.

(ii) (2x² − 3y²) − (−x² + y² + 2xy)

= 2x² − 3y² + x² − y² − 2xy

= (2 + 1)x² + (−3 − 1)y² − 2xy

= 3x² − 4y² − 2xy.

Hence, the required difference = 3x² − 4y² − 2xy.

First, add ab + 2bc − ca and 2ab − bc − ca by the column method:

$$\begin{array}{r r r r} & ab + 2bc - ca \\ + & 2ab - bc - ca \\ \hline & 3ab + bc - 2ca \\ \end{array}$$

Now, subtract 4ab + 5bc − 3ca from this sum by the column method:

$$\begin{array}{r r r r} & 3ab & +\ bc & -\ 2ca \\ - & 4ab & +\ 5bc & -\ 3ca \\ & (-) & (-) & (+) \\ \hline & -ab & -\ 4bc & +\ ca \\ \end{array}$$

The signs of the terms being subtracted are reversed, then the columns are added:

(3ab + bc − 2ca) − (4ab + 5bc − 3ca)

= 3ab + bc − 2ca − 4ab − 5bc + 3ca

= −ab − 4bc + ca.

Hence, final result = −ab − 4bc + ca.

Perimeter of a triangle = sum of its three sides.

Perimeter = (5a − 3b) + (3a + 2b) + (5b − 2a)

= (5 + 3 − 2)a + (−3 + 2 + 5)b

= 6a + 4b.

Hence, perimeter of triangle = 6a + 4b.

Perimeter of a rectangle = 2(length + breadth).

Perimeter = 2[(4x + 7y) + (3y − x)]

= 2[(4 − 1)x + (7 + 3)y]

= 2(3x + 10y)

= 6x + 20y

Hence, perimeter of rectangle = 6x + 20y.

Sum of 3x² + 2xy − 2y² and 5y² − 7xy:

= (3x² + 2xy − 2y²) + (5y² − 7xy)

= 3x² + (2 − 7)xy + (−2 + 5)y²

= 3x² − 5xy + 3y²

Now, subtracting this sum from 5x² + 2y² − 3xy:

(5x² + 2y² − 3xy) − (3x² − 5xy + 3y²)

= 5x² + 2y² − 3xy − 3x² + 5xy − 3y²

= (5 − 3)x² + (2 − 3)y² + (−3 + 5)xy

= 2x² + 2xy − y².

Hence, the required result = 2x² + 2xy − y².

Required expression = (7x³ + 7x − 5) − (5x³ − 2x² + 3x + 7)

= 7x³ + 7x − 5 − 5x³ + 2x² − 3x − 7

= (7 − 5)x³ + 2x² + (7 − 3)x + (−5 − 7)

= 2x³ + 2x² + 4x − 12.

Hence, the required expression = 2x³ + 2x² + 4x − 12.

Required difference = (4p + 3q − 5r) − (3p − 4q + r)

= 4p + 3q − 5r − 3p + 4q − r

= (4 − 3)p + (3 + 4)q + (−5 − 1)r

= p + 7q − 6r.

Hence, the required difference = p + 7q − 6r.

Required difference = (3a² − 5ab + 7b² + 3) − (2a² + 2ab + 5)

= 3a² − 5ab + 7b² + 3 − 2a² − 2ab − 5

= (3 − 2)a² + (−5 − 2)ab + 7b² + (3 − 5)

= a² − 7ab + 7b² − 2.

Hence, the required difference = a² − 7ab + 7b² − 2.

Required expression = (6x² + 7) − (5x³ + 3x² − 2x + 1)

= 6x² + 7 − 5x³ − 3x² + 2x − 1

= −5x³ + (6 − 3)x² + 2x + (7 − 1)

= −5x³ + 3x² + 2x + 6.

Hence, the required expression = −5x³ + 3x² + 2x + 6.

Sum of 12ab − 10b² − 18a² and 9ab + 12b² + 14a²:

= (12ab − 10b² − 18a²) + (9ab + 12b² + 14a²)

= (12 + 9)ab + (−10 + 12)b² + (−18 + 14)a²

= 21ab + 2b² − 4a²

Sum of ab + 2b² and 3b² − a²:

= (ab + 2b²) + (3b² − a²)

= ab + (2 + 3)b² − a²

= ab + 5b² − a²

Now, subtracting the first sum from the second sum:

(ab + 5b² − a²) − (21ab + 2b² − 4a²)

= ab + 5b² − a² − 21ab − 2b² + 4a²

= (1 − 21)ab + (5 − 2)b² + (−1 + 4)a²

= −20ab + 3b² + 3a²

= 3a² − 20ab + 3b².

Hence, the required result = 3a² − 20ab + 3b².

Given, a = 3, b = 0 and c = −2.

(i) Substituting a = 3, b = 0 and c = −2 in ab + 2bc + 3ca + 4abc, we get:

ab + 2bc + 3ca + 4abc

= (3)(0) + 2(0)(−2) + 3(−2)(3) + 4(3)(0)(−2)

= 0 + 0 − 18 + 0

∴ ab + 2bc + 3ca + 4abc = −18

(ii) Substituting a = 3, b = 0 and c = −2 in a³ + b³ + c³ − 3abc, we get:

a³ + b³ + c³ − 3abc

= (3)³ + (0)³ + (−2)³ − 3(3)(0)(−2)

= 27 + 0 − 8 − 0

∴ a³ + b³ + c³ − 3abc = 19

Perimeter of a rectangle = 2(length + breadth).

So, length + breadth = $\frac{\text{Perimeter}}{2}$

∴ breadth = $\frac{\text{Perimeter}}{2}$ − length

$$\begin{aligned}\phantom{=} \text{breadth} = \frac{7x + 8y + 17z}{2} - (3x - 4y + 6z) \\[1em] = \frac{7x}{2} + 4y + \frac{17z}{2} - 3x + 4y - 6z \\[1em] = \left(\frac{7x}{2} - 3x\right) + (4y + 4y) + \left(\frac{17z}{2} - 6z\right) \\[1em] = \frac{7x - 6x}{2} + 8y + \frac{17z - 12z}{2} \\[1em] = \frac{x}{2} + 8y + \frac{5z}{2}\end{aligned}$$

Hence, the breadth of the rectangle is $\frac{1}{2}x + 8y + \frac{5}{2}z$.

$$\begin{aligned}\phantom{=} \frac{3x}{5} + \frac{2x}{3} - \left(\frac{x}{2} + \frac{2x}{5}\right) \\[1em] = \frac{3x}{5} + \frac{2x}{3} - \frac{x}{2} - \frac{2x}{5} \\[1em] = \frac{18x + 20x - 15x - 12x}{30} \\[1em] = \frac{11x}{30}\end{aligned}$$

Hence, the simplified value is $\frac{11}{30}x$.

Given, a = 3 and b = −1.

(i) $a^b = 3^{-1} = \frac{1}{3}$.

(ii) b^a = (−1)³ = −1.

(iii) $(ab)^b = (3 \times (-1))^{-1} = (-3)^{-1} = -\frac{1}{3}$.

(iv) $(a + b)^b = (3 + (-1))^{-1} = (2)^{-1} = \frac{1}{2}$.

(v) $\left(\frac{b}{a}\right)^{b} = \left(\frac{-1}{3}\right)^{-1} = \frac{3}{-1} = -3$.

(vi) $$\begin{aligned}\left(\frac{a}{b} + \frac{b}{a}\right)^{b}\\[1em] = \left(\frac{3}{-1} + \frac{-1}{3}\right)^{-1}\\[1em] = \left(-3 - \frac{1}{3}\right)^{-1} = \left(\frac{-9 - 1}{3}\right)^{-1}\\[1em]= \left(\frac{-10}{3}\right)^{-1} = -\frac{3}{10}.\end{aligned}$$