Chapter-12 Algebraic Expressions

1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y, both squared and added.
(vi) Number 5 added to three times the product of numbers m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.

Solution:-

(i)

Here, in the question it is given that we have to subtract the value z from the value Y.

We know that for subtracting we use the symbol ”-“

So, by applying the given condition.

= y – z

(ii)

Here, in the question it is given that we have to add the value of x in the value of y, than make it one-half.

We know that to make any value equal to one-half we have to multiply that value with ½

So, by applying the given condition.

= ½ (x + y)

= (x + y)/2

(iii)

Here, in the question it is given that we have to multiply the value z with itself means we have to multiply the value z with the value z.

So, by applying the given condition.

= z × z

= z2

(iv)

Here, in the question it is given that we have to obtain the one-fourth of the product of the number p and q.

Now, to obtain the product of the number of p and q we have to multiply then and to obtain the one-fourth of the product we have to multiply that whole number by ¼.

So, by applying the given condition.

= ¼ (p × q)

= pq/4

(v)

Here, in the question it is given that we have to we have to add the squares of the number x and y.

Now, here first we need to obtain the squares of the number x and y.

So, to obtain the squares of any number we multiply that number with itself and after obtaining the squares we need to add then.

So, by applying the given condition.

= x2 + y2

(vi)

Here, in the question it is given that we have to first multiply the number m and n and after that we need to obtain the three times of that number and then we gave to add 5 to that whole number.

So, by applying the given condition.

= 3mn + 5

(vii)

Here, in the question it is given that we have to subtract the product of the number y and z from 10.

Now, we first need to obtain the product of the number y and z, so, to obtain the product we multiply, by multiplying the number y and z we obtain the product.

After that subtract that product from 10.

So, by applying the given condition.

= 10 – (y × z)

= 10 – yz

(viii)

Here, in the question it is given that we have to subtract the sum of the number a and b from the product of the number a and b.

So, first we need to obtain the product and sum of the number a and b then subtract the sum from the product.

So, by applying the given condition.

= (a × b) – (a + b)

= ab – (a + b)

2.Identify the terms and their factors in the following expressions.
Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x2
(c) y – y3
(d) 5xy2 + 7x2y
(e) – ab + 2b2 – 3a2

Solution:-

(a)

Here, in the question it is given that we are subtracting 3 from the number x.

So, there is 2 terms, one is x and the second is-3.

Terms: x, -3

Factors: x; -3

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 1

(b)

Here, in the question it is given that we have to add 1, x and x².

So, the is the terms in the expression: 1 + x + x2

Terms: 1, x, x2

For factors there is only one term which is in factorization form and that is x² which can be divided into x,x

Factors: 1; x; x,x

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 2

(c)

Here, in the question it is given that we have to subtract the cube of the number y from the number y.

So, the is 2 terms in the above expression: y – y3

Terms: y, -y3

For, factorization, their is only one term which can be factorise and that is y³.

Factors: y; -y, -y, -y

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 3

(d)

Here, in the question it is given that we have to add seven times of the product of the number x² and y to the five times of the product of the number x and y².

So, there are total 2 terms in the above expression: 5xy2 + 7x2y

Terms: 5xy2, 7x2y

For factorization we can factorize x² into x,x and y² into y,y

So, by applying the given condition.

Factors: 5, x, y, y; 7, x, x, y

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 4

(e)

Here, in the question it is given that we have to subtract the product of the number a and b and Ajay subtract the product of the number a² from the 2 times of the number b².

So, there are total three terms in the above expression: -ab + 2b2 – 3a2

Terms: -ab, 2b2, -3a2

For factorization we can factorize a² into a,a; b² into b,b and ab into a,b

Factors: -a, b; 2, b, b; -3, a, a

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Image 5

3. Identify terms and factors in the expressions given below.
(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2
(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼
(h) 0.1 p2 + 0.2 q2

Solution:-

In the equation there may be different types of number, variable and symbols.

In the Mathematics different types of the operators are available (such as +. – , × and ÷) which are used in the equations.

So, this types of the number, variable, symbol and the operators in the equations are known as expressions..

Now, when this types of the existing are separated by the operators like “+” or “-“ they are known as terms. They can be single number, single variable numbers and variables multiplied together.

Now, when we separate the terms in the form of expressions the value we obtained is known as factors.

Sl.No.	Expression	Terms	Factors
(a)	– 4x + 5	-4x 5	-4, x 5
(b)	– 4x + 5y	-4x 5y	-4, x 5, y
(c)	5y + 3y2	5y 3y2	5, y 3, y, y
(d)	xy + 2x2y2	xy 2x2y2	x, y 2, x, x, y, y
(e)	pq + q	pq q	P, q Q
(f)	1.2 ab – 2.4 b + 3.6 a	1.2ab -2.4b 3.6a	1.2, a, b -2.4, b 3.6, a
(g)	¾ x + ¼	¾ x ¼	¾, x ¼
(h)	0.1 p2 + 0.2 q2	0.1p2 0.2q2	0.1, p, p 0.2, q, q

4. Identify the numerical coefficients of terms (other than constants) in the following expressions.
(i) 5 – 3t2 (ii) 1 + t + t2 + t3 (iii) x + 2xy + 3y (iv) 100m + 1000n (v) – p2q2 + 7pq (vi) 1.2 a + 0.8 b (vii) 3.14 r2 (viii) 2 (l + b)
(ix) 0.1 y + 0.01 y2

Solution:-

In the equation there may be different types of number, variable and symbols.

In the Mathematics different types of the operators are available (such as +. – , × and ÷) which are used in the equations.

So, this types of the number, variable, symbol and the operators in the equations are known as expressions..

In the terms, the remaining part except the variable is known as the coefficient.

So, mainly coefficients are the number multiplied with the operators. If there is only one expressions in the term that means the coefficient of that term is 1.

Sl.No.	Expression	Terms	Coefficients
(i)	5 – 3t2	– 3t2	-3
(ii)	1 + t + t2 + t3	t t2 t3	1 1 1
(iii)	x + 2xy + 3y	x 2xy 3y	1 2 3
(iv)	100m + 1000n	100m 1000n	100 1000
(v)	– p2q2 + 7pq	-p2q2 7pq	-1 7
(vi)	1.2 a + 0.8 b	1.2a 0.8b	1.2 0.8
(vii)	3.14 r2	3.142	3.14
(viii)	2 (l + b)	2l 2b	2 2
(ix)	0.1 y + 0.01 y2	0.1y 0.01y2	0.1 0.01

5.Identify terms which contain x and give the coefficient of x.
(i) y2x + y (ii) 13y2 – 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25
(vii) 7x + xy2

Solution:-

In the equation there may be different types of number, variable and symbols.

In the Mathematics different types of the operators are available (such as +. – , × and ÷) which are used in the equations.

So, this types of the number, variable, symbol and the operators in the equations are known as expressions..

In the terms, the remaining part except the variable is known as the coefficient.

So, mainly coefficients are the number multiplied with the operators. If there is only one expressions in the term that means the coefficient of that term is 1.

Sl.No.	Expression	Terms	Coefficient of x
(i)	y2x + y	y2x	y2
(ii)	13y2 – 8yx	– 8yx	-8y
(iii)	x + y + 2	x	1
(iv)	5 + z + zx	x zx	1 z
(v)	1 + x + xy	xy	y
(vi)	12xy2 + 25	12xy2	12y2
(vii)	7x + xy2	7x xy2	7 y2

6. Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2

Solution:-

In the equation there may be different types of number, variable and symbols.

In the Mathematics different types of the operators are available (such as +. – , × and ÷) which are used in the equations.

So, this types of the number, variable, symbol and the operators in the equations are known as expressions..

In the first expression, 8-xy², there is 2 terms. From this two teens the first one is the number and the second one is the product of x and y². So, the coefficient of the second term is -x.
In the second expression, 5y²+7x, there is 2 terms. From this two term, the first one is the product of 5 and y² and the stuffing one is the product of the number 7 and x. So, there is only one term which contains y² and the coefficient of this term of 5.
In the third expression, 2x²y-15xy²+7y², there is 3 terms. From this three only serving and third terms contains y². So, the coefficient of this terms are -15x and 7.

(i)

8 – xy2

– xy2

– x

(ii)

5y2 + 7x

5y2

(iii)

2x2y – 15xy2 + 7y2

– 15xy2

7y2

– 15x

7. Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y2
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p2q – 4pq2
(viii) 7mn
(ix) z2 – 3z + 8
(x) a2 + b2
(xi) z2 + z
(xii) 1 + x + x2

Solution:-

Binomial.

When, the expression contains two term, the expression is known as binomial.

Here, in the above expression there is two terms.

(ii)

Monomial.

When, the expression contains only one terms, the expression is known as monomial.

Here, in the above expression there is only one term.

(iii)

Trinomial.

When, the expression contains three terms, the expression is known as trinomial.

Here, in the above expression there is three terms.

(iv)

Monomial.

When, the expression contains only one term, the expression is known as monomial.

Here, in the above expression there is only one term.

(v)

Trinomial.

When, the expression contains three terms, the expression is known as trinomial.

Here, in the above expression there is three terms.

(vi)

Binomial.

When, the expression contains two term, the expression is known as binomial.

Here, in the above expression there is two terms.

(vii)

Binomial.

When, the expression contains two term, the expression is known as binomial.

Here, in the above expression there is two terms.

(viii)

Monomial.

When, the expression contains only one term, the expression is known as monomial.

Here, in the above expression there is only one term.

(ix)

Trinomial.

When, the expression contains three terms, the expression is known as trinomial.

Here, in the above expression there is three terms.

(x)

Binomial.

When, the expression contains two term, the expression is known as binomial.

Here, in the above expression there is two terms.

(xi)

Binomial.

When, the expression contains two term, the expression is known as binomial.

Here, in the above expression there is two terms.

(xii)

Trinomial.

When, the expression contains three terms, the expression is known as trinomial.

Here, in the above expression there is three terms.

8. State whether a given pair of terms is of like or unlike terms.
1, 100

Solution:-

Like term.

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

Here, both the quantities have same algebraic factors they are known as like terms.

9. –7x, (5/2)x

Like term.

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

Here, both the quantities have same algebraic factors they are known as like terms.

10. – 29x, – 29y

Solution:-

Unlike terms.

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

Here, both the quantities not having same algebraic factors, so, they are known as unlike terms.

11. 14xy, 42yx

Solution:-

Like term.

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

Here, both the quantities have same algebraic factors they are known as like terms.

12. 4m2p, 4mp2

Solution:-

Unlike terms.

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

Here, both the quantities not having same algebraic factors, so, they are known as unlike terms.

13. 12xz, 12x2z2

Solution:-

Unlike terms.

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

Here, both the quantities not having same algebraic factors, so, they are known as unlike terms.

14. Identify like terms in the following.
– xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

They are,

– xy2, 2xy2

– 4yx2, 20x2y

8x2, – 11x2, – 6x2

7y, y

– 100x, 3x

– 11yx, 2xy

15. 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,
13p2q, qp2, 701p2

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

They are,

10pq, – 7qp, 78qp

7p, 2405p

8q, – 100q

– p2q2, 12q2p2

– 23, 41

– 5p2, 701p2

13p2q, qp2

16. Simplify combining like terms.
21b – 32 + 7b – 20b

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Hence,

= (21b + 7b – 20b) – 32

= b (21 + 7 – 20) – 32

= b (28 – 20) – 32

= b (8) – 32

= 8b – 32

17. – z2 + 13z2 – 5z + 7z3 – 15z

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Hence,

= 7z3 + (-z2 + 13z2) + (-5z – 15z)

= 7z3 + z2 (-1 + 13) + z (-5 – 15)

= 7z3 + z2 (12) + z (-20)

= 7z3 + 12z2 – 20z

18. p – (p – q) – q – (q – p)

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Hence,

= p – p + q – q – q + p

= p – q

19. 3a – 2b – ab – (a – b + ab) + 3ab + b – a

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Hence,

= 3a – 2b – ab – a + b – ab + 3ab + b – a

= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)

= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)

= a (1) + b (0) + ab (1)

= a + ab

20. 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Hence,

= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2

= x2y (5 + 3) + x2 (- 5 + 1) + y2 (-3 – 1 -3) + 8xy2

= x2y (8) + x2 (-4) + y2 (-7) + 8xy2

= 8x2y – 4x2 – 7y2 + 8xy2

21. (3y2 + 5y – 4) – (8y – y2 – 4)

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Hence,

= 3y2 + 5y – 4 – 8y + y2 + 4

= 3y2 + y2 + 5y – 8y – 4 + 4

= y2 (3 + 1) + y (5 – 8) + (-4 + 4)

= y2 (4) + y (-3) + (0)

= 4y2 – 3y

22. Add:
3mn, – 5mn, 8mn, – 4mn

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= 3mn + (-5mn) + 8mn + (- 4mn)

= 3mn – 5mn + 8mn – 4mn

= mn (3 – 5 + 8 – 4)

= mn (11 – 9)

= mn (2)

= 2mn

23. t – 8tz, 3tz – z, z – t

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= t – 8tz + (3tz – z) + (z – t)

= t – 8tz + 3tz – z + z – t

= t – t – 8tz + 3tz – z + z

= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)

= t (0) + tz (- 5) + z (0)

= – 5tz

24. – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)

= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3

= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3

= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)

= mn (- 9 + 21) + (7 – 11)

= mn (12) – 4

= 12mn – 4

25. a + b – 3, b – a + 3, a – b + 3

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= a + b – 3 + (b – a + 3) + (a – b + 3)

= a + b – 3 + b – a + 3 + a – b + 3

= a – a + a + b + b – b – 3 + 3 + 3

= a (1 – 1 + 1) + b (1 + 1 – 1) + (-3 + 3 + 3)

= a (2 -1) + b (2 -1) + (-3 + 6)

= a (1) + b (1) + (3)

= a + b + 3

26. 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

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

= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy

= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18

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

= x (7) + y (0) + xy(0) + (5)

= 7x + 5

27. 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

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

= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5

= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5

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

= m (3) + n (-4) – 3mn + (-3)

= 3m – 4n – 3mn – 3

28. 4x2y, – 3xy2, –5xy2, 5x2y

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= 4x2y + (-3xy2) + (-5xy2) + 5x2y

= 4x2y + 5x2y – 3xy2 – 5xy2

= x2y (4 + 5) + xy2 (-3 – 5)

= x2y (9) + xy2 (- 8)

= 9x2y – 8xy2

29. 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= 3p2q2 – 4pq + 5 + (- 10p2q2) + 15 + 9pq + 7p2q2

= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15

= p2q2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)

= p2q2 (0) + pq (5) + 20

= 5pq + 20

30. ab – 4a, 4b – ab, 4a – 4b

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= ab – 4a + (4b – ab) + (4a – 4b)

= ab – 4a + 4b – ab + 4a – 4b

= ab – ab – 4a + 4a + 4b – 4b

= ab (1 -1) + a (4 – 4) + b (4 – 4)

= ab (0) + a (0) + b (0)

= 0

31. x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can add the line terms. So, by adding the like terms we get,

= x2 – y2 – 1 + (y2 – 1 – x2) + (1 – x2 – y2)

= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2

= x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1

= x2 (1 – 1- 1) + y2 (-1 + 1 – 1) + (-1 -1 + 1)

= x2 (1 – 2) + y2 (-2 +1) + (-2 + 1)

= x2 (-1) + y2 (-1) + (-1)

= -x2 – y2 -1

32. Subtract:
–5y2 from y2

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= y2 – (-5y2)

= y2 + 5y2

= 6y2

33. 6xy from –12xy

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= -12xy – 6xy

= – 18xy

34. (a – b) from (a + b)

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= (a + b) – (a – b)

= a + b – a + b

= a – a + b + b

= a (1 – 1) + b (1 + 1)

= a (0) + b (2)

= 2b

35. a (b – 5) from b (5 – a)

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= b (5 -a) – a (b – 5)

= 5b – ab – ab + 5a

= 5b + ab (-1 -1) + 5a

= 5a + 5b – 2ab

36. –m2 + 5mn from 4m2 – 3mn + 8

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= 4m2 – 3mn + 8 – (- m2 + 5mn)

= 4m2 – 3mn + 8 + m2 – 5mn

= 4m2 + m2 – 3mn – 5mn + 8

= 5m2 – 8mn + 8

37. – x2 + 10x – 5 from 5x – 10

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= 5x – 10 – (-x2 + 10x – 5)

= 5x – 10 + x2 – 10x + 5

= x2 + 5x – 10x – 10 + 5

= x2 – 5x – 5

38. 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)

= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2

= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2

= 10ab – 7a2 – 7b2

39. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

Solution:-

When, the quantities in the expression are separated by the operators like “+” and “-“ they are known as terms.

The quantities have same algebraic factors they are known as like terms.

Now, we can subtract the line terms. So, by subtracting the like terms we get,

= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)

= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq

= 8p2 + 8q2 – 5pq

40. What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?

Solution:-

Here, in the above expression we need to add something to obtain the other expression.

So, let's suppose that the term added to the first expression is m.

Hence,

M+ (x2 + xy + y2) = 2x2 + 3xy

M= (2x2 + 3xy) – (x2 + xy + y2)

M = 2x2 + 3xy – x2 – xy – y2

M = 2x2 – x2 + 3xy – xy – y2

M = x2 + 2xy – y2

41.What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?

Solution:-

Here, in the above expression we need to subtract something from the field expression to obtain the second expression.

So, let's suppose that the term subtracted from the first expression is x.

Hence,

2a + 8b + 10 – x = -3a + 7b + 16

x = (2a + 8b + 10) – (-3a + 7b + 16)

x = 2a + 8b + 10 + 3a – 7b – 16

x = 2a + 3a + 8b – 7b + 10 – 16

x = 5a + b – 6

42. What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20?

Solution:-

Here, in the above expression we need to subtract something from the field expression to obtain the second expression.

So, let’s suppose that the term subtracted from the first expression is a.

Hence,

3x2 – 4y2 + 5xy + 20 – a = -x2 – y2 + 6xy + 20

a = 3x2 – 4y2 + 5xy + 20 – (-x2 – y2 + 6xy + 20)

a = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20

a = 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20

a = 4x2 – 3y2 – xy

43. From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.

Solution:-

Here, in the question it is given that we need to add the expression 3x – y + 11 and – y – 11.

So, by adding we get

= 3x – y + 11 + (-y – 11)

= 3x – y + 11 – y – 11

= 3x – y – y + 11 – 11

= 3x – 2y

After that we asked to subtract 3x – y – 11 from 3x – 2y.

So, by subtracting we get

= 3x – 2y – (3x – y – 11)

= 3x – 2y – 3x + y + 11

= 3x – 3x – 2y + y + 11

= -y + 11

44. From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and
–x2 + 2x + 5.

Solution:-

Here, in the question it is given that we need to add the expression 4 + 3x and 5 – 4x + 2x2

So, by adding we get

= 4 + 3x + (5 – 4x + 2x2)

= 4 + 3x + 5 – 4x + 2x2

= 4 + 5 + 3x – 4x + 2x2

= 9 – x + 2x2

= 2x2 – x + 9 … [equation 1]

Now in the second part of the question we required to add the other two expression. Which is 3x2 – 5x and – x2 + 2x + 5.

So, by adding them we get

= 3x2 – 5x + (-x2 + 2x + 5)

= 3x2 – 5x – x2 + 2x + 5

= 3x2 – x2 – 5x + 2x + 5

= 2x2 – 3x + 5 … [equation 2]

Now, at last we need to subtract the values obtained by adding last two expression from the first try expression. So, by subtracting equation (2) from equation (1) we get

= 2x2 – x + 9 – (2x2 – 3x + 5)

= 2x2 – x + 9 – 2x2 + 3x – 5

= 2x2 – 2x2 – x + 3x + 9 – 5

= 2x + 4

45. If m = 2, find the value of:
m – 2

Solution:-

Here, in the question, the value of m is given as 2.

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as 2

= 2 -2

= 0

46. 3m – 5

Solution:-

Here, in the question, the value of m is given as 2.

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as 2

= (3 × 2) – 5

= 6 – 5

= 1

47. 9 – 5m

Solution:-

Here, in the question, the value of m is given as 2.

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as 2

= 9 – (5 × 2)

= 9 – 10

= – 1

48. 3m2 – 2m – 7

Solution:-

Here, in the question, the value of m is given as 2.

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as 2

= (3 × 22) – (2 × 2) – 7

= (3 × 4) – (4) – 7

= 12 – 4 -7

= 12 – 11

= 1

49. (5m/2) – 4

Solution:-

Here, in the question, the value of m is given as 2.

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as 2

= ((5 × 2)/2) – 4

= (10/2) – 4

= 5 – 4

= 1

50. If p = – 2, find the value of:
4p + 7

Solution:-

Here, in the question, the value of m is given as -2

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as -2

= (4 × (-2)) + 7

= -8 + 7

= -1

51. – 3p2 + 4p + 7

Solution:-

Here, in the question, the value of m is given as -2

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as -2

= (-3 × (-2)2) + (4 × (-2)) + 7

= (-3 × 4) + (-8) + 7

= -12 – 8 + 7

= -20 + 7

= -13

52. – 2p3 – 3p2 + 4p + 7

Solution:-

Here, in the question, the value of m is given as -2

So, by putting the value of m in the above expression, we get the value of the expression.

Now, put the value of m as -2

= (-2 × (-2)3) – (3 × (-2)2) + (4 × (-2)) + 7

= (-2 × -8) – (3 × 4) + (-8) + 7

= 16 – 12 – 8 + 7

= 23 – 20

= 3

53. Find the value of the following expressions when x = –1:
2x – 7

Solution:-

Here, in the question, the value of x is given as -1.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as -1

= (2 × -1) – 7

= – 2 – 7

= – 9

54. – x + 2

Solution:-

Here, in the question, the value of x is given as -1.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as -1

= – (-1) + 2

= 1 + 2

= 3

55. x2 + 2x + 1

Solution:-

Here, in the question, the value of x is given as -1.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as -1

= (-1)2 + (2 × -1) + 1

= 1 – 2 + 1

= 2 – 2

= 0

56. 2x2 – x – 2

Solution:-

Here, in the question, the value of x is given as -1.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as -1

= (2 × (-1)2) – (-1) – 2

= (2 × 1) + 1 – 2

= 2 + 1 – 2

= 3 – 2

= 1

57. If a = 2, b = – 2, find the value of:
a2 + b2

Solution:-

Here, in the question, the value of a and b is given as 2 and -2 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 2 and b as -2

= (2)2 + (-2)2

= 4 + 4

= 8

58. a2 + ab + b2

Solution:-

Here, in the question, the value of a and b is given as 2 and -2 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 2 and b as -2

= 22 + (2 × -2) + (-2)2

= 4 + (-4) + (4)

= 4 – 4 + 4

= 4

59. a2 – b2

Solution:-

Here, in the question, the value of a and b is given as 2 and -2 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 2 and b as -2

= 22 – (-2)2

= 4 – (4)

= 4 – 4

= 0

60. When a = 0, b = – 1, find the value of the given expressions:
2a + 2b

Solution:-

Here, in the question, the value of a and b is given as 0 and -1 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 0 and b as -1

= (2 × 0) + (2 × -1)

= 0 – 2

= -2

61. 2a2 + b2 + 1

Solution:-

Here, in the question, the value of a and b is given as 0 and -1 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 0 and b as -1

= (2 × 02) + (-1)2 + 1

= 0 + 1 + 1

= 2

62. 2a2b + 2ab2 + ab

Solution:-

Here, in the question, the value of a and b is given as 0 and -1 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 0 and b as -1

= (2 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1)

= 0 + 0 +0

= 0

63. a2 + ab + 2

Solution:-

Here, in the question, the value of a and b is given as 0 and -1 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 0 and b as -1

= (02) + (0 × (-1)) + 2

= 0 + 0 + 2

= 2

64. Simplify the expressions and find the value if x is equal to 2
x + 7 + 4 (x – 5)

Solution:-

Here, in the question, the value of x is given as 2.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as 2

= x + 7 + 4x – 20

= 5x + 7 – 20

Then, substitute the value of x in the equation.

= (5 × 2) + 7 – 20

= 10 + 7 – 20

= 17 – 20

= – 3

65. 3 (x + 2) + 5x – 7

Solution:-

Here, in the question, the value of x is given as 2.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as 2

= 3x + 6 + 5x – 7

= 8x – 1

Then, substitute the value of x in the equation.

= (8 × 2) – 1

= 16 – 1

= 15

66. 6x + 5 (x – 2)

Solution:-

Here, in the question, the value of x is given as 2.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as 2

= 6x + 5x – 10

= 11x – 10

Then, substitute the value of x in the equation.

= (11 × 2) – 10

= 22 – 10

= 12

67. 4(2x – 1) + 3x + 11

Solution:-

Here, in the question, the value of x is given as 2.

So, by putting the value of x in the above expression, we get the value of the expression.

Now, put the value of x as 2

= 8x – 4 + 3x + 11

= 11x + 7

Then, substitute the value of x in the equation.

= (11 × 2) + 7

= 22 + 7

= 29

68. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.
3x – 5 – x + 9

Solution:-

Here, in the question, the value of x, a and b is given as 3, -1 and -2 respectively.

So, by putting the value of x, a and b in the above expression, we get the value of the expression.

Now, put the value of x as 3, a as -1 and b as -2

= 3x – x – 5 + 9

= 2x + 4

= (2 × 3) + 4

= 6 + 4

= 10

69. 2 – 8x + 4x + 4

Solution:-

Here, in the question, the value of x, a and b is given as 3, -1 and -2 respectively.

So, by putting the value of x, a and b in the above expression, we get the value of the expression.

Now, put the value of x as 3, a as -1 and b as -2

= 2 + 4 – 8x + 4x

= 6 – 4x

= 6 – (4 × 3)

= 6 – 12

= – 6

70. 3a + 5 – 8a + 1

Solution:-

Here, in the question, the value of x, a and b is given as 3, -1 and -2 respectively.

So, by putting the value of x, a and b in the above expression, we get the value of the expression.

Now, put the value of x as 3, a as -1 and b as -2

= 3a – 8a + 5 + 1

= – 5a + 6

= – (5 × (-1)) + 6

= – (-5) + 6

= 5 + 6

= 11

71. 10 – 3b – 4 – 5b

Solution:-

Here, in the question, the value of x, a and b is given as 3, -1 and -2 respectively.

So, by putting the value of x, a and b in the above expression, we get the value of the expression.

Now, put the value of x as 3, a as -1 and b as -2

= 10 – 4 – 3b – 5b

= 6 – 8b

= 6 – (8 × (-2))

= 6 – (-16)

= 6 + 16

= 22

72. 2a – 2b – 4 – 5 + a

Solution:-

Here, in the question, the value of x, a and b is given as 3, -1 and -2 respectively.

So, by putting the value of x, a and b in the above expression, we get the value of the expression.

Now, put the value of x as 3, a as -1 and b as -2

= 2a + a – 2b – 4 – 5

= 3a – 2b – 9

= (3 × (-1)) – (2 × (-2)) – 9

= -3 – (-4) – 9

= – 3 + 4 – 9

= -12 + 4

= -8

73. If z = 10, find the value of z3 – 3(z – 10).

Solution:-

Here, in the question, the value of z is given as 10

So, by putting the value of z in the above expression, we get the value of the expression.

Now, put the value of z as 10

= z3 – 3z + 30

= (10)3 – (3 × 10) + 30

= 1000 – 30 + 30

= 1000

74. If p = – 10, find the value of p2 – 2p – 100

Solution:-

Here, in the question, the value of p is given as -10

So, by putting the value of p in the above expression, we get the value of the expression.

Now, put the value of p as -10

= p2 – 2p – 100

= (-10)2 – (2 × (-10)) – 100

= 100 + 20 – 100

= 20

75. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?

Solution:-

Here, in the question it is given that we have to find the value of a such that the expression 2x²+x-a is equal to 5 and the value of x is given as 0.

So, by putting the value of x in the above expression, we get the expression which contains only one variable and that is a, and simplifying that expression we get the value of a.

2x2 + x – a = 5

a = 2x2 + x – 5

a = (2 × 02) + 0 – 5

a = 0 + 0 – 5

a = -5

76. Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab

Solution:-

Here, in the question, the value of a and b is given as 5 and -3 respectively.

So, by putting the value of a and b in the above expression, we get the value of the expression.

Now, put the value of a as 5 and b as -3

= 2a2 + 2ab + 3 – ab

= 2a2 + ab + 3

= (2 × 52) + (5 × (-3)) + 3

= (2 × 25) + (-15) + 3

= 50 – 15 + 3

= 53 – 15

= 38

77. Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind

Solution:-

(a)

Here, in the question, we have given that we need n digits of the same kind to form the segment is (5n+1).

Now,

For 5 digits, we required the number of segments = ((5 × 5) + 1)

= (25 + 1)

= 26

For 10 digits, we required the number of segments = ((5 × 10) + 1)

= (50 + 1)

= 51

For 100 digits, we required the number of segments = ((5 × 100) + 1)

= (500 + 1)

= 501

(b)

Here, in the question, we have given that we need n digits of the same kind to form the segment is (3n+1).

Now,

For 5 digits, we required the number of segments = ((3 × 5) + 1)

= (15 + 1)

= 16

For 10 digits, we required the number of segments = ((3 × 10) + 1)

= (30 + 1)

= 31

For 100 digits, we required the number of segments = ((3 × 100) + 1)

= (300 + 1)

= 301

(c)

Here, in the question, we have given that we need n digits of the same kind to form the number of segments is (5n+2)

Now,

For 5 digits, we required the number of segments = ((5 × 5) + 2)

= (25 + 2)

= 27

For 10 digits, we required the number of segments = ((5 × 10) + 2)

= (50 + 2)

= 52

For 100 digits, we required the number of segments = ((5 × 100) + 1)

= (500 + 2)

= 502

78. Use the given algebraic expression to complete the table of number patterns.
S. No.
Expression
Terms
1st
2nd
3rd
4th
5th
…
10th
…
100th
…
(i)
2n – 1
1
3
5
7
9
–
19
–
–
–
(ii)
3n + 2
5
8
11
14
–
–
–
–
–
–
(iii)
4n + 1
5
9
13
17
–
–
–
–
–
–
(iv)
7n + 20
27
34
41
48
–
–
–
–
–
–
(v)
n2 + 1
2
5
10
17
–
–
–
–
10001
–

Solution:-

(i)

Here, in the above table the algebraic expression is given as (2n – 1)

So for 100th term we can calculate by putting the value of n as 100 in the above expression

Then,

n = 100

= (2 × 100) – 1

= 200 – 1

= 199

(ii)

Here, in the above table the algebraic expression is given as (3n+2)

So for 5th term we can calculate by putting the value of n as 5 in the above expression

Then,

n = 5

= (3 × 5) + 2

= 15 + 2

= 17

For 10th term put n=10 in the above expression

n = 10

= (3 × 10) + 2

= 30 + 2

= 32

Foin the above expression in the above expression

n = 100

= (3 × 100) + 2

= 300 + 2

= 302

(iii)

Here, in the above table the algebraic expression is given as (4n+1)

So for 5th term we can calculate by putting the value of n as 5 in the above expression

Then,

n = 5

= (4 × 5) + 1

= 20 + 1

= 21

For 10th term put n=10 in the above expression

n = 10

= (4 × 10) + 1

= 40 + 1

= 41

For 100th term put n=100 in the above expression

n = 100

= (4 × 100) + 1

= 400 + 1

= 401

(iv)

Here, in the above table the algebraic expression is given as (7n+20)

So for 5th term we can calculate by putting the value of n as 5 in the above expression

Then,

n = 5

= (7 × 5) + 20

= 35 + 20

= 55

For 10th term put n=10 in the above expression

n = 10

= (7 × 10) + 20

= 70 + 20

= 90

For 100th term put n=100 in the above expression

n = 100

= (7 × 100) + 20

= 700 + 20

= 720

(v)

Here, in the above table the algebraic expression is given as (n²+1)

So for 5th term we can calculate by putting the value of n as 5 in the above expression

Then,

n = 5

= (52) + 1

= 25+ 1

= 26

For 10th term n=10 in the above expression

Where n = 10

= (102) + 1

= 100 + 1

= 101

So, the table is completed below.

S. No.	Expression	Terms
S. No.	Expression	1st	2nd	3rd	4th	5th	…	10th	…	100th	…
(i)	2n – 1	1	3	5	7	9	–	19	–	199	–
(ii)	3n + 2	5	8	11	14	17	–	32	–	302	–
(iii)	4n + 1	5	9	13	17	21	–	41	–	401	–
(iv)	7n + 20	27	34	41	48	55	–	90	–	720	–
(v)	n2 + 1	2	5	10	17	26	–	101	–	10001	–

Chapter-12 Algebraic Expressions

2.Identify the terms and their factors in the following expressions.Show the terms and factors by tree diagrams.(a) x – 3(b) 1 + x + x2(c) y – y3(d) 5xy2 + 7x2y(e) – ab + 2b2 – 3a2

3. Identify terms and factors in the expressions given below.(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼(h) 0.1 p2 + 0.2 q2

4. Identify the numerical coefficients of terms (other than constants) in the following expressions.(i) 5 – 3t2 (ii) 1 + t + t2 + t3 (iii) x + 2xy + 3y (iv) 100m + 1000n (v) – p2q2 + 7pq (vi) 1.2 a + 0.8 b (vii) 3.14 r2 (viii) 2 (l + b)(ix) 0.1 y + 0.01 y2

5.Identify terms which contain x and give the coefficient of x.(i) y2x + y (ii) 13y2 – 8yx (iii) x + y + 2(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25(vii) 7x + xy2

6. Identify terms which contain y2 and give the coefficient of y2.(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2

7. Classify into monomials, binomials and trinomials.(i) 4y – 7z(ii) y2(iii) x + y – xy(iv) 100(v) ab – a – b(vi) 5 – 3t(vii) 4p2q – 4pq2(viii) 7mn(ix) z2 – 3z + 8(x) a2 + b2(xi) z2 + z(xii) 1 + x + x2

8. State whether a given pair of terms is of like or unlike terms.1, 100

9. –7x, (5/2)x

10. – 29x, – 29y

11. 14xy, 42yx

12. 4m2p, 4mp2

13. 12xz, 12x2z2

14. Identify like terms in the following.– xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x

15. 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,13p2q, qp2, 701p2

16. Simplify combining like terms.21b – 32 + 7b – 20b

17. – z2 + 13z2 – 5z + 7z3 – 15z

18. p – (p – q) – q – (q – p)

19. 3a – 2b – ab – (a – b + ab) + 3ab + b – a

20. 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2

21. (3y2 + 5y – 4) – (8y – y2 – 4)

22. Add:3mn, – 5mn, 8mn, – 4mn

23. t – 8tz, 3tz – z, z – t

24. – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3

25. a + b – 3, b – a + 3, a – b + 3

26. 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy

27. 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5

28. 4x2y, – 3xy2, –5xy2, 5x2y

29. 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2

30. ab – 4a, 4b – ab, 4a – 4b

31. x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2

32. Subtract:–5y2 from y2

33. 6xy from –12xy

34. (a – b) from (a + b)

35. a (b – 5) from b (5 – a)

36. –m2 + 5mn from 4m2 – 3mn + 8

37. – x2 + 10x – 5 from 5x – 10

38. 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2

39. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

40. What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?

41.What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?

42. What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20?

43. From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.

44. From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and–x2 + 2x + 5.

45. If m = 2, find the value of:m – 2

46. 3m – 5

47. 9 – 5m

48. 3m2 – 2m – 7

49. (5m/2) – 4

50. If p = – 2, find the value of:4p + 7

51. – 3p2 + 4p + 7

52. – 2p3 – 3p2 + 4p + 7

53. Find the value of the following expressions when x = –1:2x – 7

54. – x + 2

55. x2 + 2x + 1

56. 2x2 – x – 2

57. If a = 2, b = – 2, find the value of:a2 + b2

58. a2 + ab + b2

59. a2 – b2

60. When a = 0, b = – 1, find the value of the given expressions:2a + 2b

61. 2a2 + b2 + 1

62. 2a2b + 2ab2 + ab

63. a2 + ab + 2

64. Simplify the expressions and find the value if x is equal to 2x + 7 + 4 (x – 5)

65. 3 (x + 2) + 5x – 7

66. 6x + 5 (x – 2)

67. 4(2x – 1) + 3x + 11

68. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.3x – 5 – x + 9

69. 2 – 8x + 4x + 4

70. 3a + 5 – 8a + 1

71. 10 – 3b – 4 – 5b

72. 2a – 2b – 4 – 5 + a

73. If z = 10, find the value of z3 – 3(z – 10).

74. If p = – 10, find the value of p2 – 2p – 100

75. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?

76. Simplify the expression and find its value when a = 5 and b = – 3.2(a2 + ab) + 3 – ab

2.Identify the terms and their factors in the following expressions.
Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x2
(c) y – y3
(d) 5xy2 + 7x2y
(e) – ab + 2b2 – 3a2

3. Identify terms and factors in the expressions given below.
(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2
(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼
(h) 0.1 p2 + 0.2 q2

4. Identify the numerical coefficients of terms (other than constants) in the following expressions.
(i) 5 – 3t2 (ii) 1 + t + t2 + t3 (iii) x + 2xy + 3y (iv) 100m + 1000n (v) – p2q2 + 7pq (vi) 1.2 a + 0.8 b (vii) 3.14 r2 (viii) 2 (l + b)
(ix) 0.1 y + 0.01 y2

5.Identify terms which contain x and give the coefficient of x.
(i) y2x + y (ii) 13y2 – 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25
(vii) 7x + xy2

6. Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2

7. Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y2
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p2q – 4pq2
(viii) 7mn
(ix) z2 – 3z + 8
(x) a2 + b2
(xi) z2 + z
(xii) 1 + x + x2

8. State whether a given pair of terms is of like or unlike terms.
1, 100

14. Identify like terms in the following.
– xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x

15. 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,
13p2q, qp2, 701p2

16. Simplify combining like terms.
21b – 32 + 7b – 20b

22. Add:
3mn, – 5mn, 8mn, – 4mn

32. Subtract:
–5y2 from y2

44. From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and
–x2 + 2x + 5.

45. If m = 2, find the value of:
m – 2

50. If p = – 2, find the value of:
4p + 7

53. Find the value of the following expressions when x = –1:
2x – 7

57. If a = 2, b = – 2, find the value of:
a2 + b2

60. When a = 0, b = – 1, find the value of the given expressions:
2a + 2b

64. Simplify the expressions and find the value if x is equal to 2
x + 7 + 4 (x – 5)

68. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.
3x – 5 – x + 9

76. Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab