Algebraic Expressions
Class 7th Mathematics CBSE Solution
- Get the algebraic expressions in the following cases using variables, constants…
- Identify the terms and their factors in the following expressions. Show the…
- Identify terms and factors in the expression given below: (a) -4x + 5 (b) -4x…
- Identify the numerical coefficients of terms (other than constants) in the…
- Identify terms which contain x and give the coefficient of x. (i) y^2 x + y…
- Identify terms which contain y^2 and give the coefficient of y^2 . (i) 8 - xy^2…
- Classify into monomials, binomials and trinomials. (i) 4y - 7z (ii) y^2 (iii) x…
- State whether a given pair of terms is of like or unlike terms. (i) 1, 100 (ii)…
- Identify like terms in the following: (a) - xy^2 , - 4yx^2 , 8x^2 , 2xy^2 , 7y,…
- Simplify combining like terms: (i) 21b - 32 +7b + 20b (ii) -z^2 + 13z^2 - 5z +…
- Add: (i) 3mn, -5mn, 8mn -4mn (ii) t - 8tz, 3tz - z, z - t (iii) -7mn + 5, 12mn…
- Subtract: (i) -5y^2 - from y^2 (ii) 6xyfrom -12xy (iii) (a - b) from (a + b)…
- What should be added to x2+ xy + y^2 to obtain 2x2+ 3xy ?
- What should be subtracted from 2a+8b+10 to get -3a+7b+16?
- What should be taken away form 3x^2 - 4y^2 + 5xy + 20 to obtain -x^2 - y^2 +…
- From the sum of 3x - y + 11 and -y - 11, subtract 3x - y - 11.
- From the sum of 4 + 3x and 5 - 4x + 2x^2 , subtract the sum of 3x^2 5x and…
- If m = 2, find the value of : (i) m - 2 (ii) 3m - 5 (iii) 9 - 5m (iv) 3m^2 - 2m…
- If p=-2 find the value of: (i) 4p + 7 (ii) -3p^2 + 4p + 7 (iii) -2p^3 - 3p^2 +…
- Find the values of the following expressions when x=-1: (i) 2x - 7 (ii) -x + 2…
- If a = 2, b = - 2, find the value of : (i) a^2 + b^2 (ii) a^2 + ab + b^2 (iii)…
- When a=0, b=-1 find the value of the given expressions: (i) 2a + 2b (ii) 2a^2 +…
- Simplify the expressions and find the value if x is equal to 2 (i) x + 7 + 4(x…
- Simplify these expressions and find their values, if x=3,a=-1,b=-2 (i) 3x - 5 -…
- If z = 10, find the value of z^3 - 3(z - 10)
- If p = -10, find the value of p^2 - 2p - 100
- What should be the value of a if the value of 2x^2 + x - a equals to 5, when x…
- Simplify the expression and find its value when a=5 and b = -3, then 2(a^2 +…
Exercise 12.1
Question 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 number y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Answer:
(i) y - z
(ii) sum of numbers x and y = x + y
One half of the sum of numbers x and y = 
(iii) z × z = z2
(iv) Products of two numbers p and q = p × q = pq
So one - fourth of the above quantity is 
(v) Square of no x : x22
Square of no y : y2
Addition of squares of x and y = x2 + y2
(vi) Product of m and n = m × n = mn
Three times of product of m and n = 3 × mn = 3mn
Five added to three times of product of m and n = 5 + 3mn
(vii) Product of number y and z = xy
Product of number y and z substracted from 10 = 10 - xy
(viii) Sum of numbers a and b = a + b
Products of numbers = ab
Subtraction of sum from product = ab - (a + b) or ab - a - b
Question 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
Answer:
(a) 
(b) 
(c) 
(d) 
(e) 
Question 3.
Identify terms and factors in the expression given below:
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y2
(d) xy + 2x2y2
(e) pq + q
(f) 1.2ab - 2.4b + 3.6a
(g) 
(h) 0.1p2 + 0.2q2
Answer:
Question 4.
Identify the numerical coefficients of terms (other than constants) in the following expression:
(i) 5 - 3t2
(ii) 1 + t + t2 + t3
(iii) x + 2xy + 3y
(iv) 100m + 100n
(v) -p2q2 + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14r2
(viii) 2(l + b)
(ix) 0.1y + 0.01y2
Answer:
Question 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
Answer:
Question 6.
Identify terms which contain 
and give the coefficient of y2.
(i) 8 - xy2
(ii) 5y2 + 7x
(iii) 2x2y - 15xy2 + 7y2
Answer:
Question 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
Answer:
(i) 4y – 7z
As the expression contains two terms expression is Binomial.
(ii) y2
As the expression contains one term expression is Monomial
(iii) x + y – xy
As the expression contains three terms expression is Trinomial
(iv) 100 is a constant polynomial.
As the expression contains one term expression is Monomial
(v) ab – a – b
As the expression contains three terms expression is Trinomial
(vi) 5 – 3t
As the expression contains two terms expression is Binomial
(vii) 4p2q – 4pq2.
As the expression contains two terms expression is Binomial
(viii) 7mn
As the expression contains one term expression is Monomial
(ix) z2 -3z + 8
As the expression contains three terms expression is Trinomial.
(x) a2 + b2
As the expression contains two terms expression is Binomial
(xi) z2 + z
As the expression contains two terms expression is Binomial
(xii) 1 + x + x2
As the expression contains three terms expression is Trinomial
Question 8.
State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) -7x, 
(iii) -29x, -29y
(iv) 14xy, 42yx
(v) 4m2p, 4mp2
(vi) 12xz, 12x2z2
Answer:
(i) 1, 100
As it contains only single term it is simply a like term
(ii) -7x, 
As both term have same algebraic factor as x, the terms are like.
(iii) -29x, -29y
As both term don’t have same algebraic factor (As -29x has x and -29 has y), the terms are unlike.
(iv) 14xy, 42yx
As both term have same algebraic factors as x and y, the terms are like.
(v) 4m2p, 4mp2
As term 4m2p has factors m, m and p but term 4mp2 has factors m, p and p. so both terms are unlike
(vi) 12xz, 12x2z2
As term 12xz has factors x and z but term 12x2y2 has factors x, x, y and y. so both terms are unlike.
Question 9.
Identify like terms in the following:
(a) – xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x
(b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2
Answer:
(a)
So, From the above table we conclude that sets of like terms are
1. -xy2, 2xy2 : As both have common variable factors as x, y and y
2. -4yx2, 20x2y : As both have common variable factors as x, x and y
3. 8x2 , -11x2, -6x2 : As both have common variable factors as x and x
4. -11yx, 2xy : As both have common variable factors as x and y
5. -100x, 3x : As both have common variable factor as x
6. 7y, y : As both have common variable factor as y
(b)
So, From the above table we conclude that sets of like terms are
1. 10pq, –7qp, 78qp : As both have common variable factors as p and q
2. 7p, 2405p: As both have common variable factor as p
3. 8q, – 100q : As both have common variable factor as q
4. –p2q2, 12q2p2 : As both have common variable factors as p, q, q and q
5. –23, 41 : As both terms are constant and don’t have any variable factor .
6. –5p2, 701p2 :As both have common variable factors as p and p.
7. 13p2q, qp2 : As both have common variable factors as p, p and q.
Exercise 12.2
Question 1.Simplify combining like terms:
(i) 21b - 32 +7b + 20b
(ii) -z2 + 13z2 - 5z + 7z3-15z
(iii) p – (p – q) - q – (q - p)
(iv) 3a - 2b - ab – (a - b + ab) + 3ab + b - a
(v) 5x2y - 5x2 + 3xy2 - 3x2 + x2 - y2 + 8xy2 - 3y2
(vi) (3y2 + 5y – 4) – (8y - y2 – 4)
Answer:
Like terms are terms with the same variables and exponents.
(i) 21b + 7b - 20b - 32
= (21b + 7b - 20b )- 32
=8b - 32
(ii) 7z3 + 13z2 - z2 - 15z - 5z
= 7z3 + (13z2 - z2 )- (15z + 5z)=7z3 + 12z2 - 20z
(iii) p – (p – q) - q – (q - p)
= p - p + p + q - q -q
= p - q
(iv) 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)
= ( 3a - 2a) + ( -2b + 2b) + (-2ab + 3ab)
= a + 0 + ab
= a + ab
(v)5x2y + 3x2y + 8xy2 - 5x2 + x2 - 3y2 - y2 - 3y2
= (5x2y + 3x2y) + 8xy2 +( - 5x2 + x2 ) + ( - 3y2 - y2 - 3y2)= 8x2y + 8xy2- 4x2 + ( - 4y2 - 3y2)
= 8x2y + 8xy2- 4x2 - 7y2
(vi)(3y2 + 5y – 4) – (8y - y2 – 4)
= 3y2 + 5y – 4– 8y + y2 + 4= (3y2 + y2) +( 5y - 8y) - 4 + 4
= 4y2 - 3y
Question 2.
Add:
(i) 3mn, -5mn, 8mn -4mn
(ii) t - 8tz, 3tz – z, z - t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn -3
(iv) a + b -3, b - a + 3, a - b + 3
(v) 14x + 10y - 12xy – 13, 18 - 7x - 10y + 8xy,4xy
(vi) 5m - 7n, 3n - 4m + 2, 2m - 3mn -5
(vii) 4x2y, -3xy2, -5xy2, 5x2y
(viii) 3p2q2 - 4pq + 5, -10p2q2, 15 + 9pq + 7p2q2
(ix) ab - 4a, 4b – ab, 4a - 4b
(x) x2 - y2 – 1, y2 - 1 - x2, 1 - x2 - y2
Answer:
(i) 3mn + (-5mn) + 8mn + (-4mn)
= 3mn - 5mn + 8mn - 4mn
= 11 mn - 9mn
= 2mn
(ii) t - 8tz, 3tz – z, z - t
= t - 8tz + 3tz - z + z - t
=t - t - z + z - 8tz + 3tz
= -5tz
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn -3
= (-7mn + 5) + (12mn + 2) + (9mn - 8)+(- 2mn - 3)
= -7mn + 12mn + 9mn - 2mn + 5 + 2 - 8 - 3
= - 9mn + 21 mn + 7 - 11
= 12mn - 4
(iv) 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 + b + 3
(v) 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
= 7x + 5
(vi) ( 5m - 7n) + (3n - 4m + 2) + (2m - 3mn -5 )
= 5m - 7n + 3n - 4m + 2 + 2m - 3mn -5
= 3m - 4n - 3mn - 3
(vii) 4x2y + (-3xy2) + (-5xy2) + 5x2y
=4x2y - 3xy2 - 5xy2 +5x2y
=4x2y + 5x2y - 3xy2 - 5xy2
= 9x2y - 8xy2
(viii) (3p2q2 - 4pq + 5) + (-10p2q2) + (15 + 9pq + 7p2q2 )
= 3p2q2 - 4pq + 5 - 10p2q2 + 15 + 9pq + 7p2q2
= 3p2q2 - 10p2 q2 + 7p2q2 - 4pq + 9pq + 5 + 15
= 5pq + 20
(ix) (ab - 4a) + (4b - ab) + (4a - 4b)
= ab - 4a + 4b - ab + 4a - 4b
=ab - ab - 4a + 4a + 4b - 4b
= 0
(x) (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
= -x2 - y2 - 1
Question 3.
Subtract:
(i) -5y2– from y2
(ii) 6xyfrom -12xy
(iii) (a - b) from (a + b)
(iv) a(b - 5) from b(5 - a)
(v) -m2 + 5mn from 4m2 – 3mn + 8
(vi) -x2 + 10x – 5 from 5x -10 x2
(vii) 5a2 - 7ab + 5b2 from 3ab – 2a2 -2b2
(viii) 4pq - 5q2 - 3p2 from 5p2 + 3q2 - pq
Answer:
(i) y2- (-5y2)
= y2 + 5y2
= 6y2
(ii) 6xy from -12xy
-12xy - 6xy = -18xy
(iii) (a - b) from (a + b)
(a + b) - (a - b)
= a + b - a + b
= a - a + b + b
=2b
(iv) a(b - 5) from b(5 - a)
b(5 - a) - a(b - 5)
= 5b - ab - ab + 5a
=5a + 5b - ab - ab
= 5a + 5b - 2ab
(v) -m2 + 5mn from 4m2 – 3mn + 8
4m2 - 3mn + 8 - (-m2 + 5mn )
= 4m2 - 3mn + 8 + m2 - 5mn
= 4m2 + m2 - 8mn + 8
= 5m2 - 8mn + 8
(vi) -x2 + 10x – 5 from 5x -10 x2
5x - 10x2 - (-x2 + 10x - 5 )
= 5x - 10x2 + x2 - 10x + 5
= x2 - 10x2 - 10x + 5x + 5
= - 9x2 - 5x + 5
(vii) 5a2 - 7ab + 5b2 from 3ab – 2a2 -2b2
3ab - 2a2 - 2b2 - (5a2 - 7ab + 5b2)
= 3ab - 2a2 - 2b2 - 5a2 + 7ab - 5b2
= -2a2 - 5a2 + 3ab + 7ab - 2b2 - 5b2
= -7a2 + 10ab - 7b2
(viii) 4pq - 5q2 - 3p2 from 5p2 + 3q2 - pq
5p2 + 3q2 - pq - (4pq - 5q2 - 3p2)
= 5p2 + 3q2 - pq - 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 - pq - 4pq
= 8p2 + 8q2 - 5pq
Question 4.
What should be added to x2 + xy + y2 to obtain 2x2 + 3xy ?
Answer:
To Find: Expression which is to be added to x2 + xy + y2 to obtain 2x2 + 3xy
We need to subtract x2 + xy + y2 from 2x2 + 3xy to obtain the required fraction.
Therefore, the expression = 2x2 + 3xy - (x2 + xy + y2)
Required Expression = 2x2 + 3xy - x2 - xy - y2
Required Expression = x2 + 2xy - y2
Hence, x2 + 2xy - y2 is to be added to x2 + xy + y2 to obtain 2x2 + 3xy.
Question 5.
What should be subtracted from 2a+8b+10 to get -3a+7b+16?
Answer:
Let "k" should be subtracted
2a + 8b + 10 - k = -3a + 7b + 16
Then,
k = 2a + 8b + 10 - ( -3a + 7b + 16)
k = 2a + 8b + 10 + 3a - 7b - 16
k = 5a + b - 6
Question 6.
What should be taken away form 3x2 - 4y2 + 5xy + 20 to obtain -x2 - y2 + 6xy + 20?
Answer:
Let k should be taken away from 3x2 - 4y2 + 5xy + 20
3x2 - 4y2 + 5xy + 20 - k = -x2 - y2 + 6xy + 20
k = 3x2 - 4y2 + 5xy + 20 - (-x2 - y2 + 6xy + 20)
k = 3x2 - 4y2 + 5xy + 20 + x2 + y2 - 6xy - 20
k = 4x2 - 3y2 - xy
Hence value of k is 4x2 - 3y2 - xy.
Question 7.
From the sum of 3x - y + 11 and -y – 11, subtract 3x - y - 11.
Answer:
The algebraic equation for above problem will be as
[(3x - y + 11)+ (-y - 11)] - ( 3x - y - 11)]
= 3x - y + 11 - y - 11 - 3x + y + 11
= 3x - 3x - y - y + y +11 - 11 + 11
= -y + 11
Hence the value is 11-y.
Question 8.
From the sum of 4 + 3x and 5 - 4x + 2x2, subtract the sum of 3x2 – 5x and -x2 + 5.
Answer:
The equation for the problem is
[(4 + 3x) + (5 - 4x + 2x2 ) - [ ( 3x2 - 5x) + ( -x2 +5)]
= [ 4 + 3x + 5 - 4x + 2x2 ] - [3x2 - 5x - x2 + 5]
= 4 + 3x + 5 - 4x + 2x2 - 3x2 + 5x + x2 - 5
= 2x2 - 3x2 + x2 + 3x - 4x + 5x + 4 + 5 - 5
= 4x + 4
Exercise 12.3
Question 1.If m = 2, find the value of :
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m2 - 2m - 7
(v) 
Answer:
(i) m - 2
Put m = 2
= 2 -2
=0
(ii) 3m - 5
Put m = 2
=3(2) - 5
=6 - 5
= 1
(iii) 9 - 5m
Put m = 2
= 9 - 5(2)
= 9 - 10
= -1
(iv) 3m2 - 2m - 7
Put m = 2
= 3(2)2 - 2(2) - 7
=3(4) - 4 - 7
=12 - 11
=1
(v) 
Put m = 2
= 5 - 4
= 1
Question 2.
If p=-2 find the value of:
(i) 4p + 7
(ii) -3p2 + 4p + 7
(iii) -2p3 - 3p2 + 4p +7
Answer:
(i) 4p + 7
Put p = -2
=4(-2) + 7
= -8 + 7
= -1
(ii) -3p2 + 4p + 7
Put p = -2
= -3(-2)2 + 4(-2) + 7
= -3(4) - 8 + 7
= -12 - 8 + 7
= -20 + 7
= -13
(iii) -2p3 - 3p2 + 4p + 7
Put p = -2
=-2(-2)3 - 3(-2)2 + 4(-2) + 7
= -2(-8) -3(4) - 8 + 7
= 16 -12 - 8 + 7
= 4 - 1= 3
Question 3.
Find the values of the following expressions when x=-1:
(i) 2x - 7
(ii) -x + 2
(iii) x2 + 2x +1
(iv) 2x2- x -2
Answer:
(i) 2x - 7
Put x = -1
= 2(-1) - 7
= -2 - 7
= -9
(ii) -x + 2
Put x = -1
= -(-1) + 2
= 1 + 2
= 3
(iii) x2 + 2x + 1
Put x = -1
=(-1)2 + 2(-1) + 1
=1 - 2 + 1
= 2 - 2
=0
(iv) 2x2 - x - 2
Put x = -1
=2(-1)2 - (-1) -2
= 2(1) + 1 - 3
= 2 + 1 - 3
=3 - 3
= 0
Question 4.
If a = 2, b = - 2, find the value of :
(i) a2 + b2
(ii) a2 + ab + b2
(iii) a2 - b2
Answer:
(i) a2 + b2
Put a = 2 and b = -2
=(2)2 + (-2)2
=4 + 4
=8
(ii) a2 + ab + b2
Put a = 2 and b = -2
= (2)2 + 2(-2) + (-2)2
= 4 - 4 + 4
= 4
(iii)a2 - b2
we know that,a2 - b2 = (a + b)(a - b)
Put a = 2 and b = -2
a2 - b2 = (2 + (-2)) (2 - (-2))a2 - b2 = (2 - 2)(2 + 2)
a2 - b2 = 0 . 4 = 0
a2 - b2 = 0
Question 5.
When a=0, b=-1 find the value of the given expressions:
(i) 2a + 2b
(ii) 2a2 + b2+1
(iii) 2a2b + 2ab2 + ab
(iv) a2 + ab +2
Answer:
(i) 2a + 2b
Put a = 0 and b = -1
= 2(0) + 2(-1)
= 0 - 2
= -2
(ii) 2a2 + b2 +1
Put a = 0 and b = -1
= 2(0)2 + (-1)2 + 1
=2(0) + 1 + 1
=0 + 1 + 1
=2
(iii)2a2b + 2ab2 + ab
Put a = 0 and b = -1
= 2(0)2(-1) + 2(0)(-1)2 + 0(-1)
=2(0)(-1) + 2(0)(1) + 0
=0 + 0 + 0
=0
(iv) a2 + ab + 2
Put a = 0 and b = -1
=(0)2 + 0(-1) + 2
= 0 + 0 + 2
=2
Question 6.
Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4(x - 5)
(ii) 3(x + 2) + 5x - 7
(iii) 6x + 5(x - 2)
(iv) 4(2x - 1) + 3x +11
Answer:
(i) x + 7 + 4(x - 5)
Opening the brackets we get,= x + 7 + 4x - 20
= x + 4x + 7 - 20
= 5x - 13
Put x = 2
= 5(2) - 13
= 10 - 13
= -3
(ii)3(x+2) + 5x - 7
Opening the brackets we get,= 3x + 6 + 5x - 7
= 3x + 5x + 6 - 7
= 8x - 1
Put x = 2
= 8(2) - 1
= 16 - 1
= 15
(iii) 6x + 5(x - 2)
Opening the brackets we get,= 6x + 5x - 10
= 11x - 10
Put x = 2
= 11(2) - 10
= 22 - 10
= 12
(iv) 4(2x - 1) + 3x + 11
Opening the brackets we get,= 8x - 4 + 3x + 11
= 8x + 3x -4 + 11
= 11x + 7
Put x = 2
=11(2) + 7
= 22 + 7
= 29
Question 7.
Simplify these expressions and find their values, if x=3,a=-1,b=-2
(i) 3x – 5 – x + 9
(ii) 2 - 8x + 4x + 4
(iii) 3a + 5 - 8a + 1
(iv) 10 - 3b- 4 - 5b
(v) 2a - 2b – 4 – 5 + a
Answer:
(i) 3x - 5 - x + 9
=3x - x - 5 + 9
= 2x - 5 + 9
= 2x + 4
Put x = 3
= 2(3) + 4
= 6 + 4
=10 Ans.
(ii)2 - 8x + 4x + 4
= -8x + 4x + 2 + 4
= -4x + 2 + 4
= -4x + 6
Put x = 3
= -4(3) + 6
=-12+6
= -6 Ans.
(iii) 3a + 5 - 8a + 1
On rearranging the terms,
= 3a - 8a + 5 + 1
= -5a + 5 + 1
= -5a + 6
Put a = -1
= -5(-1) + 6
= 5 + 6
= 11 Ans.
(iv) 10 - 3b - 4 - 5b
= 10 - 4 - 3b - 5b
= 6 - 3b - 5b
= 6 - 8b
Put b = -2
= 6 - 8(-2)
= 6 + 16
= 22 Ans.
(v) 2a - 2b - 4 - 5 + a
On rearranging the terms,
= 2a + a - 2b - 4 - 5
= 3a - 2b - 4 - 5
= 3a - 2b - 9
Put a = -1 & b = -2
= 3(-1) - 2(-2) - 9
= -3 + 4 - 9
= -8 Ans.
Question 8.
If z = 10, find the value of z3 – 3(z – 10)
Answer:
z3 – 3(z – 10)
As z = 10
z3 - 3(z - 10)
= (10)3 - 3(10 - 10)
= 1000 - 3(0)
= 1000 - 0
= 1000
Question 9.
If p = -10, find the value of p2 – 2p – 100
Answer:
As p = -10
p2 - 2p - 100
= (-10)2 - 2(-10) - 100
= (-10)(-10) + 20 - 100= 100 + 20 - 100
= 20
Question 10.
What should be the value of a if the value of 2x2 + x - a equals to 5, when x = 0?
Answer:
As x = 0
And
2x2 + x - a = 5
2(0)2 + 0 - a = 5
2(0) - a = 5
0 - a = 5
-a = 5
a = -5
Question 11.
Simplify the expression and find its value when a=5 and b = -3, then 2(a2 + ab) +3 - ab
Answer:
To Simplify: 2(a2 + ab) + 3 - ab
2(a2 + ab) + 3 - abOpening the brackets we get,
= 2a2 + 2ab + 3 - ab
Taking like terms on one side we get,= 2a2 + 2ab - ab + 3
= 2a2+ ab + 3
Put a = 5 and b = -3
= 2(5)2 + 5(-3) + 3
= 2(25) - 15 + 3
= 50 - 12
= 38
Exercise 12.4
Question 1.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.
(a) 
(b) 
(c) 
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 
Answer:
(a) For digit
Expression : 5n + 1
Where n = no of digits
For 5 digits
No of segments required = 5(5) + 1 = 25 + 1 = 26
For 10 digits
No of segments required = 5(10) + 1 = 50 + 1 = 51
For 100 digits
No of segments required = 5(100) + 1 = 500 + 1 = 501
(b) For digit
Expression : 3n + 1
Where n = no of digits
For 5 digits
No of segments required = 3(5) + 1 = 15 + 1 = 16
For 10 digits
No of segments required = 3(10) + 1 = 30 + 1 = 31
For 100 digits
No of segments required = 3(100) + 1 = 300 + 1 = 301
(c) For digit
Expression : 5n + 2
Where n = no of digits
For 5 digits
No of segments required = 5(5) + 2 = 25 + 2 = 27
For 10 digits
No of segments required = 5(10) + 2 = 50 + 2 = 52
For 100 digits
No of segments required = 5(100) + 2 = 500 + 2 = 502
Question 2.
Use the given algebraic expression to complete the table of number patterns.
Answer:
(i)Expression = 2n - 1
100th term (i.e. n = 100)
= 2(100) - 1
= 200 - 1
= 199
(ii)Expression = 3n + 2
5th term( i.e. n = 5)
= 3(5) + 2
= 15 + 2
= 17
10th terms(i.e.. n =10)
=3(10) + 2
=30 + 2
= 32
100th term( i.e. n = 100)
=3(100) + 2
= 300 + 2
= 302
(iii)Expression = 4n + 1
5th term( i.e. n = 5)
= 4(5) + 1
= 20 + 1
= 21
10th terms(i.e.. n =10)
=4(10) + 1
=40 + 1
= 41
100th term( i.e. n = 100)
=4(100) + 1
= 400 + 1
= 401
(iv)Expression = 7n + 20
5th term( i.e. n = 5)
= 7(5) + 20
= 35 + 20
= 55
10th terms(i.e.. n =10)
=7(10) + 20
=70 + 20
= 90
100th term( i.e. n = 100)
=7(100) + 20
= 700 + 20
= 720
(v)Expression = n2 + 1
5th term( i.e. n = 5)
= (5)2 + 1
= 25 + 1
= 26
10th terms(i.e.. n =10)
=(10)2 + 1
=100 + 1
= 101
So the Table is
