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Exercise 12.1
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 m and n.
(vii) A product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Answer
(i) y - z
(ii) (x + y)/2
(iii) z2
(iv) pq/4
(v) x2 + y2
(vi) 3mn + 5
(vii) 10 - yz
(viii) ab - (a + b)
2. (i) Identify the terms and their factors in the following expressions, show the terms and factors by tree diagram:
(a) x – 3
(b) 1 + x + x2
(c) y – y3
(d) 5xy2 + 7x2y
(e) -ab + 2b2 – 3a2
Answer
(ii) Identify the terms and factors in the expressions given below:
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y2
(d) xy + 2x2y2
(e) pq + q
(f) 1.2ab - 2.4b + 3.6a
(h) 0.1 p2 + 0.2q2
Answer
(a) -4x+ 5
Terms: -4x,5
Factors: -4,x ; 5
(b) -4x + 5y
Terms: -4x, 5y
Factors: -4,x ; 5,y
(c) 5y + 3y2
Terms: 5y,3y2
Factors: 5, y ; 3,y,y
(d) xy+2x2y2
Terms: xy,2x2y2
Factors: x,y ; 2x,x,y,y
(e) pq+q
Terms: pq,q
Factors: p,q ; q
(f) 1.2ab-2.4b+3.6a
Terms: 1,2ab.-2.4b,3 6a
Factors: 1.2.a.b ; -2.4,6 ; 3.6,a
(h) 0.1p2+0.2q2
Terms: 0.1 p2,0.2q2
Factors: 0. 1,p,p, ; 0.2, q,q
3. Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 - 3t2
(ii) 1 + t + t2 + t2
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) -p2q2 + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14 r2
(viii) 2(l+b)
(ix) 0.1y + 0.01y2
Answer
| S.No. | Expression | Terms | Numerical Coefficient |
| (i) | 5-3t2 | -3t1 | -3 |
| (ii) | 1+t+t2+t3 | t | 1 |
| t2 | 1 | ||
| t3 | 1 | ||
| (iii) | x + 2xy + 3y | x | 1 |
| 2xy | 2 | ||
| 3y | 3 | ||
| (iv) | 100m+1000n | 100 m | 100 |
| 1000 n | 1000 | ||
| (v) | -p2q2+7 pq | -p2q2 | -1 |
| 7 pq | 7 | ||
| (vi) | 1.2a+0.8b | 1.2 a | 1.2 |
| 0.8b | 0.8 | ||
| (vii) | 3.14 r2 | 3.14 r2 | 3.14 |
| (viii) | 2 (l + b) = 2l+ 2b | 2l | 2 |
| 2b | 2 | ||
| (ix) | 0.1y + 0.01y2 | 0.1y | 0.1 |
| 0.01y2 | 0.01 |
4. (a) 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 + x25
(vii) 7x + xy2
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 - xy2
(ii) 5y2 + 7x
(iii) 2x2y - 15xy2 + 7y2
Answer
| S.No. | Expression | Term with factor x | Coefficient of x |
| (i) | y2x + y | y2 x | y |
| (ii) | 13y2 -8yx | -8 yx | -8 y |
| (iii) | x + y + 2 | x | 1 |
| (iv) | 5 + z + zx | zx | z |
| (v) | 1 + x + xy | x | 1 |
| xy | 1 | ||
| (vi) | 12xy2 +25 | 12 xy2 | 12y2 |
| (vii) | 7x+xy2 | xy2 | y2 |
| 7x | 7 |
| S. No. | Expression | Term containing y2 | Coefficient of y2 |
| (i) | 8-xy2 | -xy2 | -x |
| (ii) | 5y+7x | 5y2 | 5 |
| (iii) | 2x2y-15xy2 + 7y2 | -15xy2 | -15x |
| 7y2 | 7 |
5. Classify into monomials, binomials and trinomials:
(i) 4y - 7x
(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
| S.No. | Expression | Type of Polynomial |
| (i) | 4y-7z | Binomial |
| (ii) | y2 | Monomial |
| (iii) | x+y-xy | Trinomial |
| (iv) | 100 | Monomial |
| (v) | ab-a-b | Trinomial |
| (vi) | 5-3t | Binomial |
| (vii) | 4p2q-4pq2 | Binomial |
| (viii) | 7mn | Monomial |
| (ix) | z2-3z + 8 | Trinomial |
| (x) | a2 + b2 | Binomial |
| (xi) | z2 +z | Binomial |
| (xii) | 1 + x + x2 | Trinomial |
6. State whether a given pair of terms is of like or unlike terms:
(i) 1,100
(ii) 
(iii) -29x, -29y
(iv) 14xy, 42 yx
(v) 4m2p, 4mp2
(vi) 12xz, 12x2 z2
Answer
| S.No. | Pair of terms | Like / Unlike terms |
| (i) | 1, 100 | Like terms |
| (ii) | | Like terms |
| (iii) | -29x,-29y | Unlike terms |
| (iv) | 14xy,42yx | Like terms |
| (v) | 4m2p,4mp2 | Unlike terms |
| (vi) | 12xz,12x2z2 | Unlike terms |
7. Identify like terms in the following:
(a) -xy2, -4yx2, 8x2, 2xy2, 7y, -11x2 - 100x, - 11yx, 20x2y, -6x2, y, 2xy, 3x
Answer
(i) -xy2,2 xy2
(ii) -4yx2 , 20x2y
(iii) 8x2,-11x2,-6x2
(iv) 7y, y
(v) -100x, 3x
(vi) -11yx, 2xy
(b) 10pq, 7p, 8q, -p2q2, -7qp, -100q, -23, 12q2p2, -5p2, 41,2405 p, 78qp, 13p2q, qp2, 701p2
Answer
(i) 10 pq - 7 pq,78 pq
(ii) 7p, 2405 p
(iii) 8q,- 100q
(iv) -p2q2, 12p2q2
(v) -12,41
(vi) -5p2,701p2
(vii) 13 p2q,qp2
Exercise 12.2
1. Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
Answer
When term have the same algebraic factors, they are like terms.
Then,
= (21b + 7b – 20b) – 32
= b (21 + 7 – 20) – 32
= b (28 - 20) – 32
= b (8) - 32
= 8b - 32
(ii) – z2 + 13z2 – 5z + 7z3 – 15z
Answer
When term have the same algebraic factors, they are like terms.
Then,
= 7z3 + (-z2 + 13z2) + (-5z – 15z)
= 7z3 + z2 (-1 + 13) + z (-5 - 15)
= 7z3 + z2 (12) + z (-20) + 7z3
= 7z3 + 12z2 – 20z + 7z3
(iii) p – (p – q) – q – (q – p)
Answer
When term have the same algebraic factors, they are like terms.
Then,
= p – p + q – q – q + p
= p – q
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
Answer
When term have the same algebraic factors, they are like terms.
Then,
= 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
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
Answer
When term have the same algebraic factors, they are like terms.
Then,
= 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
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
Answer
When term have the same algebraic factors, they are like terms.
Then,
= 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.
2. Add:
(i) 3mn, – 5mn, 8mn, – 4mn
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= mn (3 – 5 + 8 - 4)
= mn (11 - 9)
= mn (2)
= 2mn
(ii) t – 8tz, 3tz – z, z – t
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= - 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
(iv) a + b – 3, b – a + 3, a – b + 3
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
(vii) 4x2y, – 3xy2, –5xy2, 5x2y
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 4x2y + (-3xy2) + (-5xy2) + 5x2y
= 4x2y + 5x2y – 3xy2 – 5xy2
= x2y (4 + 5) + xy2 (-3 - 5)
= x2y (9) + xy2 (- 8)
= 9x2y – 8xy2
(viii) 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
(ix) ab – 4a, 4b – ab, 4a – 4b
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2
Answer
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 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
3. Subtract:
(i) -5y2 from y2
Answer
y2 - (-5y2) = y2 + 5y2 = 6y2
(ii) 6xy from -12xy
Answer
-12xy -(6xy) = -12xy - 6xy = -18xy
(iii) (a - b) from (a + b)
Answer
(a + b)-(a -b) = a + b -a + b
= a - a + b + b = 2b
(iv) a (b - 5) from b (5 - a)
Answer
= b (5 - a)-a (b -5)
= 5b - ab - ab + 5a
= 5b - 2ab+5a
= 5a + 5b -2ab
(v) -m2 + 5mn from 4m2 - 3mn + 8
Answer
= 4m2 - 3mn + 8 - (- m2 + 5mn)
= 4m2 - 3mn + 8 + m2 - 5mn
= 4m2 + m2 - 3mn - 5mn + 8
= 5m2 - 8mn + 8
(vi) -x2 +10x - 5 from 5x-10
Answer
= 5x – 10 – (-x2 + 10x - 5)
= 5x – 10 + x2 – 10x + 5
= x2 + 5x – 10x – 10 + 5
= x2 – 5x - 5
(vii) 5a2 - 7ab + 5b2 from 3ab - 2a2 - 2b2
Answer
= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2
= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2
= 10ab – 7a2 – 7b2
(viii) 4pq - 5q2 - 3p2 from 5p2 + 3q2 - pq
Answer
= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq
= 8p2 + 8q2 – 5pq
4. (a) What should be added to x2 +xy+y2 to obtain 2x2 +3xy ?
Answer
Let p should be added.
Then according to question,
x2 + xy + y2 + p = 2x2 + 3xy
⇒ p = 2x2 + 3xy - (x2 + xy + y2)
⇒ p = 2x2 + 3xy - x2 - xy - y2
⇒ p = 2x2 - x2- y2 +3xy - xy
⇒ p = x2 - y2 + 2xy
Hence, x2 - y2 + 2xy should be added.
(b) What should be subtracted from 2a + 8b+10 to get -3a + 7b + 16?
Answer
Let q should be subtracted.
Then according to question, 2a + 8b + 10-q = -3a + 7b + 16
⇒ -q = -3a +7b + 16 - (2a + 8b + 10)
⇒ -q = -3a + 7b + 16 - 2a - 8b - 10
⇒ - q = -3a - 2a + 7b - 8b + 16 - 10
⇒ -q = -5a - b + 6
⇒ q = - (- 5a - b + -6)
⇒ q = 5a + b - 6
5. What should be taken away from 3x2- 4y2 + 5xy + 20 to obtain - x2 - y2 + 6xy + 20 ?
Answer
Let q should be subtracted.
Then according to question,
3x2 - 4y2 + 5xy + 20 -q = -x2 - y2 + 6xy + 20
⇒ q = 3x2 - 4y2 + 5xy + 20 - (-x2 - y2 + 6xy + 20)
⇒ q = 3x2 - 4y2 + 5xy+ 20 + x2 + y2 - 6xy - 20
⇒ q = 3x2 + x2 - 4y2 +y2 + 5xy - 6xy + 20-20
⇒ q = 4x2 - 3y2 - xy + 0
Hence, 4x2 -3y2 -xy should be subtracted.
6. (a) From the sum of 3x - y + 11 and - y - 11, subtract 3x - y - 11.
Answer
First we have to find out the sum of 3x – y + 11 and – y – 11
= 3x – y + 11 + (-y - 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
Now, subtract 3x – y – 11 from 3x – 2y
= 3x – 2y – (3x – y - 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11
(b) From the sum of 4 + 3x and 5 - 4x + 2x2, subtract the sum of 3x2 - 5x and -x2 + 2x + 5.
Answer
First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2
= 4 + 3x + (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2
= 4 + 5 + 3x – 4x + 2x2
= 9 - x + 2x2
= 2x2 – x + 9 … [equation 1]
Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5.
Exercise 12.3
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 = 2 - 2 [Putting m = 2]
= 0
(ii) 3m - 5 = 3 x 2 - 5 [Putting m = 2]
= 6 - 5 = 1
(iii) 9 - 5m = 9 - 5 x 2 [Putting m = 2]
= 9 - 10 = - 1
(iv) 3m2 - 2m - 7
= 3(2)2 - 2 (2) - 7 [Putting m = 2]
=3 × 4 - 2 × 2 - 7
= 12-4-7
= 12- 11 = 1
(v) 
[Putting m = 2]
[Putting m = 2]= 5 - 4 = 1
2. If p = -2, find the value of:
(i) 4p + 7
(ii) - 3p2 + 4p + 7
(iii) -2p3 - 3p2 +4/7 + 7
Answer
(i) 4p + 7 = 4 (- 2) + 7 [Putting p= -2]
= -8 + 7 = -1
(ii) -3p2+4p + 7
= -3 (-2)2+ 4 (-2) + 7 [Putting p = - 2]
= - 3 × 4 - 8 + 7
= - 12 - 8 + 7
= -20 + 7 = -13
(iii) - 2p3 - 3p2 +4p + 7
= - 2 (-2)3 - 3(-2)2 + 4 (-2) + 7 [Putting p = - 2]
= -2 ×(-8)-3 ×4 -8 + 7
= 16-12-8 + 7
= -20 + 23 = 3
3. Find the value 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 = 2 (-1) - 7 [Putting x= - 1]
= - 2 - 7 = - 9
(ii) - x + 2 = - (-1) + 2 [Putting x= - 1]
= 1 + 2 = 3
(iii) x2 + 2 x + 1 = (-1)2 + 2 (-1) + 1 [Putting x= - 1]
= 1 - 2 + 1
= 2 - 2 = 0
(iv) 2x2- x - 2 = 2 (-1)2 - (-1) - 2 [Putting x= - 1]
= 2x 1 + 1-2
= 2 + 1 - 2
= 3 - 2 = 1
4. If a = 2,b = -2, find the value of:
(i) a2 + b2
(ii) a2+ab + b2
(iii) a2 - b2
Answer
(i) a2 + b2 ( 2)2 + (- 2)2 [Putting a = 2. b = - 2 ]
= 4 + 4 = 8
(ii) a2+ab + b2
= (2) + ( 2) (- 2) +(-2)2 [Putting a = 2. b = - 2 ]
= 4 - 4 + 4 = 4
(iii) a2 - b2 = (2)2 - (-2)2 [Putting a = 2,b = - 2]
= 4 - 4 = 0
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 = 2 (0) + 2 (-1) [Putting a - 0,b = - 1]
= 0 - 2 = -2
(ii) 2a2 + b2 + 1 = 2 (0)2 + (-1)2 + 1 [Putting a - 0,b = - 1]
= 2 x 0 + 1+ 1 = 0 + 2 = 2
(iii) 2a2b + 2ab2 + ab = 2(0)2 (-1) + 2 (0 )(-1)2 + (0 )(-1) [Putting a - 0,b = - 1]
= 0 + 0 + 0 = 0
(iv) a2 +ab + 2 - (0)2 + (0) (-1) + 2 [Putting a - 0,b = - 1]
= 0 + 0 + 2 = 2
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) = x + 7 + 4x - 20 = x + 4 x + 7 - 20
= 5 x - 13 = 5 x 2 - 13 [Putting x = 2]
= 10-13 = -3
(ii) 3 (x+ 2) + 5x - 7 = 3x + 6 + 5x -7 = 3x + 5x + 6 - 7
= 8x - 1 = 8 x 2-1 [Putting x = -1]
= 16 - 1 = 15
(iii) 6x + 5 (x - 2) = 6x + 5x -10 = 11x - 10
= 11 x 2 - 10 [Putting x = -1]
= 22 - 10 = 12
(iv) 4(2x - 1) + 3x + 11 = 8x - 4 + 3x +11 = 8x + 3a - 4 + 11
= 11a + 7 = 11 x 2 + 7 [Putting x = - 1]
= 22+7 = 29
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) 3a - 5 - x + 9 = 3x - x - 5 + 9 = 2x + 4
= 2x3+4 [Putting a = 3]
= 6 + 4 = 10
(ii) 2 - 8x + 4x + 4 = - 8x + 4x + 2 + 4 = -4x + 6
= - 4 x 3 + 6 [Putting a = 3]
= -12 + 6 =12
(iii) 3a + 5 - 8a + 1 = 3a - 8a + 5 + 1 = - 5a + 6
= -5(- 1) + 6 [Putting a = - 1]
= 5 + 6 = 11
(iv) 10 - 3b - 4 - 5b = - 3b - 5b + 10 - 4 = -8b+6
= -8 (-2)+ 6 [Putting b = -2]
= 16 + 6 = 22
(v) 2a - 2b - 4 - 5 + a = 2a + a - 2b - 4 - 5
= 3a - 2b - 9 = 3 (-1)-2 (-2) -9 [Putting a = -1 , b = - 2]
= -3 + 4 -9 = -8
8. (i) If z = 10, find the value of z3 - 3 (z - 10).
(ii) If p = - 10, find the value of p2 - 2p - 100
Answer
(i) z3 -3(z-10) = (10)3-3(10 - 10) [Putting z = 10]
= 1000 - 3 x 0 = 1000- 0
= 1000
(ii) p2 - 2p - 100 = (-10)2 - 2 (-10) - 100 (Putting p = - 10]
= 100+ 20 - 100 = 20
9. What should be the value of a if the value of 2x2 + x - a equals to 5, when x = 0 ?
Answer
Given: 2x2 + x - a = 5
⇒ 2 (0)2 + 0 - a = 5 [Putting x = 0]
⇒ 0 + 0 - a = 5
⇒ a = -5
Hence, the value of a is -5.
10. Simplify the expression and find its value when a = 5 and b = - 3: 2 (a2 + ab) + 3 - ab
Answer
Given 2 (a2 + ab) + 3 - ab
⇒ 2a2 + 2ab + 3 - ab
⇒ 2a2 + 2ab - ab + 3
⇒ 2a2 + ab + 3
⇒ 2 (5)2 + (5) (-3) + 3 [Putting a = 5 , b = -3]
⇒ 2 x 25 - 15 + 3
⇒ 50 - 15 + 3
⇒ 38.
Go Back To NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions
Chapter 12 NCERT Solutions are accurate and detailed which will encourage students to learn new topics. Expressions that contain constants and variables, or just variables, are called algebraic expressions.
• A term that contains only a number is called a constant term.
• The constants and the variables whose product makes a term of an algebraic expression, are called the factors of the term. The factors of a constant term in an algebraic expression are not considered.
• The numerical factor of a variable term is called its coefficient. The variable factors of a term are called its algebraic factors.
There are total 4 exercises in the Chapter 12 which will help you in finding specific questions that you're looking for by visiting the link below.
- Exercise 12.1 Chapter 12 Class 7 Maths NCERT Solutions
- Exercise 12.2 Chapter 12 Class 7 Maths NCERT Solutions
- Exercise 12.3 Chapter 12 Class 7 Maths NCERT Solutions
- Exercise 12.4 Chapter 12 Class 7 Maths NCERT Solutions
NCERT Solutions will improve your retention memory and studying habits. Students can develop their understanding of the chapter and obtain maximum marks in the exams.
NCERT Solutions for Class 7 Maths Chapters:
FAQ on Chapter 12 Algebraic Expressions
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NCERT Solutions for Class 7 can help you in figuring out the latest marking scheme and devise your own strategy. These NCERT Solutions are prepared as per the accordance of latest CBSE guidelines so you can score maximum marks.
Ritu spends ₹x daily and saves ₹ y per day. What is her income after 3 weeks?
₹21 (x + y).
Simplify the expression and find its value when a = 5 and b = –3. 2(a + ab) + 3 – ab.
2a2 + ab + 3, 38.
A bag contains 25 paise and so paise coins whose total values is ₹30. If the total number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.
50 Paise coins = 20, 25 Paise coins = 80.