Chapter – 12
Algebraic Expression
Exercise – 12.1
Q1. Get the algebraic expression in the following cases using variables, constant and arithmetic operation.
- Subtraction of z from y.
Solution:-
= Y – z
- One – half of the sum of numbers x and y.
Solution:-
= 1/2 ( x + y )
= (x + y ) / 2
- The number z multiplied by itself.
Solution:-
= z x z
= z2
- One – fourth of the product of number p and q.
Solution:-
= 1/4 ( p x q )
= pq/4
- Number x and y both squared and added .
Solution:-
= x2 + y2
- Number 5 added to three times the product of numbers m and n.
Solution:-
= 3mn + 5
- Product of numbers y and z subtracted from 10.
Solution:-
= 10 – ( y x z )
= 10 – yz
- Sum of numbers a and b subtracted from their product.
Solution:-
= ( a x b ) – ( a + b )
= ab – ( a + b )
Q2. (i) Identify the terms and their factor in the following expression
Show the terms the factors by tree diagrams.
- X – 3
Solution:-
Expression: x – 3
Terms: x, -3
Factor: x; -3
- 1 + X + X2
Solution:-
NCERT solutions for class 7 science chapter -16
Expression: 1 + x + x2
Term: 1,x , x2
Factors: 1; x; x,x
- Y – y3
Solution:-
Expression: y – y3
Terms: y, -y3
Factors: y; -y , -y , -y
- 5xy2 + 7x2y
Solution:-
Expression: 5x2y + 7x2y
Terms: 5xy2 , 7x2y
Factors: 5,x , y , y; 7 , x , x , y
- -ab + 2b2 – 3a2
Solution:-
Expression: -ab + 2b2 – 3a2
Terms: -ab 2b2-3a2
Factors: -a , b; 2, b, b; -3 , a, a
(ii) Identify terms and factors in the expression gien 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) 3/4 x + 1/4
(h) 0.1 p2 + 0.2 q2
Solution:-
S.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 | 1.2 ab , -2.4b , 3.6 a | 1.2 , a, b, -2.4 , b
3.6 , a |
(g)
|
3/4 x + 1/4 | 3/4 x 1/4 | 3/4 x 1/4 |
(h) | 0.1 p2 + 0.2 q2 | 0.1 p2,0.212 | 0.1, p ,p , 0.2 , q , q |
Q-3 identify the numerical coefficients of terms in the following expressions :
- 5-3t2(ii) 1 + t + t2(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)
- 1 y + 0.01 y2
Solution:-
SI No. | Expression | Terms | Coefficients |
(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 + 0.8b | 1.2a
0.8b |
1.2
0.8 |
(vii) | 3.14r2 | 3.14r2 | 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 |
Q4. (A) Identify terms which contain x and give the coefficient of x.
- Y2x + y (ii) 13y2 – 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy2 + 25 (vii) 7x + xy2
Solution:-
SI. No. | Expression | Terms | Coefficient of x |
(i) | Y2x + y | Y2x | Y2 |
(ii) | 13y2 – 8yx | -8yz | -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 |
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2(ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2
Solution:-
SI No. | Expression | Terms | Coefficient of y2 |
(i) | 8 – xy2 | -xy2 | -x |
(ii) | 5y2 + 7x | 5y2 | 5 |
(iii) | 2x2y – 15xy2 + 7y2 | -15xy2
7y2 |
15x
7 |
Q5. Classify into monomials binomials and trinomials.
- 4y – 7z
Solution:-
Binomial
(ii)y2
Solution:-
Monomial
(iii) x + y -xy
Solution:-
Trinomial.
(iv) 100
Solution:-
Monomial
(v ) ab – a – b
Solution:-
Trinomial.
(vi) 5 – 3t
Solution:-
Binomial
(vii) 4p2q – 4pq2
Solution:-
Binomial.
(viii) 7mm
Solution:-
Monomial.
(ix ) z2 – 3z + 8
Solution:-
Trinomial
- A2 + b2
Solution:-
BIONOMIAL
(XI) Z2 + Z
Solution:-
Binomial
(xii ) 1 + x + x2
Solution:-
Trinomial
Q6. State whether a given pair of terms is like or unlike terms.
- 1,100
Solution:-
Like term
- -7x , (5/2)x
Solution:-
Like term
- -29 x , – -29y
Solution:-
Like terms
- 14xy, 42yx.
Solution:-
Like term
- 4m2p , 4mp2
Solution:-
Unlike terms.
- 12xz , 12x2z2
Solution:-
Unlike term.
Q7. Identify like terms in the following :
- -xy2 , -4yx2 ,8x2, 2xy2, 7y, – 11x2 – 100x, – 11yx, 20x2y, – 6x2 , y , 2xy, 3x
Solution:-
When the term have the same algebraic factor they are like terms.
- 10pq , 7p , 8q , -p2q2, -7qp, , -100q, -23 , 12q2p2 , -5p2 , 41, 2450p , 78 qp, 13p2q , qp2, 710p2
Solution:-
When they are same algebraic factors, they are like terms.
they, are
10pq, – 7 qp , 78qp
8q, – 100 q
-p2q2 , 12q2p2
– 23,41
-13p2q, qp2