NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions
Exercise 10.1
1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
Solution:-
= Y – z
(ii) One-half of the sum of numbers x and y.
Solution:-
= ½ (x + y)
= (x + y)/2
(iii) The number z multiplied by itself.
Solution:-
= z × z
= z2
(iv) One-fourth of the product of numbers p and q.
Solution:-
= ¼ (p × q)
= pq/4
(v) Numbers x and y, both squared and added.
Solution:-
= x2 + y2
(vi) Number 5 added to three times the product of numbers m and n.
Solution:-
= 3mn + 5
(vii) Product of numbers y and z subtracted from 10.
Solution:-
= 10 – (y × z)
= 10 – yz
(viii) Sum of numbers a and b subtracted from their product.
Solution:-
= (a × b) – (a + b)
= ab – (a + b)
2. (i) Identify the terms and their factors in the following expressions.
Show the terms and factors by tree diagrams.
(a) x – 3
Solution:-
Expression: x – 3
Terms: x, -3
Factors: x; -3
(b) 1 + x + x2
Solution:-
Expression: 1 + x + x2
Terms: 1, x, x2
Factors: 1; x; x,x
(c) y – y3
Solution:-
Expression: y – y3
Terms: y, -y3
Factors: y; -y, -y, -y
(d) 5xy2 + 7x2y
Solution:-
Expression: 5xy2 + 7x2y
Terms: 5xy2, 7x2y
Factors: 5, x, y, y; 7, x, x, y
(e) – 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 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:-
Expressions are defined as numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.
In algebra, a term is either a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.
Factors are defined as numbers we can multiply together to get another number.
3. 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:-
Expressions are defined as numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.
In algebra, a term is either a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.
A coefficient is a number used to multiply a variable (2x means 2 times x, so 2 is a coefficient). Variables on their own (without a number next to them) actually have a coefficient of 1 (x is really 1x).
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 + 25
(vii) 7x + xy2
Solution:-
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2 (ii) 5y2 + 7x (iii) 2x2y – 15xy2 + 7y2
Solution:-
5. Classify into monomials, binomials and trinomials.
(i) 4y – 7z
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(ii) y2
Solution:-
Monomial.
An expression with only one term is called a monomial.
(iii) x + y – xy
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(iv) 100
Solution:-
Monomial.
An expression with only one term is called a monomial.
(v) ab – a – b
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(vi) 5 – 3t
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(vii) 4p2q – 4pq2
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(viii) 7mn
Solution:-
Monomial.
An expression with only one term is called a monomial.
(ix) z2 – 3z + 8
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(x) a2 + b2
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(xi) z2 + z
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(xii) 1 + x + x2
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
6. State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
Solution:-
Like term.
When terms have the same algebraic factors, they are like terms.
(ii) –7x, (5/2)x
Solution:-
Like term.
When terms have the same algebraic factors, they are like terms.
(iii) – 29x, – 29y
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(iv) 14xy, 42yx
Solution:-
Like term.
When terms have the same algebraic factors, they are like terms.
(v) 4m2p, 4mp2
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(vi) 12xz, 12x2z2
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
7. Identify like terms in the following.
(a) – xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x
Solution:-
When terms have the same algebraic factors, they are like terms.
They are,
– xy2, 2xy2
– 4yx2, 20x2y
8x2, – 11x2, – 6x2
7y, y
– 100x, 3x
– 11yx, 2xy
(b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,
13p2q, qp2, 701p2
Solution:-
When terms have the same algebraic factors, they are like terms.
They are,
10pq, – 7qp, 78qp
7p, 2405p
8q, – 100q
– p2q2, 12q2p2
– 23, 41
– 5p2, 701p2
13p2q, qp2
Exercise 10.2
1. If m = 2, find the value of:
(i) m – 2
Solution:-
From the question, it is given that m = 2
Then, substitute the value of m in the question.
= 2 -2
= 0
(ii) 3m – 5
Solution:-
From the question, it is given that m = 2
Then, substitute the value of m in the question.
= (3 × 2) – 5
= 6 – 5
= 1
(iii) 9 – 5m
Solution:-
From the question, it is given that m = 2
Then, substitute the value of m in the question.
= 9 – (5 × 2)
= 9 – 10
= – 1
(iv) 3m2 – 2m – 7
Solution:-
From the question, it is given that m = 2
Then, substitute the value of m in the question.
= (3 × 22) – (2 × 2) – 7
= (3 × 4) – (4) – 7
= 12 – 4 -7
= 12 – 11
= 1
(v) (5m/2) – 4
Solution:-
From the question, it is given that m = 2
Then, substitute the value of m in the question.
= ((5 × 2)/2) – 4
= (10/2) – 4
= 5 – 4
= 1
2. If p = – 2, find the value of:
(i) 4p + 7
Solution:-
From the question, it is given that p = -2
Then, substitute the value of p in the question.
= (4 × (-2)) + 7
= -8 + 7
= -1
(ii) – 3p2 + 4p + 7
Solution:-
From the question, it is given that p = -2
Then, substitute the value of p in the question.
= (-3 × (-2)2) + (4 × (-2)) + 7
= (-3 × 4) + (-8) + 7
= -12 – 8 + 7
= -20 + 7
= -13
(iii) – 2p3 – 3p2 + 4p + 7
Solution:-
From the question, it is given that p = -2
Then, substitute the value of p in the question.
= (-2 × (-2)3) – (3 × (-2)2) + (4 × (-2)) + 7
= (-2 × -8) – (3 × 4) + (-8) + 7
= 16 – 12 – 8 + 7
= 23 – 20
= 3
3. Find the value of the following expressions when x = –1:
(i) 2x – 7
Solution:-
From the question, it is given that x = -1
Then, substitute the value of x in the question.
= (2 × -1) – 7
= – 2 – 7
= – 9
(ii) – x + 2
Solution:-
From the question, it is given that x = -1
Then, substitute the value of x in the question.
= – (-1) + 2
= 1 + 2
= 3
(iii) x2 + 2x + 1
Solution:-
From the question, it is given that x = -1
Then, substitute the value of x in the question.
= (-1)2 + (2 × -1) + 1
= 1 – 2 + 1
= 2 – 2
= 0
(iv) 2x2 – x – 2
Solution:-
From the question, it is given that x = -1
Then, substitute the value of x in the question.
= (2 × (-1)2) – (-1) – 2
= (2 × 1) + 1 – 2
= 2 + 1 – 2
= 3 – 2
= 1
4. If a = 2, b = – 2, find the value of:
(i) a2 + b2
Solution:-
From the question, it is given that a = 2, b = -2
Then, substitute the value of a and b in the question.
= (2)2 + (-2)2
= 4 + 4
= 8
(ii) a2 + ab + b2
Solution:-
From the question, it is given that a = 2, b = -2
Then, substitute the value of a and b in the question.
= 22 + (2 × -2) + (-2)2
= 4 + (-4) + (4)
= 4 – 4 + 4
= 4
(iii) a2 – b2
Solution:-
From the question, it is given that a = 2, b = -2
Then, substitute the value of a and b in the question.
= 22 – (-2)2
= 4 – (4)
= 4 – 4
= 0
5. When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
Solution:-
From the question, it is given that a = 0, b = -1
Then, substitute the value of a and b in the question.
= (2 × 0) + (2 × -1)
= 0 – 2
= -2
(ii) 2a2 + b2 + 1
Solution:-
From the question, it is given that a = 0, b = -1
Then, substitute the value of a and b in the question.
= (2 × 02) + (-1)2 + 1
= 0 + 1 + 1
= 2
(iii) 2a2b + 2ab2 + ab
Solution:-
From the question, it is given that a = 0, b = -1
Then, substitute the value of a and b in the question.
= (2 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1)
= 0 + 0 +0
= 0
(iv) a2 + ab + 2
Solution:-
From the question, it is given that a = 0, b = -1
Then, substitute the value of a and b in the question.
= (02) + (0 × (-1)) + 2
= 0 + 0 + 2
= 2
6. Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4 (x – 5)
Solution:-
From the question, it is given that x = 2
We have,
= 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
(ii) 3 (x + 2) + 5x – 7
Solution:-
From the question, it is given that x = 2
We have,
= 3x + 6 + 5x – 7
= 8x – 1
Then, substitute the value of x in the equation.
= (8 × 2) – 1
= 16 – 1
= 15
(iii) 6x + 5 (x – 2)
Solution:-
From the question, it is given that x = 2
We have,
= 6x + 5x – 10
= 11x – 10
Then, substitute the value of x in the equation.
= (11 × 2) – 10
= 22 – 10
= 12
(iv) 4(2x – 1) + 3x + 11
Solution:-
From the question, it is given that x = 2
We have,
= 8x – 4 + 3x + 11
= 11x + 7
Then, substitute the value of x in the equation.
= (11 × 2) + 7
= 22 + 7
= 29
7. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.
(i) 3x – 5 – x + 9
Solution:-
From the question, it is given that x = 3
We have,
= 3x – x – 5 + 9
= 2x + 4
Then, substitute the value of x in the equation.
= (2 × 3) + 4
= 6 + 4
= 10
(ii) 2 – 8x + 4x + 4
Solution:-
From the question, it is given that x = 3
We have,
= 2 + 4 – 8x + 4x
= 6 – 4x
Then, substitute the value of x in the equation.
= 6 – (4 × 3)
= 6 – 12
= – 6
(iii) 3a + 5 – 8a + 1
Solution:-
From the question, it is given that a = -1
We have,
= 3a – 8a + 5 + 1
= – 5a + 6
Then, substitute the value of a in the equation.
= – (5 × (-1)) + 6
= – (-5) + 6
= 5 + 6
= 11
(iv) 10 – 3b – 4 – 5b
Solution:-
From the question, it is given that b = -2
We have,
= 10 – 4 – 3b – 5b
= 6 – 8b
Then, substitute the value of b in the equation.
= 6 – (8 × (-2))
= 6 – (-16)
= 6 + 16
= 22
(v) 2a – 2b – 4 – 5 + a
Solution:-
From the question, it is given that a = -1, b = -2
We have,
= 2a + a – 2b – 4 – 5
= 3a – 2b – 9
Then, substitute the value of a and b in the equation.
= (3 × (-1)) – (2 × (-2)) – 9
= -3 – (-4) – 9
= – 3 + 4 – 9
= -12 + 4
= -8
8. (i) If z = 10, find the value of z3 – 3(z – 10).
Solution:-
From the question, it is given that z = 10
We have,
= z3 – 3z + 30
Then, substitute the value of z in the equation.
= (10)3 – (3 × 10) + 30
= 1000 – 30 + 30
= 1000
(ii) If p = – 10, find the value of p2 – 2p – 100
Solution:-
From the question, it is given that p = -10
We have,
= p2 – 2p – 100
Then, substitute the value of p in the equation.
= (-10)2 – (2 × (-10)) – 100
= 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?
Solution:-
From the question, it is given that x = 0
We have,
2x2 + x – a = 5
a = 2x2 + x – 5
Then, substitute the value of x in the equation.
a = (2 × 02) + 0 – 5
a = 0 + 0 – 5
a = -5
10. Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab
Solution:-
From the question, it is given that a = 5 and b = -3
We have,
= 2a2 + 2ab + 3 – ab
= 2a2 + ab + 3
Then, substitute the value of a and b in the equation.
= (2 × 52) + (5 × (-3)) + 3
= (2 × 25) + (-15) + 3
= 50 – 15 + 3
= 53 – 15
= 38
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