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NCERT Solutions for class 7 Maths chapter 12 – Algebraic Expressions

Back Exercise

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.
Solution:
y – z

(ii) One-half of the sum of numbers x and y.
Solution:
$\frac{1}{2}$  (x -y)

(iii) The number z multiplied by itself.
Solution:
z × z i.e., z²

(iv) One-fourth of the product of numbers p and q.
Solution:
$\frac{1}{4}$ pq

(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 – yz

(viii) Sum of numbers a and 6 subtracted from their product.
Solution:
ab – (a + b)

Question 2.
(i) 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

(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) $\frac { 3 }{ 4 }$ x + $\frac { 1 }{ 4 }$
(h) 0.1 p² + 0.2 q²

Solution:

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions :

1. 5 – 3t²
2. 1 + t + t² + t³
3. x + 2xy + 3y
4. 100m + 1000n
5. -p²q² + 7pq
6. 1.2 a + 0.8 b
7. 3.14 r²
8. 2 (l + b)
9. 0.1 y + 0.01 y².

Solution:

Question 4.
(a) Identify terms which contain x and give the coefficient of x.

1. y²x + y
2. 13y² – 8yx
3. x + y + 2
4. 5 + z + zx
5. 1 + x + xy
6. 12xy² + 25
7. 7x + xy².

(b) Identify terms which contain y² and give the coefficient of y².

1. 8 – xy²
2. 5y² + 7x
3. 2x²y – 15xy² + 7y²

Solution:

Question 5.
Classify into monomials, binomials and trinomials.

1. 4y – 7z
2. y²
3. x + y – xy
4. 100
5. ab – a – b
6. 5 – 3t
7. 4p²q – 4pq²
8. 7mn
9. z² – 3z + 8 a² + b²
10. z² + z
11. 1 + x+ x²

Solution:
We know that an algebraic expression containing only one term is called a monomial. So, the monomials are : (ii), (iv), and (viii).

We know that an algebraic expression containing two terms is called a binomial. So, the binomials are : (i), (vi), (vii), (x) and (xi).

We know that an algebraic expression containing three terms is called a trinomial. So, the trinomial are : (iii), (v), (ix) and (xii).

Question 6.
State whether a given pair of terms is of like or unlike terms :
(i) 1, 100
Solution:
Like

(ii) -7x, $\frac{5}{2}$x
Solution:
Like

(iii) – 29x, – 29y
Solution:
Unlike

(iv)14xy, 42yx
Solution:
Like

(v) 4m²p, 4mp²
Answer:
Unlike

(vi) 12xz, 12x²z²
Solution:
Unlike

(i) 4y – 7z.
This expression is a binomial because it contains two terms: 4y and – Iz.
(ii) y².
This expression is a monomial because it contains only one term: y²
(iii) x + y – xy.
This expression is a trinomial because it contains three terms: x, y, and – xy.
(iv) 100.
This expression is a monomial because it contains only one term: 100
(v) ab – a – b.
This expression is a trinomial because it contains three terms: ab, -a, and -b
(vi) 5 – 3t.
This expression is a binomial because it contains two terms : 5 and – 31.
(vii) 4p²q – 4pq².
This expression is a binomial because it contains two terms: 4p²q and – 4pq².
(viii) 7mn.
This expression is a monomial because it contains only one term : 7mn.
(ix) z² – 3z + 8.
This expression is a trinomial because it contains three terms : z², – 3z and 8.
(x) a² + b².
This expression is a binomial because it contains two terms: a² and b².
(xi) z² + z.
This expression is a binomial because it contains two terms : z² and z.
(xii) 1 + x + x².
This expression is a trinomial because it contains three terms: 1, x, and x².

Question 6.
State whether a given pair of terms is of like or unlike terms :
(i) 1, 100
Solution:
Like

(ii) -7x, $\frac{5}{2}$x
Solution:
Like

(iii) – 29x, – 29y
Solution:
Unlike

(iv)14xy, 42yx
Solution:
Like

(v) 4m²p, 4mp²
Answer:
Unlike

(vi) 12xz, 12x²z²
Solution:
Unlike

Question 7.
Identify like terms in the following :

Solution:

Exercise 12.2

Question 1.
Simplify combining like terms:

1. 21b – 32 + 7b – 20b
2. – z² + 13z² – 5z + 7z³ – 15z
3. p – (p – q) – q – (q – p)
4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
5. 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
6. (3y² + 5y – 4) – (8y – y² – 4).

Solution:

Question 2.
Add:

Solution:

Question 3.
Subtract:

1. -5y² from y²
2. 6xy from – 12xy
3. (a – b) from (a + b)
4. a (b – 5) from b (5 – a)
5. -m² + 5mn from 4m² – 3mn + 8
6. -x² + 10x – 5 from 5x – 10
7. 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
8. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq.

Solution:

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 76 + 16?
Solution:

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

Question 6.
(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

Exercise 12.3

Question 1.
If m = 2, find the value of :

1. m – 2
2. 3m – 5
3. 9 – 5m
4. 3m² – 2m – 7
5. $\frac { 5m }{ 2 }$ – 4

Solution:

Question 2.
If p = – 2, find the value of :

1. 4p + 7
2. – 3p² + 4p + 7
3. – 2p³ – 3p² + 4p + 7.

Solution:

Question 3.
Find the value of the following expressions, when x = – 1 :

1. 2x – 7
2. – x + 2
3. x² + 2x + 1
4. 2x² – x – 2.

Solution:

Question 4.
If a = 2, b = – 2, find the value of :

1. a² + b²
2. a² + ab + b²
3. a² – b².

Solution:

Question 5.
When a = 0, b = – 1, find the value of the given expressions :

1. 2a + 2b
2. 2a² + b² + 1
3. 2a²b + 2ab² + ab
4. a² + ab + 2.

Solution:

Question 6.
Simplify the expressions and find the value ifx is equal to 2.

1. x + 7 + 4 (x – 5)
2. 3 (x + 2) + 5x – 7
3. 6x + 5 (x – 2)
4. 4 (2x – 1) + 3x + 11.

Solution:

Question 7.
Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

1. 3x – 5 – x + 9
2. 2 – 8x + 4x + 4
3. 3a + 5 – 8a + 1
4. 10 – 3b – 4 – 5b
5. 2a – 2b – 4 – 5 + a.

Solution:

Question 8.
(i) If z = 10, find the value of z³ – 3(z – 10).
(ii) If p = -10, find the value of p² – 2p – 100.
Solution:

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3 2(a² + ab) + 3 – ab.
Solution:

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.

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:

Let the number of digits formed be n, Then, the number of segments required to form n digits is given by the algebraic expression 5n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 1 = 25 + 1 = 26
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 1 = 50 + 1 = 51
3. the number of segments required to form 100 digits of this kind = 5 × 100 + 1 = 500 + 1 = 501.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 3n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 3 × 5 + 1 = 15 + 1 = 16
2. the number of segments required to form 10 digits of this kind = 3 × 10 + 1 = 30 + 1 = 31
3. The number of segments required to form 100 digits of this kind = 3 × 100 + 1 = 300 + 301.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 5n + 2.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 2 = 25 + 2 = 27
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 2 = 50 + 2 = 52
3. the number of segments required to form loo digits of this kind = 5 × 100 + 2 = 500 + 2 = 502.

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution: