NCERT solutions for Exercise 9.1 Class 8 Maths Chapter 9 Algebraic Expressions And Identities

1.Identify the terms, their coefficients for each of the following expressions :
(i) \(5xyz^2 – 3zy\)

(ii) \(1 + x + x^2\)

(iii) \(4x^2y^2 – 4x^2y^2z^2 + z^2\)

(iv) \(3 – pq + qr -rp\)

(v) \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\)

(vi)\( 0.3a – 0.6ab + 0.5b\)


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Answer :

(i) We have the expression \(5xyz^2 – 3zy\), the terms are \(5xyz^2\) and \(-3zy\).
Coefficient of \(xyz^2\) in the term \(5xyz^2\) is 5.

Coefficient of \(zy\) in the term \(– 3yz\) is \(– 3\).

(ii)We have the expression \(1 + x + x^2\), the terms are 1, x and \(x^2\).

Coefficient of the term 1 is 1.

Coefficient of x in the term x is 1.

Coefficient of \(x^2\) in the term \(x^2\) is 1.

(iii) We have the expression \(4xy – 4xyz + z\) , the terms are \(4x^2y^2, – 4x^2y^2z^2\) and \(z^2\).

Coefficient of \(x^2y^2\) in the term \(4x^2y^2\) is 4.

Coefficient of \(x^2y^2z^2\) in the term \(– 4x^2y^2z^2\) is – 4.

Coefficient of \(z^2\) in the term \(z^2\) is 1.

(iv) We have the expression \(3 – pq + qr – rp\), the terms are 3, – pq, qr and – rp.

Coefficient of the term 3 is 3.

Coefficient of pq in the term – pq is -1.

Coefficient of qr in the term qr is 1.

Coefficient of rp in the term – rp is -1.

(v) We have the expression \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\), the terms are \(\frac { x }{ 2 } ,\frac { 7 }{ 2 }\) and \(– xy\).

Coefficient of x in the term \(\frac { x }{ 2 }\) is \(\frac { 1 }{ 2 }\)

Coefficient of y in the term \(\frac { y }{ 2 }\) is \(\frac { 1 }{ 2 }\)

Coefficient of xy in the term \(– xy\) is -1.

(vi) In the expression \(0.3a -0.6 ab + 0.5 b\), the terms are \(0.3a, – 006ab \)and \(0.5b\).

Coefficient of a in the term 0.3a is 0.3.

Coefficient of ab in the term – 0.6ab is – 0.6.

Coefficient of b in the term 0.5b is 0.5.

2.Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories? \(x + y, 1000, x + x^2 + x^3 + x^4, 7 + y + 5x, 2y – 3y^2, 2y – 3y^2 + 4y^3\),\( 5x – 4y + 3xy, 4z – 15z^2, ab + bc+cd + da, pqr, p^2q + pq^2, 2p + 2q.\)


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Answer :

The calssification of the given expressions are:

Monomials : 1000, pqr

Binomials : \(x + y, 2y – 3y^2, 4z -15z^2, p^2q + pq^2, 2p + 2q.\)

Trinomials : \(7 + y + 5x, 2y – 3y^2 + 4y^3, 5x – 4y + 3x.\)

And the polynomials which don't have any category:

\(x + x^2 + x^3 + x^4, ab + be + cd + da.\)

3.Add the following :
(i) \(ab – be, be – ca, ea – ab\)

(ii)\( a – b + ab, b-c + be, c – a + ac\)

(iii) \(2p^2q^2 – 3pq + 4, 5 + 7pq – 3p^2q^2\)

(iv)\( l^2 + m^2, m^2 + n^2, n^2 + l^2, 2lm + 2mn + 2nl\)


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Answer :

(i) We have the expressions as:\( ab – bc, bc – ca, ca – ab\)
Adding them :

\((ab – bc) + (bc – ca) + (ca – ab) \)

\(\Rightarrow ab – bc + bc – ca + ca – ab \)

\(\Rightarrow (ab – ab) + (bc – bc) + (ca – ca)\quad\) [Placing the like terms together]

\(\Rightarrow 0 + 0 + 0\)

\(\Rightarrow 0\)

(ii) We have the expressions as:\(a – b + ab, b – c + bc, c – a + ac\)

Adding these we have:

\((a – b + ab) + (b – c + bc) + (c – a + ac) \)

\(\Rightarrow a – b + ab + b – c + bc + c – a + ac\)

\(\Rightarrow (a – a) + (b – b) + (c – c) + ab + bc + ac\quad \)[Placing all the like terms together]

\(\Rightarrow 0 + 0 + 0 + ab + bc + ac\)

\(\Rightarrow ab + bc + ac\)

(iii)We have the expressions as:\(2p^2q^2 – 3pq + 4, 5 + 7pq – 3p^2q^2\)

Adding these we get: \(2p^2q^2 – 3pq + 4 + 5 + 7pq – 3p^2q^2\)

\(\Rightarrow (2p^2q^2– 3p^2q^2)+(– 3pq+ 7pq)+ (4 + 5)\quad\)[Placing like terms together]

\(\Rightarrow -p^2q^2 + 4pq+ 9\)

(iv)We have the expressions as:\( l^2 + m^2, m^2 + n^2, n^2 + l^2, 2lm + 2mn + 2nl\)
Adding these we get: \((l^2 + m^2)+( m^2 + n^2)+( n^2 + l^2)+ (2lm + 2mn + 2nl)\)

\(\Rightarrow (l^2+l^2)+ (m^2+m^2)+ (n^2+n^2)+(2lm + 2mn + 2nl)\quad\)[Placing like terms together]

\(\Rightarrow (2l^2)+ (2m^2)+ (2n^2)+(2lm + 2mn + 2nl\)

\(\Rightarrow 2(l^2+ m^2+n^2+lm + mn + nl)\)

4.(a) Subtract \(4a – 7ab + 3b + 12\) from \(12a – 9ab + 5b – 3\)
(b) Subtract \(3xy + 5yz – 7zx\) from \(5xy – 2yz – 2zx + 10xyz\)

(c) Subtract \(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10 \) from \(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q\)


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Answer :

(a)We have the expressions to subtract:\(4a – 7ab + 3b + 12\) from \(12a – 9ab + 5b – 3\)
Subtracting these we get: \((12a – 9ab + 5b – 3)-(4a – 7ab + 3b + 12 )\)

\((12a – 9ab + 5b – 3)+(-4a +7ab - 3b - 12 )\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (12a-4a)+ (– 9ab+7ab)+ (+ 5b- 3b )+( – 3 - 12 )\quad\)[Placing like terms together]

\(\Rightarrow (8a)+ (-2ab)+ (2b)+(-15)\)

\(\Rightarrow (8a-2ab+ 2b-15)\)

(b)We have the expressions to subtract:\(3xy + 5yz – 7zx\) from \(5xy – 2yz – 2zx + 10xyz\)
Subtracting these we get: \((5xy – 2yz – 2zx + 10xyz)-(3xy + 5yz – 7zx)\)

\((5xy – 2yz – 2zx + 10xyz)+(-3xy - 5yz + 7zx )\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (5xy-3xy )+ (– 2yz - 5yz)+ (– 2zx+ 7zx )+( + 10xyz )\quad\)[Placing like terms together]

\(\Rightarrow (2xy)+ (-7yz)+ (5zx)+(10xyz)\)

\(\Rightarrow (2xy-7yz+5zx+10xyz)\)

(c)We have the expressions to subtract:\(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10 \) from \(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q\)
Subtracting these we get: \((18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q )-(4p^2q – 3pq + 5pq^2 – 8p + 7q – 10)\)

\(\Rightarrow(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q)+(-4p^2q +3pq-5pq^2 +8p-7q+ 10)\quad\)[Taking (-) inside the second bracket]

\(\Rightarrow (+18+10)+(-3p+8p)+(-11q-7q)+(5pq+3pq)+(-2pq^2-5pq^2)+(+ 5p^2q-4p^2q) \quad\)[Placing like terms together]

\(\Rightarrow (+28)+ (+5p)+(-18q)+(8pq)+(-7pq^2)+(+p^2q)\)

\(\Rightarrow (+28+5p-18q+8pq-7pq^2+p^2q)\)

4.Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) \(5a, 3a^2, 7a^4\)

(ii) \(2p, 4q, 8r\)

(iii) \(xy, 2x^2y, 2xy^2\)

(iv) \(a, 2b, 3c\)


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Answer :

(i)Given that, length = 5a, breadth = \(3a^2\), height = \(7a^4\)

∴Volume of the box =\( l \times b \times h = 5a \times 3a^2 \times 7a^4 = 105 a^7 \;\)cu. units

(ii)Given that, length = 2p, breadth = 4q, height = 8r

∴Volume of the box =\( l \times b \times h = 2p \times 4q \times 8r = 64pqr\;\) cu. units

(iii)Given that, length = xy, breadth =\( 2x^2y\), height = 2xy^2\)

Volume of the box =\( l \times b \times h = xy \times 2x^2y \times 2xy^2 = (1 \times 2 \times 2) \times xy \times x^2y \times xy^2 = 4x^4y^4 \;\)cu. units

(iv)Given that, length = a, breadth = 2b, height = 3c

∴Volume of the box = \(l \times b \times h = a \times 2b \times 3c = (1 \times 2 \times 3)abc = 6 abc\;\) cu. units




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