RD Sharma Solutions Class 7 Algebraic Expressions Exercise 7.2

RD Sharma Solutions Class 7 Chapter 7 Exercise 7.2

RD Sharma Class 7 Solutions Chapter 7 Ex 7.2 PDF Free Download

Exercise 7.2

Q1) Add the following:

(i) 3x and 7x

(ii) -5xy and 9xy

Solution:

We have

(i) 3x + 7x = (3 + 7)x = 10x

(ii) -5xy + 9xy = (-5 + 9)xy = 4xy

Q2) Simplify each of the following:

\((i) 7x^{3}y+9yx^{3}\)

\((ii) 12a^{2}b+3ba^{2}\)

Solution:

Simplifying the given expressions, we have

\((i) 7x^{3}y+9yx^{3}=(7+9)x^{3}y=16x^{3}y\)

\((ii) 12a^{2}b+3ba^{2}=(12+3)a^{2}b=15a^{2}b\)

Q3) Add the following:

(i) 7abc, -5abc, 9abc, -8abc

\((ii) 2x^{2}y,\;-4x^{2}y,\;6x^{2}y,\;-5x^{2}y\)

Solution:

Adding the given terms, we have

(i) 7abc + (-5abc) + (9abc) + (-8abc)

= 7abc – 5abc + 9abc – 8abc

= (7 – 5 + 9 – 8)abc

= (16 – 13)abc

= 3abc

\((ii) 2x^{2}y+(-4x^{2}y)+(6x^{2}y)+(-5x^{2}y)\)

\(= 2x^{2}y-4x^{2}y+6x^{2}y-5x^{2}y\)

\(= (2-4+6-5)x^{2}y\)

\(= (8-9)x^{2}y\)

\(= -x^{2}y\)

Q4) Add the following expressions:

\((i) x^{3}-2x^{2}y+3xy^{2}-y^{3},\;2x^{3}-5xy^{2}+3x^{2}y-4y^{3}\)

\((ii) a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4},\;-2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4}\)

Solution:

Adding the given expressions, we have

\((i) x^{3}-2x^{2}y+3xy^{2}-y^{3}+2x^{3}-5xy^{2}+3x^{2}y-4y^{3}\)

Collecting positive and negative like terms together, we get

\(x^{3}+2x^{3}-2x^{2}y+3x^{2}y +3xy^{2}-5xy^{2}-y^{3}-4y^{3}\)

\(= 3x^{3}+x^{2}y-2xy^{2}-5y^{3}\)

\((ii) (a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4})+(-2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4})\)

\(a^{4}-2a^{3}b+3ab^{3}+4a^{2}b^{2}+3b^{4}-2a^{4}-5ab^{3}+7a^{3}b-6a^{2}b^{2}+b^{4}\)

Collecting positive and negative like terms together, we get

\(a^{4}-2a^{4}-2a^{3}b+7a^{3}b+3ab^{3}-5ab^{3}+4a^{2}b^{2}-6a^{2}b^{2}+3b^{4}+b^{4}\)

\(= -a^{4}+5a^{3}b-2ab^{3}-2a^{2}b^{2}+4b^{4}\)

Q5) Add the following expressions:

(i) 8a – 6ab + 5b, –6a – ab – 8b and –4a + 2ab + 3b

\((ii) 5x^{3}+7+6x-5x^{2},\;2x^{2}-8-9x,\;4x-2x^{2}+3x^{3},\;3x^{3}-9x-x^{2}\;and\;x-x^{2}-x^{3}-4\)

Solution:

(i) Required expression = (8a – 6ab + 5b) + (–6a – ab – 8b) + (–4a + 2ab + 3b)

Collecting positive and negative like terms together, we get

8a – 6a – 4a – 6ab – ab + 2ab + 5b – 8b + 3b

= 8a – 10a – 7ab + 2ab + 8b – 8b

= –2a – 5ab

(ii) Required expression = \((5x^{3}+7+6x-5x^{2})+(2x^{2}-8-9x)+(4x-2x^{2}+3x^{3})+(3x^{3}-9x-x^{2})+(x-x^{2}-x^{3}-4)\)

Collecting positive and negative like terms together, we get

\(5x^{3}+3x^{3}+3x^{3}-x^{3}-5x^{2}+2x^{2}-2x^{2}-x^{2}-x^{2}+6x-9x+4x-9x+x+7-8-4\)

\(= 10x^{3}-7x^{2}-7x-5\)

Q6) Add the following:

(i) x – 3y – 2z

    5x + 7y – 8z

    3x – 2y + 5z

(ii) 4ab – 5bc + 7ca

–3ab + 2bc – 3ca

5ab – 3bc + 4ca

Solution:

(i) Required expression = (x – 3y – 2z) + (5x + 7y – 8z) + (3x – 2y + 5z)

Collecting positive and negative like terms together, we get

x + 5x + 3x – 3y + 7y – 2y – 2z – 8z + 5z

= 9x – 5y + 7y – 10z + 5z

= 9x + 2y – 5z

(ii) Required expression = (4ab – 5bc + 7ca) + (–3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)

Collecting positive and negative like terms together, we get

4ab – 3ab + 5ab – 5bc + 2bc – 3bc + 7ca – 3ca + 4ca

= 9ab – 3ab – 8bc + 2bc + 11ca – 3ca

= 6ab – 6bc + 8ca

Q7) Add \(2x^{2}-3x+1\) to the sum of \(3x^{2}-2x\) and 3x + 7.

Solution:

Sum of \(3x^{2}-2x\) and 3x + 7

\(= (3x^{2}-2x) + (3x + 7)\)

\(= 3x^{2}-2x+3x+7\)

\(= (3x^{2}+x+7)\)

Now, required expression = \(2x^{2}-3x+1\) + \((3x^{2}+x+7)\)

\(= 2x^{2}+3x^{2}-3x+x+1+7\)

\(= 5x^{2}-2x+8\)

Q8) Add \(x^{2}+2xy+y^{2}\) to the sum of \(x^{2}-3y^{2}\) and \(2x^{2}-y^{2}+9\).

Solution:

Sum of \(x^{2}-3y^{2}\) and \(2x^{2}-y^{2}+9\)

\(= (x^{2}-3y^{2})+(2x^{2}-y^{2}+9)\)

\(= x^{2}+2x^{2}-3y^{2}-y^{2}+9\)

\(= 3x^{2}-4y^{2}+9\)

Now, required expression = \((x^{2}+2xy+y^{2})\) + \(3x^{2}-4y^{2}+9\)

\(= x^{2}+3x^{2}+2xy+y^{2}-4y^{2}+9\)

\(= 4x^{2}+2xy-3y^{2}+9\)

Q9) Add \(a^{3}+b^{3}-3\) to the sum of \(2a^{3}-3b^{3}-3ab+7\) and \(-a^{3}+b^{3}+3ab-9\).

Solution:

First, we need to find the sum of \(2a^{3}-3b^{3}-3ab+7\) and \(-a^{3}+b^{3}+3ab-9\)

\(= (2a^{3}-3b^{3}-3ab+7)+(-a^{3}+b^{3}+3ab-9)\)

Collecting positive and negative like terms together, we get

\(= 2a^{3}-a^{3}-3b^{3}+b^{3}-3ab+3ab+7-9\)

\(= a^{3}-2b^{3}-2\)

Now, the required expression = \((a^{3}+b^{3}-3)\) + \((a^{3}-2b^{3}-2)\)

\(= a^{3}+a^{3}+b^{3}-2b^{3}-3-2\)

\(= 2a^{3}-b^{3}-5\)

Q10) Subtract:

\((i) 7a^{2}b\) from \(3a^{2}b\)

(ii) 4xy from -3xy

Solution:

(i) Required expression = \(3a^{2}b-7a^{2}b\)

\(= (3-7)a^{2}b\)

\(= -4a^{2}b\)

(ii) Required expression = –3xy – 4xy

= –7xy

Q11) Subtract:

(i) -4x from 3y

(ii) -2x from -5y

Solution:

(i) Required expression = (3y) – (–4x)

= 3y + 4x

(ii) Required expression = (-5y) – (–2x)

= –5y + 2x

Q12) Subtract:

\((i) 6x^{3}-7x^{2}+5x-3\) from \(4-5x+6x^{2}-8x^{3}\)

\((ii) -x^{2}-3z\) from \(5x^{2}-y+z+7\)

\((iii) x^{3}+2x^{2}y+6xy^{2}-y^{3}\) from \(y^{3}-3xy^{2}-4x^{2}y\)

Solution:

(i) Required expression = \((4-5x+6x^{2}-8x^{3})-(6x^{3}-7x^{2}+5x-3)\)

\(= 4-5x+6x^{2}-8x^{3}-6x^{3}+7x^{2}-5x+3\)

\(= -8x^{3}-6x^{3}+7x^{2}+6x^{2}-5x-5x+3+4\)

\(= -14x^{3}+13x^{2}-10x+7\)

(ii) Required expression = \((5x^{2}-y+z+7)-(-x^{2}-3z)\)

\(= 5x^{2}-y+z+7+x^{2}+3z\)

\(= 5x^{2}+x^{2}-y+z+3z +7\)

\(= 6x^{2}-y+4z +7\)

(iii) Required expression = \((y^{3}-3xy^{2}-4x^{2}y)-(x^{3}+2x^{2}y+6xy^{2}-y^{3})\)

\(= y^{3}-3xy^{2}-4x^{2}y-x^{3}-2x^{2}y-6xy^{2}+y^{3}\)

\(y^{3}+y^{3}-3xy^{2}-6xy^{2}-4x^{2}y-2x^{2}y-x^{3}\)

\(= 2y^{3}-9xy^{2}-6x^{2}y-x^{3}\)

Q13) From

(i) \(p^{3}-4+3p^{2}\), take away \(5p^{2}-3p^{3}+p-6\)

(ii) \(7+x-x^{2}\), take away \(9+x+3x^{2}+7x^{3}\)

(iii) \(1-5y^{2}\), take away \(y^{3}+7y^{2}+y+1\)

(iv) \(x^{3}-5x^{2}+3x+1\), take away \(6x^{2}-4x^{3}+5+3x\)

Solution:

(i) Required expression = \((p^{3}-4+3p^{2})-(5p^{2}-3p^{3}+p-6)\)

\(= p^{3}-4+3p^{2}-5p^{2}+3p^{3}-p+6\)

\(= p^{3}+3p^{3}+3p^{2}-5p^{2}-p-4+6\)

\(= 4p^{3}-2p^{2}-p+2\)

(ii) Required expression = \((7+x-x^{2})-(9+x+3x^{2}+7x^{3})\)

\(= 7+x-x^{2}-9-x-3x^{2}-7x^{3}\)

\(= -7x^{3}-x^{2}-3x^{2}+7-9\)

\(= -7x^{3}-4x^{2}-2\)

(iii) Required expression = \((1-5y^{2})-(y^{3}+7y^{2}+y+1)\)

\(= 1-5y^{2}-y^{3}-7y^{2}-y-1\)

\(= -y^{3}-5y^{2}-7y^{2}-y\)

\(= -y^{3}-12y^{2}-y\)

(iv) Required expression = \((x^{3}-5x^{2}+3x+1)-(6x^{2}-4x^{3}+5+3x)\)

\(= x^{3}-5x^{2}+3x+1-6x^{2}+4x^{3}-5-3x\)

\(= x^{3}+4x^{3}-5x^{2}-6x^{2}+1-5\)

\(= 5x^{3}-11x^{2}-4\)

Q14) From the sum of \(3x^{2}-5x+2\) and \(-5x^{2}-8x+9\) subtract \(4x^{2}-7x+9\).

Solution:

Required expression = \([(3x^{2}-5x+2)+(-5x^{2}-8x+9)]-(4x^{2}-7x+9)\)

\(= [3x^{2}-5x+2-5x^{2}-8x+9]-(4x^{2}-7x+9)\)

\(= [3x^{2}-5x^{2}-5x-8x+2+9]-(4x^{2}-7x+9)\)

\(= [-2x^{2}-13x+11]-(4x^{2}-7x+9)\)

\(= -2x^{2}-13x+11-4x^{2}+7x-9\)

\(= -2x^{2}-4x^{2}-13x+7x+11-9\)

\(= -6x^{2}-6x+2\)

Q15) Subtract the sum of 13x – 4y + 7z and –6z + 6x + 3y from the sum of 6x – 4y – 4z and   2x + 4y – 7.

Solution:

Sum of (13x – 4y + 7z) and (–6z + 6x + 3y)

= (13x – 4y + 7z) + (–6z + 6x + 3y)

= (13x – 4y + 7z – 6z + 6x + 3y)

= (13x + 6x – 4y + 3y + 7z – 6z)

= (19x – y + z)

Sum of (6x – 4y – 4z) and (2x + 4y – 7)

= (6x – 4y – 4z) + (2x + 4y – 7)

= (6x – 4y – 4z + 2x + 4y – 7)

= (6x + 2x – 4z – 7)

= (8x – 4z – 7)

Now, required expression = (8x – 4z – 7) – (19x – y + z)

= 8x – 4z – 7 – 19x + y – z

= 8x – 19x + y – 4z – z – 7

= –11x + y – 5z – 7

Q16) From the sum of \(x^{2}+3y^{2}-6xy,\;2x^{2}-y^{2}+8xy,\;y^{2}+8\;and\;x^{2}-3xy\) subtract \(-3x^{2}+4y^{2}-xy+x-y+3\).

Solution:

Sum of \((x^{2}+3y^{2}-6xy),\;(2x^{2}-y^{2}+8xy),\;(y^{2}+8)\;and\;(x^{2}-3xy)\)

\(= (x^{2}+3y^{2}-6xy)+(2x^{2}-y^{2}+8xy)+(y^{2}+8)+(x^{2}-3xy)\)

\(= (x^{2}+3y^{2}-6xy+2x^{2}-y^{2}+8xy+y^{2}+8+x^{2}-3xy)\)

\(= (x^{2}+2x^{2}+x^{2}+3y^{2}-y^{2}+y^{2}-6xy+8xy-3xy+8)\)

\(= (4x^{2}+3y^{2}-xy+8)\)

Now, required expression = \((4x^{2}+3y^{2}-xy+8)-(-3x^{2}+4y^{2}-xy+x-y+3)\)

\(= 4x^{2}+3y^{2}-xy+8+3x^{2}-4y^{2}+xy-x+y-3\)

\(= 4x^{2}+3x^{2}+3y^{2}-4y^{2}-xy+xy-x+y-3+8\)

\(= 7x^{2}-y^{2}-x+y+5\)

Q17) What should be added to xy – 3yz + 4zx to get 4xy – 3zx + 4yz + 7?

Solution:

The required expression can be got by subtracting xy – 3yz + 4zx from 4xy – 3zx + 4yz + 7.

Therefore, required expression = (4xy – 3zx + 4yz + 7) – (xy – 3yz + 4zx)

= 4xy – 3zx + 4yz + 7 – xy + 3yz – 4zx

= 4xy – xy – 3zx – 4zx + 4yz + 3yz + 7

= 3xy – 7zx + 7yz + 7

Q18) What should be subtracted from \(x^{2}-xy+y^{2}-x+y+3\) to obtain \(-x^{2}+3y^{2}-4xy+1\)?

Solution:

Let ‘M’ be the required expression. Then, we have

\(x^{2}-xy+y^{2}-x+y+3-M = -x^{2}+3y^{2}-4xy+1\)

Therefore,

\(M = (x^{2}-xy+y^{2}-x+y+3)-(-x^{2}+3y^{2}-4xy+1)\)

\(= x^{2}-xy+y^{2}-x+y+3+x^{2}-3y^{2}+4xy-1\)

Collecting positive and negative like terms together, we get

\(x^{2}+x^{2}-xy+4xy+y^{2}-3y^{2}-x+y+3-1\)

\(= 2x^{2}+3xy-2y^{2}-x+y+2\)

Q19) How much is x – 2y + 3z greater than 3x + 5y – 7?

Solution:

Required expression = (x – 2y + 3z) – (3x + 5y – 7)

= x – 2y + 3z – 3x – 5y + 7

Collecting positive and negative like terms together, we get

x – 3x – 2y + 5y + 3z + 7

= –2x – 7y + 3z + 7

Q20) How much is \(x^{2}-2xy+3y^{2}\) less than \(2x^{2}-3y^{2}+xy\)?

Solution:

Required expression = \((2x^{2}-3y^{2}+xy)-(x^{2}-2xy+3y^{2})\)

\(= 2x^{2}-3y^{2}+xy-x^{2}+2xy-3y^{2}\)

Collecting positive and negative like terms together, we get

\(2x^{2}-x^{2}-3y^{2}-3y^{2}+xy+2xy\)

\(x^{2}-6y^{2}+3xy\)

Q21) How much does \(a^{2}-3ab+2b^{2}\) exceed \(2a^{2}-7ab+9b^{2}\)?

Solution:

Required expression = \((a^{2}-3ab+2b^{2})-(2a^{2}-7ab+9b^{2})\)

\(= a^{2}-3ab+2b^{2}-2a^{2}+7ab-9b^{2}\)

Collecting positive and negative like terms together, we get

\(= a^{2}-2a^{2}-3ab+7ab+2b^{2}-9b^{2}\)

\(= -a^{2}+4ab-7b^{2}\)

Q22) What must be added to \(12x^{3}-4x^{2}+3x-7\) to make the sum \(x^{3}+2x^{2}-3x+2\)?

Solution:

Let ‘M’ be the required expression. Thus, we have

\(12x^{3}-4x^{2}+3x-7+M=x^{3}+2x^{2}-3x+2\)

Therefore,

\(M = (x^{3}+2x^{2}-3x+2) – (12x^{3}-4x^{2}+3x-7)\)

\(M = x^{3}+2x^{2}-3x+2-12x^{3}+4x^{2}-3x+7\)

Collecting positive and negative like terms together, we get

\(M = x^{3}-12x^{3}+2x^{2}+4x^{2}-3x-3x+7+2\)

\(x^{3}-12x^{3}+2x^{2}+4x^{2}-3x-3x+7+2\)

\( = -11x^{3}+6x^{2}-6x+9\)

Q23) If P = \(7x^{2}+5xy-9y^{2}\), Q = \(4y^{2}-3x^{2}-6xy\) and R = \(-4x^{2}+xy+5y^{2}\), show that P + Q + R = 0.

Solution:

We have

P + Q + R = \((7x^{2}+5xy-9y^{2})\) + \((4y^{2}-3x^{2}-6xy)\) + \((-4x^{2}+xy+5y^{2})\)

\(= 7x^{2}+5xy-9y^{2}+4y^{2}-3x^{2}-6xy-4x^{2}+xy+5y^{2}\)

Collecting positive and negative like terms together, we get

\(7x^{2}-3x^{2}-4x^{2}+5xy-6xy+xy-9y^{2}+4y^{2} +5y^{2}\)

\(= 7x^{2}-7x^{2}+6xy-6xy-9y^{2}+9y^{2}\)

= 0

Q24) If P = \(a^{2}-b^{2}+2ab\), Q = \(a^{2}+4b^{2}-6ab\), R = \(b^{2}+b\), S = \(a^{2}-4ab\) and T = \(-2a^{2}+b^{2}-ab+a\). Find P + Q + R + S – T.

Solution:

We have

\(P + Q + R + S – T = [(a^{2}-b^{2}+2ab)+(a^{2}+4b^{2}-6ab)+(b^{2}+b)+(a^{2}-4ab)]-(-2a^{2}+b^{2}-ab+a)\)

\(= [a^{2}-b^{2}+2ab+a^{2}+4b^{2}-6ab+b^{2}+b+a^{2}-4ab]-(-2a^{2}+b^{2}-ab+a)\)

\(= [3a^{2}+4b^{2}-8ab+b]-(-2a^{2}+b^{2}-ab+a)\)

\(= 3a^{2}+4b^{2}-8ab+b+2a^{2}-b^{2}+ab-a\)

Collecting positive and negative like terms together, we get

\(3a^{2}+2a^{2}+4b^{2}-b^{2}-8ab+ab-a+b\)

\(= 5a^{2}+3b^{2}-7ab-a+b\)


Practise This Question

In an experiment of tossing a coin once, what is the probability of getting a head?