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RD Sharma Solutions for Class 7 Chapter 7 Decimals Ex 7.2 Download PDF

Chapter 7: Algebraic Expressions Exercise – 7.2

Question: 1

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

Question: 2

Simplify each of the following:

(i) 7x³y +9yx³

(ii) 12a²b + 3ba²

Solution:

Simplifying the given expressions, we have

(i) 7x³y + 9yx³ = (7 + 9)x³y = 16x³y

(ii) 12a²b + 3ba² = (12 + 3)a²b =15a²b

Question: 3

Add the following:

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

(ii) 2x²y, – 4x²y, 6x²y, -5x²y

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) 2x²y +(-4x²y) + (6x²y) + (-5x²y)

= 2x²y – 4x²y + 6x²y – 5x2y

= (2- 4 + 6 – 5) x 2y

= (8 – 9) x 2y

= -x²y

Question: 4

Add the following expressions:

(i) x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

(ii) a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Solution:

Adding the given expressions, we have

(i) x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

Collecting positive and negative like terms together, we get

x³ +2x³ – 2x²y + 3x²y + 3xy² – 5xy² – y³– 4y³

= 3x³ + x²y – 2xy² – 5y³

(ii) a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴ – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Collecting positive and negative like terms together, we get

a⁴ – 2a⁴– 2a³b + 7a³b + 3ab³ – 5ab³ + 4a²b² – 6a²b² + 3b⁴ + b⁴

= – a⁴ + 5a³b – 2ab³ – 2a²b² + 4b⁴

Question: 5

Add the following expressions:

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

(ii) 5x³ + 7 + 6x – 5x², 2x² – 8 – 9x, 4x – 2x² + 3 x 3, 3 x 3 – 9x – x² and x – x² – x³ – 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 = (5 x 3 + 7+ 6x – 5x²) + (2 x 2 – 8 – 9x) + (4x – 2x² + 3 x 3) + (3 x 3 – 9x-x²) + (x – x² – x³ – 4)

Collecting positive and negative like terms together, we get

5x³ + 3x³ + 3x³ – x³ – 5x² + 2x² – 2x²– x² – x² + 6x – 9x + 4x – 9x + x + 7 – 8 – 4

= 10x³ – 7x² – 7x – 5

Question: 6

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

Question: 7

Add 2x² – 3x + 1 to the sum of 3x² – 2x and 3x + 7.

Solution:

Sum of 3x² – 2x and 3x + 7

= (3x² – 2x) + (3x +7)

=3x² – 2x + 3x + 7

= (3x² + x + 7)

Now, required expression = 2x² – 3x + 1+ (3x² + x + 7)

= 2x² + 3x² – 3x + x + 1 + 7

= 5x² – 2x + 8

Question: 8

Add x² + 2xy + y² to the sum of x² – 3y²and 2x² – y² + 9.

Solution:

Question: 9

Add a³+ b³ – 3 to the sum of 2a³ – 3b³ – 3ab + 7 and -a³ + b³ + 3ab – 9.

Solution:

Question: 10

Subtract:

(i) 7a²b from 3a²b

(ii) 4xy from -3xy

Solution:

(i) Required expression = 3a²b -7a²b

= (3 -7)a²b

= – 4a²b

(ii) Required expression = –3xy – 4xy

= –7xy

Question: 11

Subtract:

(i) – 4x from 3y

(ii) – 2x from – 5y

Solution:

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

= 3y + 4x

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

= –5y + 2x

Question: 12

Subtract:

(i) 6x³ −7x² + 5x − 3 from 4 − 5x + 6x² − 8x³

(ii) − x² −3z from 5x² – y + z + 7

(iii) x³ + 2x²y + 6xy² − y³ from y³−3xy²−4x²y

Solution:

Question: 13

From

(i) p3 – 4 + 3p², take away 5p² − 3p³ + p − 6

(ii) 7 + x − x², take away 9 + x + 3x² + 7x³

(iii) 1− 5y², take away y³ + 7y² + y + 1

(iv) x³ − 5x² + 3x + 1, take away 6x² − 4x³ + 5 + 3x

Solution:

Question: 14

From the sum of 3x² − 5x + 2 and − 5x² − 8x + 9 subtract 4x² − 7x + 9.

Solution:

Question: 15

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

Question: 16

From the sum of x² + 3y² − 6xy, 2x² − y² + 8xy, y² + 8 and x² − 3xy subtract −3x² + 4y² – xy + x – y + 3.

Solution:

Question: 17

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

Question: 18

What should be subtracted from x² – xy + y² – x + y + 3 to obtain −x² + 3y² − 4xy + 1?

Solution:

Question: 19

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

Question: 20

How much is x² − 2xy + 3y² less than 2x² − 3y² + xy?

Solution:

Question: 21

How much does a² − 3ab + 2b² exceed 2a² − 7ab + 9b²?

Solution:

Question: 22

What must be added to 12x³ − 4x² + 3x − 7 to make the sum x³ + 2x² − 3x + 2?

Solution:

Question: 23

If P = 7x² + 5xy − 9y², Q = 4y² − 3x² − 6xy and R = −4x² + xy + 5y², show that P + Q + R = 0.

Solution:

Question: 24

If P = a² − b² + 2ab, Q = a² + 4b² − 6ab, R = b² + b, S = a² − 4ab and T = −2a² + b² – ab + a. Find P + Q + R + S – T.

Solution:

All Chapter RD Sharma Solutions For Class 7 Maths

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