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CBSEClass 8RD Sharma

RD Sharma Class 8 Math 6th Chapter Algebraic Expressions and identities Exercise 6.2 Solution

By Mimi Das Last updated: November 5, 2019
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Exercise 6.2

 

Contents
Exercise 6.21. Add the following algebraic expressions: Solution:Solution: Solution:Solution:Solution:2. Subtract:Solution:4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and –4x + 9y – 11zSolution:6. Subtract the sum 2x – x2 + 5 and –4x – 3 + 7x2 from 5.Solution:7. Simplify each of the following:Solution:

1. Add the following algebraic expressions:

 (i) 3a2b, −4a2b, 9a2b

 Solution:

To add the like terms, we proceed as follows:

3a2b + (-4a2b) + 9a2b

= 3a2b – 4a2b + 9a2b                (Distributive Law)

= 8a2b

(iii) 4xy2–7x2y,12x2y–6xy2, −3x2y+5xy2

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Solution:

To add, we proceed as follows:

(4xy2–7x2y)+(12x2y)+(−6xy2)+(−3x2y+5xy2)

= 4xy2–7x2y+12x2y–6xy2–3x2y+5xy2

= 4xy2–6xy2+5xy2–7x2y+12x2y–3x2y      (Collecting like terms)

= 3xy2+2x2y                 (Combining like terms)

 

(iv) 32a – 54b+25c, 23a–72b+72c, 53a+52b–54c

 Solution:

(v) 112xy+125y+137x, −112y–125x–137xy

Solution:

(vi) 72x3–12x2+53, 32x3+74x2–x+13 ,32x2–52x–2

Solution:


2. Subtract:

 (i) -5xy from 12xy

Solution:

12xy – (–5xy)

= 12xy + 5xy = 17xy

(ii) 2a2 from -7a 2

Solution:

–7a2 – (2a2)

= –7a2 – 2a2 = –9a2

(iii) 2a – b from 3a – 5b

Solution:

(3a – 5b) – (2a – b)

= (3a – 5b) – 2a + b

= 3a – 5b – 2a + b


= 3a – 2a – 5b + b = a – 4b

(iv) 2x3 – 4x2 + 3x + 5 from4x3 + x2 + x + 6

Solution:

(4×3+x2+x+6)–(2×3–4×2+3x+5)

= 4×3+x2+x+6–2×3+4×2–3x–5

= 4×3–2×3+x2+4×2+x–3x+6–5         (Collecting like terms)

= 2×3+5×2–2x+1                 (Combining like terms)

3. Take away:

4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and –4x + 9y – 11z

Solution:

First add the expressions x – 3y + 2z and –4x + 9y – 11z we get:

(x – 3y + 2z ) + (–4x + 9y – 11z)

= x – 3y + 2z – 4x + 9y – 11z

= x – 4x – 3y + 9y + 2z – 11z      (Collecting like terms)

= –3x + 6y – 9z         (Combining like terms)

Now, Subtracting the expression 3x – 4y – 7z from the above sum, we get:

(–3x + 6y – 9z) – (3x – 4y – 7z)

= – 3x + 6y – 9z – 3x + 4y + 7z

= – 3x – 3x + 6y + 4y – 9z + 7z       (Collecting like terms)

= – 6x + 10y – 2z          (Combining like terms)

Thus, the answer is – 6x + 10y – 2z.

6. Subtract the sum 2x – x2 + 5 and –4x – 3 + 7x2 from 5.

Solution:

We have to subtract the sum of (2x – x2 + 5) and (–4x – 3 + 7x2) from 5.

5 – {(2x – x2 + 5) + (–4x – 3 + 7x2)}

= 5 – (2x – 4x – x2 + 7x2 + 5 – 3)

= 5 – 2x + 4x + x2 – 7x2 – 5 + 3

= 5 – 5 + 3 – 2x + 4x + x2 – 7x2       (Collecting like terms)

= 3 + 2x – 6x2    (Combining like terms)

Thus, the answer is 3 + 2x – 6x2.

7. Simplify each of the following:

 

(ii) [5-3x+2y-(2x-y)] – (3x-7y+9)

Solution:

[5 – 3x +2y – (2x – y)] – (3x -7y + 9)

= [5 – 3x + 2y – 2x + y] –(3x – 7y + 9)

= [5- 5x + 3y ] – (3x – 7y + 9)

= 5 – 5x + 3y – 3x + 7y = -4 – 8x + 10y

TAGGED:Algebraic Expressions and identities Class 8Algebraic Expressions and identities Class 8 Exercise 6.2 HomeworkAlgebraic Expressions and identities Class 8 Exercise 6.2 SolutionAlgebraic Expressions and identities Class 8 SolutionCBSECBSE Algebraic Expressions and identities Ex 6.2Mathematics Class 8Mathematics Class VIII RD SharmaMathematics solution Class 8RD Sharma Algebraic Expressions and identities Exercise 6.2 HometaskRD Sharma Class 8 6th Chapter SolutionRD Sharma Class 8 solution
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