# RD Sharma Solutions Class 8 Algebraic Expressions And Identities Exercise 6.2

## RD Sharma Solutions Class 8 Chapter 6 Exercise 6.2

Q.1: Add the following algebraic expressions:

(i) $3a^{2}b,\;-4a^{2}b,\;9a^{2}b$

(ii) $\frac{2}{3}a,\;\frac{3}{5}a,\;-\frac{6}{5}a$

(iii) $4xy^{2} – 7x^{2}y,\; 12x^{2}y – 6xy^{2},\; -3x^{2}y + 5xy^{2}$

(iv) $\frac{3}{2}a – \frac{5}{4}b + \frac{2}{5}c,\; \frac{2}{3}a – \frac{7}{2}b + \frac{7}{2}c,\; \frac{5}{3}a + \frac{5}{2}b – \frac{5}{4}c$

(v) $\frac{11}{2}xy + \frac{12}{5}y + \frac{13}{7}x,\; -\frac{11}{2}y – \frac{12}{5}x – \frac{13}{7}xy$

(vi) $\frac{7}{2}x^{3} – \frac{1}{2}x^{2} + \frac{5}{3},\; \frac{3}{2}x^{3} + \frac{7}{4}x^{2} – x + \frac{1}{3},\; \frac{3}{2}x^{2} – \frac{5}{2}x – 2$

Solution:

(i) To add the like terms, we proceed as follows:

3a2b + (-4a2b) + 9a2b

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

= 8a2b

(ii) To add like terms, we proceed as follows:

$\frac{2}{3}a + \frac{3}{5}a + (-\frac{6}{5}a)$

= $\frac{2}{3}a + \frac{3}{5}a – \frac{6}{5}a$

= $(\frac{2}{3} + \frac{3}{5} – \frac{6}{5})a$              (Distributive Law)

= $\frac{1}{15}a$

(iii) To add, we proceed as follows:

$(4xy^{2} – 7x^{2}y) + (12x^{2}y) + (-6xy^{2}) + (-3x^{2}y + 5xy^{2})$

= $4xy^{2} – 7x^{2}y + 12x^{2}y – 6xy^{2} – 3x^{2}y + 5xy^{2}$

= $4xy^{2} – 6xy^{2} + 5xy^{2} – 7x^{2}y + 12x^{2}y – 3x^{2}y$             (Collecting like terms)

= $3xy^{2}+2x^{2}y$                 (Combining like terms)

(iv) To add, we proceed as follows:

$(\frac{3}{2}a – \frac{5}{4}b + \frac{2}{5}c) + (\frac{2}{3}a – \frac{7}{2}b + \frac{7}{2}c) + (\frac{5}{3}a + \frac{5}{2}b – \frac{5}{4}c)$

= $\frac{3}{2}a – \frac{5}{4}b + \frac{2}{5}c + \frac{2}{3}a – \frac{7}{2}b + \frac{7}{2}c + \frac{5}{3}a + \frac{5}{2}b – \frac{5}{4}c$

= $\frac{3}{2}a + \frac{2}{3}a + \frac{5}{3}a – \frac{5}{4}b – \frac{7}{2}b + \frac{5}{2}b + \frac{2}{5}c + \frac{7}{2}c – \frac{5}{4}c$

(Collecting like terms)

= $\frac{23}{6}a – \frac{9}{4}b + \frac{53}{20}c$       (Combining like terms)

(v) To add, we proceed as follows:

$(\frac{11}{2}xy + \frac{12}{5}y + \frac{13}{7}x) + (-\frac{11}{2}y – \frac{12}{5}x – \frac{13}{7}xy)$

= $\frac{11}{2}xy + \frac{12}{5}y + \frac{13}{7}x – \frac{11}{2}y – \frac{12}{5}x – \frac{13}{7}xy$

= $\frac{11}{2}xy – \frac{13}{7}xy + \frac{12}{5}y – \frac{11}{2}y + \frac{13}{7}x – \frac{12}{5}x$                                    (Collecting like terms)

= $\frac{51}{14}xy – \frac{31}{10}y – \frac{19}{35}x$         (Combining like terms)

(vi) To add, we proceed as follows:

$(\frac{7}{2}x^{3} – \frac{1}{2}x^{2} + \frac{5}{3}) + (\frac{3}{2}x^{3} + \frac{7}{4}x^{2} – x + \frac{1}{3}) + (\frac{3}{2}x^{2} – \frac{5}{2}x – 2)$

= $\frac{7}{2}x^{3} – \frac{1}{2}x^{2} + \frac{5}{3} + \frac{3}{2}x^{3} + \frac{7}{4}x^{2} – x + \frac{1}{3} + \frac{3}{2}x^{2} – \frac{5}{2}x – 2$

= $\frac{7}{2}x^{3} + \frac{3}{2}x^{3} – \frac{1}{2}x^{2} + \frac{7}{4}x^{2} + \frac{3}{2}x^{2} – x – \frac{5}{2}x + \frac{5}{3} + \frac{1}{3} – 2$

(Collecting like terms)

= $5x^{3} + \frac{11}{4}x^{2} – \frac{7}{2}x$                     (Combining like terms)

Q2) Subtract:

(i) -5xy from 12xy

(ii) 2a2 from -7a 2

(iii) 2a – b from 3a – 5b

(iv) $2x^{3} – 4x^{2} + 3x + 5\; from \; 4x^{3} + x^{2} + x + 6$

(v) $\frac{2}{3}y^{3} – \frac{2}{7}y^{2} – 5\; from \; \frac{1}{3}y^{3} + \frac{5}{7}y^{2} + y – 2$

(vi) $\frac{3}{2}x – \frac{5}{4}y – \frac{7}{2}z\; from \; \frac{2}{3}x + \frac{3}{2}y – \frac{4}{3}z$

(vii) $x^{2}y – \frac{4}{5}xy^{2} + \frac{4}{3}xy\; from \; \frac{2}{3}x^{2}y + \frac{3}{2}xy^{2} – \frac{1}{3}xy$

(viii) $\frac{ab}{7} – \frac{35}{3}bc + \frac{6}{5}ac\; from \; \frac{3}{5}bc – \frac{4}{5}ac$

Solution:

(i) 12xy – (–5xy)

= 12xy + 5xy = 17xy

(ii) –7a2 – (2a2)

= –7a2 – 2a= –9a2

(iii) (3a – 5b) – (2a – b)

= (3a – 5b) – 2a + b

= 3a – 5b – 2a + b

= 3a – 2a – 5b + b = a – 4b

(iv) $(4x^{3} + x^{2} + x + 6) – (2x^{3} – 4x^{2} + 3x + 5)$

= $4x^{3} + x^{2} + x + 6 – 2x^{3} + 4x^{2} – 3x – 5$

= $4x^{3} – 2x^{3} + x^{2} + 4x^{2} + x – 3x + 6 – 5$            (Collecting like terms)

= $2x^{3} + 5x^{2} – 2x + 1$                  (Combining like terms)

(v) $(\frac{1}{3}y^{3} + \frac{5}{7}y^{2} + y – 2) – (\frac{2}{3}y^{3} – \frac{2}{7}y^{2} – 5)$

= $\frac{1}{3}y^{3} + \frac{5}{7}y^{2} + y – 2 – \frac{2}{3}y^{3} + \frac{2}{7}y^{2} + 5$

= $\frac{1}{3}y^{3} – \frac{2}{3}y^{3} + \frac{5}{7}y^{2} + \frac{2}{7}y^{2} + y – 2 + 5$                     (Collecting like terms)

= $-\frac{1}{3}y^{3} + y^{2} + y + 3$                          (Combining like terms)

(vi) $(\frac{2}{3}x + \frac{3}{2}y – \frac{4}{3}z) – (\frac{3}{2}x – \frac{5}{4}y – \frac{7}{2}z)$

= $\frac{2}{3}x + \frac{3}{2}y – \frac{4}{3}z – \frac{3}{2}x + \frac{5}{4}y + \frac{7}{2}z$

= $\frac{2}{3}x – \frac{3}{2}x + \frac{3}{2}y + \frac{5}{4}y – \frac{4}{3}z + \frac{7}{2}z$                                 (Collecting like terms)

= $-\frac{5}{6}x + \frac{11}{4}y + \frac{13}{6}z$                (Combining like terms)

(vii) $(\frac{2}{3}x^{2}y + \frac{3}{2}xy^{2} – \frac{1}{3}xy) – (x^{2}y – \frac{4}{5}xy^{2} + \frac{4}{3}xy)$

= $\frac{2}{3}x^{2}y + \frac{3}{2}xy^{2} – \frac{1}{3}xy – x^{2}y + \frac{4}{5}xy^{2} – \frac{4}{3}xy$

= $\frac{2}{3}x^{2}y – x^{2}y + \frac{3}{2}xy^{2} + \frac{4}{5}xy^{2} – \frac{1}{3}xy – \frac{4}{3}xy$                          (Collecting like terms)

= $-\frac{1}{3}x^{2}y + \frac{23}{10}xy^{2} – \frac{5}{3}xy$         (Combining like terms)

(viii) $(\frac{3}{5}bc – \frac{4}{5}ac) – (\frac{ab}{7} – \frac{35}{3}bc + \frac{6}{5}ac)$

= $\frac{3}{5}bc – \frac{4}{5}ac – \frac{ab}{7} + \frac{35}{3}bc – \frac{6}{5}ac$

= $\frac{3}{5}bc + \frac{35}{3}bc – \frac{4}{5}ac – \frac{6}{5}ac – \frac{ab}{7}$               (Collecting like terms)

= $\frac{184}{15}bc – 2ac – \frac{ab}{7}$                     (Combining like terms)

Q3) Take away:

(i) $\frac{6}{5}x^{2} – \frac{4}{5}x^{3} + \frac{5}{6} + \frac{3}{2}x\; from \; \frac{ x^{3}}{3} – \frac{5}{2}x^{2} + \frac{3}{5}x + \frac{1}{4}$

(ii) $\frac{7}{4}x^{3} + \frac{3}{5}x^{2} + \frac{1}{2}x + \frac{9}{2}\; from \; \frac{7}{2} – \frac{x}{3} – \frac{x^{2}}{5}$

(iii) $\frac{y^{3}}{3} + \frac{7}{3}y^{2} + \frac{1}{2}y + \frac{1}{2}\; from \; \frac{1}{3} – \frac{5}{3}y^{2}$

(iv) $\frac{2}{3}ac – \frac{5}{7}ab + \frac{2}{3}bc\; from \; \frac{3}{2}ab – \frac{7}{4}ac – \frac{5}{6}bc$

Solution:

(i) The difference is given by:

$(\frac{ x^{3}}{3} – \frac{5}{2}x^{2} + \frac{3}{5}x + \frac{1}{4}) – (\frac{6}{5}x^{2} – \frac{4}{5}x^{3} + \frac{5}{6} + \frac{3}{2}x)$

= $\frac{ x^{3}}{3} – \frac{5}{2}x^{2} + \frac{3}{5}x + \frac{1}{4} – \frac{6}{5}x^{2} + \frac{4}{5}x^{3} – \frac{5}{6} – \frac{3}{2}x$

= $\frac{x^{3}}{3} + \frac{4}{5}x^{3} – \frac{5}{2}x^{2} – \frac{6}{5}x^{2} + \frac{3}{5}x – \frac{3}{2}x + \frac{1}{4} – \frac{5}{6}$                  (Collecting like terms)

= $(\frac{5+12}{15})x^{3} + (\frac{-25-12}{10})x^{2} + (\frac{6-15}{10}x) + (\frac{6-20}{24})$

= $\frac{17}{15}x^{3} – \frac{37}{10}x^{2} – \frac{9}{10}x – \frac{7}{12}$                       (Combining like terms)

(ii) The difference is given by:

$(\frac{7}{2} – \frac{x}{3} – \frac{x^{2}}{5}) – (\frac{7}{4}x^{3} + \frac{3}{5}x^{2} + \frac{x}{2} + \frac{9}{2})$

= $\frac{7}{2} – \frac{x}{3} – \frac{x^{2}}{5} – \frac{7}{4}x^{3} – \frac{3}{5}x^{2} – \frac{x}{2} – \frac{9}{2}$

= $\frac{7}{2} – \frac{9}{2} – \frac{x}{3} – \frac{x}{2} – \frac{x^{2}}{5} – \frac{3x^{2}}{5} – \frac{7x^{3}}{4}$                    (Collecting like terms)

= $(\frac{7-9}{2}) + (\frac{-2-3}{6})x + (\frac{-1-3}{5})x^{2} – \frac{7x^{3}}{4}$

= $-1 – \frac{5x}{6} – \frac{4x^{2}}{5} – \frac{7x^{3}}{4}$                           (Combining like terms)

(iii) The difference is given by:

$(\frac{1}{3} – \frac{5}{3}y^{2}) – (\frac{y^{3}}{3} + \frac{7}{3}y^{2} + \frac{1}{2}y + \frac{1}{2})$

= $\frac{1}{3} – \frac{5}{3}y^{2} – \frac{y^{3}}{3} – \frac{7}{3}y^{2} – \frac{1}{2}y – \frac{1}{2}$

= $\frac{1}{3} – \frac{1}{2} – \frac{y}{2} – \frac{5}{3}y^{2} – \frac{7}{3}y^{2} – \frac{y^{3}}{3}$                        (Collecting like terms)

= $(\frac{2-3}{6}) – \frac{y}{2} + (\frac{-5-7}{3})y^{2} – \frac{7}{3}y^{2} – \frac{y^{3}}{3}$

= $-\frac{1}{6} – \frac{y}{2} – 4y^{2} – \frac{y^{3}}{3}$                   (Combining like terms)

(iv) The difference is given by:

$(\frac{3}{2}ab – \frac{7}{4}ac – \frac{5}{6}bc) – (\frac{2}{3}ac – \frac{5}{7}ab + \frac{2}{3}bc)$

= $\frac{3}{2}ab – \frac{7}{4}ac – \frac{5}{6}bc – \frac{2}{3}ac + \frac{5}{7}ab – \frac{2}{3}bc$

= $\frac{3}{2}ab + \frac{5}{7}ab – \frac{7}{4}ac – \frac{2}{3}ac – \frac{5}{6}bc – \frac{2}{3}bc$                         (Collecting like terms)

= $(\frac{21+10}{14})ab + (\frac{-21-8}{12})ac + (\frac{-5-4}{6})bc$

= $\frac{31}{14}ab – \frac{29}{12}ac – \frac{3}{2}bc$                        (Combining like terms)

Q4: 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.

Q5) Subtract the sum of 3l – 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3n2 and –3l + m + 4n2.

Solution:

We have to subtract the sum of (3l – 4m – 7n2) and (2l + 3m – 4n2) from the sum of (9l + 2m – 3n2) and (–3l + m + 4n2)

{(9l + 2m – 3n2) + (–3l + m + 4n2)} – {(3l – 4m – 7n2) + (2l + 3m – 4n2)}

= (9l – 3l + 2m + m – 3n2 + 4n2) – (3l + 2l – 4m + 3m – 7n2 – 4n2)

= (6l + 3m + n2) – (5l – m – 11n2)                   (Combining like terms inside the parenthesis)

= 6l + 3m + n2 – 5l + m + 11n2

= 6l – 5l + 3m + m + n2 + 11n2                           (Collecting like terms)

= l + 4m + 12n2                                               (Combining like terms)

Thus, the required solution is l + 4m + 12n2.

Q6) 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.

Q7) Simplify each of the following:

(i) $x^{2} – 3x + 5 – \frac{1}{2}(3x^{2} – 5x + 7)$

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

(iii) $\frac{11}{2}x^{2}y – \frac{9}{4}xy^{2} + \frac{1}{4}xy – \frac{1}{14}y^{2}x + \frac{1}{15}yx^{2} + \frac{1}{2}xy$

(iv) $(\frac{1}{3}y^{2} – \frac{4}{7}y + 11) – (\frac{1}{7}y – 3 + 2y^{2}) – (\frac{2}{7}y – \frac{2}{3}y^{2} + 2)$

(v) $-\frac{1}{2}a^{2}b^{2}c + \frac{1}{3}ab^{2}c – \frac{1}{4}abc^{2} – \frac{1}{5}cb^{2}a^{2} + \frac{1}{6}cb^{2}a – \frac{1}{7}c^{2}ab + \frac{1}{8}ca^{2}b$.

Solution:

(i) $x^{2} – 3x + 5 – \frac{1}{2}(3x^{2} – 5x + 7)$

= $x^{2} – 3x + 5 – \frac{3x^{2}}{2} + \frac{5x}{2} – \frac{7}{2}$

= $x^{2} – \frac{3x^{2}}{2} – 3x + \frac{5x}{2} + 5 – \frac{7}{2}$               (Collecting like terms)

= $(\frac{1-3}{2})x^{2} + (\frac{-3+5}{2})x + (\frac{10-7}{2})$

= $\frac{-x^{2}}{2} – \frac{x}{2} + \frac{3}{2}$

Thus, the answer is $\frac{-x^{2}}{2} – \frac{x}{2} + \frac{3}{2}$.

(ii) [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 – 9

= 5 – 9 – 5x – 3x + 3y + 7y = -4 – 8x + l0y

(iii) $\frac{11}{2}x^{2}y – \frac{9}{4}xy^{2} + \frac{1}{4}xy – \frac{1}{14}y^{2}x + \frac{1}{15}yx^{2} + \frac{1}{2}xy$

= $\frac{11}{2}x^{2}y + \frac{1}{15}yx^{2} – \frac{9}{4}xy^{2} – \frac{1}{14}y^{2}x + \frac{1}{4}xy + \frac{1}{2}xy$                          (Collecting like terms)

= $(\frac{165+2}{30})x^{2}y + (\frac{-63-2}{28})xy^{2} + (\frac{1+2}{4})xy$

= $\frac{167}{30}x^{2}y – \frac{65}{28}xy^{2} + \frac{3}{4}xy$                (Combining like terms)

(iv) $(\frac{1}{3}y^{2} – \frac{4}{7}y + 11) – (\frac{1}{7}y – 3 + 2y^{2}) – (\frac{2}{7}y – \frac{2}{3}y^{2} + 2)$

= $\frac{1}{3}y^{2} – \frac{4}{7}y + 11 – \frac{1}{7}y + 3 – 2y^{2} – \frac{2}{7}y + \frac{2}{3}y^{2} – 2$

= $\frac{1}{3}y^{2} + \frac{2}{3}y^{2} – 2y^{2} – \frac{4}{7}y – \frac{1}{7}y – \frac{2}{7}y + 11 + 3 – 2$                                   (Collecting like terms)

= $(\frac{1-6+2}{3})y^{2} + (\frac{-4-1-2}{7})y + 12$

= –y2 – 7y + 12                                               (Combining like terms)

(v) $-\frac{1}{2}a^{2}b^{2}c + \frac{1}{3}ab^{2}c – \frac{1}{4}abc^{2} – \frac{1}{5}cb^{2}a^{2} + \frac{1}{6}cb^{2}a – \frac{1}{7}c^{2}ab + \frac{1}{8}ca^{2}b$

= $-\frac{1}{2}a^{2}b^{2}c – \frac{1}{5}cb^{2}a^{2} + \frac{1}{3}ab^{2}c + \frac{1}{6}cb^{2}a – \frac{1}{4}abc^{2} – \frac{1}{7}c^{2}ab + \frac{1}{8}ca^{2}b$               (Collecting like terms)

= $(\frac{-5-2}{10})a^{2}b^{2}c + (\frac{2+1}{6})ab^{2}c + (\frac{-7-4}{28})c^{2}ab + \frac{1}{8}ca^{2}b$

= $-\frac{7}{10}a^{2}b^{2}c + \frac{1}{2}ab^{2}c – \frac{11}{28}abc^{2} + \frac{1}{8}a^{2}bc$         (Combining like terms)

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