RD Sharma Solutions Class 8 Linear Equation In One Variable Exercise 9.2

RD Sharma Solutions Class 8 Chapter 9 Exercise 9.2

RD Sharma Class 8 Solutions Chapter 9 Ex 9.2 PDF Free Download

RD Sharma Solutions Class 8 Chapter 9 Exercise 9.2

Solve each of the following equations and also verify your solutions:

Q1 \(\frac{2x + 5}{3}\) = 3x – 10

Sol:

\(\frac{2x + 5}{3}\) = 3x – 10

=> 2x + 5 = 9x – 30

=> 9x – 2x = 5 + 30

=> 7x = 35

=> x = \(\frac{35}{7}\)

=> x = 5

Verification

L.H.S = \(\frac{10 + 5}{3}\)

= \(\frac{15}{3}\)

= 5

R.H.S = 15 – 10

= 5

Hence, L.H.S = R.H.S

Q2 \(\frac{a – 8}{3}\) = \(\frac{a – 3}{2}\)

Sol:

\(\frac{a – 8}{3}\) = \(\frac{a – 3}{2}\)

=> 2a – 16 = 3a – 9

=> 3a – 2a = 9 – 16

=> a = -7

Verification

L.H.S = \(\frac{-7 – 8}{3}\)

= \(\frac{-15}{3}\)

= -5

R.H.S = \(\frac{-7 – 3}{2}\)

= \(\frac{-10}{2}\)

= -5

Hence, L.H.S = R.H.S

Q3 \(\frac{7y + 2}{5}\) = \(\frac{6y – 5}{11}\)

Sol:

\(\frac{7y + 2}{5}\) = \(\frac{6y – 5}{11}\)

=> 77y + 22 = 30y – 25

=> 77y – 30y = -25 – 22

=>    47y = -47

=>        y = 1

Verification

L.H.S = \(\frac{-7 + 2}{5}\)

= \(\frac{-5}{5}\)

= -1

R.H.S = \(\frac{-6 – 5}{5}\)

= \(\frac{-11}{11}\)

= -1

Hence, L.H.S = R.H.S

Q4. x – 2x + 2 – \(\frac{16}{3}\)x + 5 = 3 – \(\frac{7}{2}\)x

Sol:

x – 2x + 2 – \(\frac{16}{3}\)x + 5 = 3 – \(\frac{7}{2}\)x

=> \(\frac{3x – 6x + 6 – 16x + 15 }{3}\) = \(\frac{6 – 7x}{2}\)

=> \(\frac{-19x + 21}{3}\) = \(\frac{6 – 7x}{2}\)

=> -38x + 42 = 18 – 21x

=> 38x – 21x = 42 – 18

=>  17x = 24

=>     x = \(\frac{24}{17}\)

Verification

L.H.S = \(\frac{24}{17}\) – 2(\(\frac{24}{17}\)) + 7 – \(\frac{16}{3}\)( \(\frac{24}{17}\))

= \(\frac{-33}{17}\)

R.H.S = 3 – \(\frac{7}{2}\)( \(\frac{24}{17}\))

= \(\frac{-33}{17}\)

Hence, L.H.S = R.H.S

Q5. \(\frac{1}{2}\)x + 7x – 6 = 7x + \(\frac{1}{4}\)

Sol:

\(\frac{1}{2}\)x + 7x – 6 = 7x + \(\frac{1}{4}\)

=> \(\frac{1}{2}\)x + 7x – 7x = \(\frac{1}{4}\) + 6

=> \(\frac{x}{2}\) = \(\frac{1 + 24}{4}\)

=> \(\frac{x}{2}\) = \(\frac{25}{4}\)

=> x = \(\frac{25}{2}\)

Verification

L.H.S = \(\frac{1}{2}\)( \(\frac{25}{2}\)) + 7(\(\frac{25}{2}\)) – 6

= \(\frac{351}{4}\)

R.H.S = \(\frac{351}{4}\)

Hence, L.H.S = R.H.S

Q6. \(\frac{3}{4}\)x + 4x = \(\frac{7}{8}\) + 6x – 6

Sol:

\(\frac{3}{4}\)x + 4x = \(\frac{7}{8}\) + 6x – 6

=> \(\frac{3}{4}\)x – 2x = \(\frac{7}{8}\) – 6

=> \(\frac{3x – 8x}{4}\) = \(\frac{7 – 48}{8}\)

=> \(\frac{-5x}{4}\) = \(\frac{-41}{8}\)

=>  -40x = -164

=>    x = \(\frac{164}{40}\)

=>     x = \(\frac{41}{10}\)

Verification

LH.S = \(\frac{3}{4}\)( \(\frac{41}{10}\)) + 4(\(\frac{41}{10}\))

= \(\frac{123}{40}\) + \(\frac{164}{10}\)

= \(\frac{123 + 656}{40}\)

= \(\frac{779}{40}\)

R.H.S = \(\frac{7}{8}\) + 6(\(\frac{41}{10}\)) – 6

= \(\frac{7}{8}\) + \(\frac{246}{10}\) – 6

= \(\frac{35 + 984 – 240}{40}\)

= \(\frac{779}{40}\)

Hence, L.H.S = R.H.S

Q7 \(\frac{7}{2}\)x – \(\frac{5}{2}\)x = \(\frac{20}{3}\)x + 10

Sol:

\(\frac{7}{2}\)x – \(\frac{5}{2}\)x = \(\frac{20}{3}\)x + 10

=> \(\frac{7x – 5x}{2}\) = \(\frac{20x + 30}{3}\)

=> 40x + 60 = 6x

=>  40x – 6x = 60

=>    34x = -60

=>        x = \(\frac{-60}{34}\)

=>        x = \(\frac{-30}{17}\)

Verification

L.H.S = \(\frac{7}{2}\)( \(\frac{-30}{7}\)) – \(\frac{5}{2}\)( \(\frac{-30}{17}\))

= \(\frac{-30}{17}\)

R.H..S = \(\frac{20}{3}\)( \(\frac{-30}{17}\)) + 10

= \(\frac{-30}{17}\)

Hence, L.H.S = R.H.S

Q8 \(\frac{6x + 1}{2}\) + 1 = \(\frac{7x – 3}{3}\)

Sol:

\(\frac{6x + 1}{2}\) + 1 = \(\frac{7x – 3}{3}\)

=> \(\frac{6x + 1 + 2}{2}\) = \(\frac{7x – 3}{3}\)

=> 18x + 9 = 14x – 6

=> 18x – 14x = -6 – 9

=> 4x = -15

=>    x = \(\frac{-15}{4}\)

Verification

L.H.S = \(\frac{6(\frac{-15}{4} + 1)}{2}\)

= \(\frac{-45 + 2 + 4}{4}\)

= \(\frac{-39}{4}\)

R.H.S = \(\frac{7(\frac{-15}{4} – 3)}{3}\)

= \(\frac{-105 – 12}{12}\)

= \(\frac{-39}{4}\)

Hence, L.H.S = R.H.S

Q9. \(\frac{3a – 2}{3}\) + \(\frac{2a + 3}{2}\) = a + \(\frac{7}{6}\)

Sol:

\(\frac{3a – 2}{3}\) + \(\frac{2a + 3}{2}\) = a + \(\frac{7}{6}\)

=> \(\frac{6a – 4a + 6a + 9}{6}\) = a + \(\frac{7}{6}\)

=> 12a + 5 = 6a + 7

=> 12a – 6a = 7 – 5

=>        6a = 2

=>          a = \(\frac{2}{6}\)

=>           a = \(\frac{1}{3}\)

Verification

L.H.S = \(\frac{3(\frac{1}{3} – 2)}{3}\) + \(\frac{2(\frac{1}{3} + 3)}{2}\)

= \(\frac{-1}{3}\) + \(\frac{11}{6}\)

= \(\frac{9}{6}\)

= \(\frac{3}{2}\)

R.H.S = \(\frac{1}{3}\) + \(\frac{7}{6}\)

= \(\frac{9}{6}\)

= \(\frac{3}{2}\)

Q10. x – \(\frac{x – 1}{2}\) = 1 – \(\frac{x – 2}{3}\)

Sol:

x – \(\frac{x – 1}{2}\) = 1 – \(\frac{x – 2}{3}\)

=> \(\frac{2x – x + 1}{2}\) = \(\frac{3 – x + 2}{3}\)

=> \(\frac{x + 1}{2}\) = \(\frac{5 – x}{3}\)

=> 3x + 3 = 10 – 2x

=> 3x + 2x = 10 – 3

=>         5x = 7

=>           x = \(\frac{7}{5}\)

L.H.S = \(\frac{7}{5}\)\(\frac{\frac{7}{5} – 1}{2}\)

= \(\frac{7}{5}\)\(\frac{1}{5}\)

= \(\frac{6}{5}\)

R.H.S = 1 – \(\frac{\frac{7}{5} – 2}{3}\)

= 1 – \(\frac{-3}{5}\)

= \(\frac{6}{5}\)

Hence, L.H.S = R.H.S

Q11 \(\frac{3}{4}\)x – \(\frac{x – 1}{2}\) = \(\frac{x – 2}{3}\)

Sol:

\(\frac{3}{4}\)x – \(\frac{x – 1}{2}\) = \(\frac{x – 2}{3}\)

=> \(\frac{3x – 2x + 2}{4}\) = \(\frac{x – 2}{3}\)

=> 4x – 8 = 3x + 6

=> 4x – 3x = 6 + 8

=>       x = 14

Verification

L.H.S = \(\frac{3 \times 14}{4}\)\(\frac{14 – 1}{2}\)

= \(\frac{21}{2}\)\(\frac{13}{2}\)

= \(\frac{8}{2}\)

= 4

R.H.S = \(\frac{14 – 2}{3}\)

= \(\frac{12}{3}\)

= 4

Hence, L.H.S = R.H.S

Q12 \(\frac{5x}{3}\)\(\frac{x – 1}{4}\) = \(\frac{x – 3}{5}\)

=> \(\frac{20x – 3x + 3}{12}\) = \(\frac{x – 3}{5}\)

=> \(\frac{17x + 3}{12}\) = \(\frac{x – 3}{5}\)

=> 85x + 15 = 21x – 36

=> 85x – 12x = -36 – 15

=> 73x = -51

=>       x = \(\frac{-51}{73}\)

Verification

L.H.S = \(\frac{5(\frac{-51}{73})}{3}\)\(\frac{\frac{-51}{73} – 1}{4}\)

= \(\frac{-225}{219}\)\(\frac{-124}{292}\)

= \(\frac{-54}{73}\)

R.H.S = \(\frac{\frac{-51}{73} – 3}{5}\)

= \(\frac{-54}{73}\)

Hence, L.H.S = R.H.S

Q13 \(\frac{3x + 1}{16}\) + \(\frac{2x – 3}{7}\) = \(\frac{x + 3}{8}\) + \(\frac{3x – 1}{14}\)

Sol:

\(\frac{3x + 1}{16}\) + \(\frac{2x – 3}{7}\) = \(\frac{x + 3}{8}\) + \(\frac{3x – 1}{14}\)

=> \(\frac{3x + 1}{16}\)\(\frac{x + 3}{8}\) = \(\frac{3x – 1}{14}\)\(\frac{2x – 3}{7}\)

=> \(\frac{3x + 1 – 2x – 6}{16}\) = \(\frac{3x – 1 – 4x + 6}{14}\)

=> \(\frac{x – 5}{8}\) = \(\frac{-x + 5}{7}\)

=> 7x – 35 = -8x + 40

=> 7x + 8x = 40 + 35

=>  15x = 75

=>        x = \(\frac{75}{15}\)

= 5

Verification

L.H.S = \(\frac{3(5) + 1}{16}\) + \(\frac{2(5) – 3}{7}\)

\(\frac{16}{16}\) + \(\frac{7}{7}\)

= 2

R.H.S = \(\frac{5 + 3}{8}\) + \(\frac{3(5) – 1}{14}\)

= \(\frac{8}{8}\) + \(\frac{14}{14}\)

= 2

Hence, L.H.S = R.H.S

Q14 \(\frac{1 – 2x}{7}\)\(\frac{2 – 3x}{8}\) = \(\frac{3}{2}\) + \(\frac{x}{4}\)

Sol:

\(\frac{1 – 2x}{7}\)\(\frac{2 – 3x}{8}\) = \(\frac{3}{2}\) + \(\frac{x}{4}\)

=> \(\frac{1 – 2x}{7}\) = \(\frac{3}{2}\) + \(\frac{x}{4}\) + \(\frac{2 – 3x}{8}\)

=> \(\frac{1 – 2x}{7}\) = \(\frac{12 + 2x + 2 – 3x}{8}\)

=> \(\frac{1 – 2x}{7}\) = \(\frac{14 – x}{8}\)

=> 8 – 16x = 98 – 7x

=>   16x – 7x = 8 – 98

=>       9x = -90

=>         x = \(\frac{-90}{9}\)

Verification

L.H.S = \(\frac{1 – 2(-10)}{7}\)\(\frac{2 – 3(-10)}{8}\)

= \(\frac{1 + 20}{7}\)\(\frac{2 + 30}{8}\)

= 3 – 4

= -1

R.H.S = \(\frac{3}{2}\) + \(\frac{-10}{4}\)

= \(\frac{3}{2}\) + \(\frac{-5}{2}\)

= \(\frac{3 – 5}{2}\)

= -1

Hence, L.H.S = R.H.S

Q15 \(\frac{9x + 7}{2}\) – (x – \(\frac{x – 2}{7}\) = 36

Sol:

\(\frac{9x + 7}{2}\) – (x – \(\frac{x – 2}{7}\) = 36

=> \(\frac{63x + 49 – 14x + 2x – 4}{14}\) = 36

=> \(\frac{51x + 45}{14}\) = 36

=> 51x + 45 = 504

=>    51x = 504 – 45

=>       51x = 459

=>           x = \(\frac{459}{51}\)

= 9

Verification

L.H.S = \(\frac{9(9) + 7}{7} – (9 – \frac{9 – 2}{7})\)

= \(\frac{88}{2}\) – 9 + \(\frac{7}{7}\)

= 44 – 9 + 1

= 36

R.H.S = 36

Hence, L.H.S = R.H.S

Q16 0.18(5x – 4) = 0.5x + 0.8

Sol:

0.18(5x – 4) = 0.5x + 0.8

=> 0.9x – 0.72 = 0.5x + 08

=> 0.9x – 0.5x = 0.8 + 0.72

=> 0.4x = 1.52

=>       x = \(\frac{1.52}{0.4}\)

= 3.8

Verification

L.H.S = 0.18(5(3.8) – 4)

= 0.18 \(\times\) 15

= 2.7

R.H.S = 0.5(3.8) + 0.8

= 2.7

Hence, L.H.S = R.H.S

Q17 \(\frac{2}{3x}\)\(\frac{3}{2x}\) = \(\frac{1}{2}\)

Sol:

\(\frac{2}{3x}\)\(\frac{3}{2x}\) = \(\frac{1}{2}\)

=> \(\frac{4 – 9}{6x}\) = \(\frac{1}{12}\)

=> \(\frac{-5}{6x}\) = \(\frac{1}{12}\)

=> 6x = -60

=>    x = \(\frac{-60}{6}\)

=>     x = -10

Verification

L.H.S = \(\frac{2}{3(-10)}\)\(\frac{3}{2(-10)}\)

= \(\frac{2}{-30}\)\(\frac{3}{-20}\)

= \(\frac{-4 + 9}{60}\)

= \(\frac{5}{60}\)

= \(\frac{1}{12}\)

R.H.S = \(\frac{1}{12}\)

Hence, L.H.S = R.H.S

Q18 \(\frac{4x}{9}\) + \(\frac{1}{3}\) + \(\frac{13}{108}\)x = \(\frac{8x + 19}{18}\)   

Sol:

\(\frac{4x}{9}\) + \(\frac{1}{3}\) + \(\frac{13}{108}\)x = \(\frac{8x + 19}{18}\)

=> \(\frac{48x + 36 + 13x}{10}\) = \(\frac{8x + 19}{18}\)

=> \(\frac{61x + 36}{108}\) = \(\frac{8x + 19}{18}\)

Multiply both sides by 108

=> 61x + 36 = 48x + 114

=> 61x – 48x = 114 – 36

=>    13x = 78

=>        x = \(\frac{78}{13}\)

=>        x = 6

Verification

L.H.S = \(\frac{4(6)}{9}\) + \(\frac{1}{3}\) + \(\frac{13}{108}\)(6)

= \(\frac{24}{9}\) + \(\frac{1}{3}\) + \(\frac{13}{18}\)

= \(\frac{48 + 6 + 13}{18}\)

= \(\frac{67}{18}\)

R.H.S = \(\frac{8(6) + 19}{18}\)

= \(\frac{67}{18}\)

Q19 \(\frac{45 – 2x}{15}\)\(\frac{4x + 10}{5}\) = \(\frac{15 – 14x}{9}\)

Sol:

\(\frac{45 – 2x}{15}\)\(\frac{4x + 10}{5}\) = \(\frac{15 – 14x}{9}\)

Multiply by ‘3’

=> \(\frac{45 – 2x – 12x – 30}{15}\) = \(\frac{15 – 14x}{3}\)

=> \(\frac{15 – 14x}{5}\) = \(\frac{15 – 14x}{3}\)

=> 45 – 42x = 75 – 70x

=> 70x – 42x = 75 – 45

=> 28x = 30

=>    x = \(\frac{30}{28}\)

=>    x = \(\frac{15}{24}\)

Verification

L.H.S = \(\frac{45 – 2(\frac{15}{14})}{15}\)\(\frac{45(\frac{15}{14}) + 10}{5}\)

= \(\frac{45(7) – 15}{105}\)\(\frac{30 + 70}{35}\)

= \(\frac{300}{105}\)\(\frac{100}{35}\)

= 0

R.H.S = \(\frac{15 – 14(\frac{15}{14})}{9}\)

= 0

Hence, L.H.S = R.H.S

Q20 5(\(\frac{7x + 5}{3}\)) – \(\frac{23}{3}\) = 13 – \(\frac{4x – 2 }{3}\)

Sol:

5(\(\frac{7x + 5}{3}\)) – \(\frac{23}{3}\) = 13 – \(\frac{4x – 2 }{3}\)

=> \(\frac{35x + 25}{3}\) + \(\frac{4x – 2}{3}\) = 13 + \(\frac{23}{3}\)

=> \(\frac{35x + 25 + 4x – 2}{3}\) = \(\frac{39}{23}\)

Multiply by ‘3’

=> 39x + 23 = 62

=> 39x = 62 – 23

=> 39x = 39

=>     x = 1

Verification

L.H.S = 15(\(\frac{7(5) + 5}{3}\)\(\frac{23}{3}\)

= \(\frac{60}{3}\)\(\frac{23}{3}\)

= \(\frac{37}{3}\)

R.H.S = 13 – \(\frac{4(1) – 2}{3}\)

= \(\frac{39 – 2}{3}\)

= \(\frac{37}{3}\)

Hence, L.H.S = R.H.S

Q21 \(\frac{7x – 1}{4}\)\(\frac{1}{3}\)(2x – \(\frac{1 – x}{2}\) = \(\frac{10}{3}\)

Sol:

\(\frac{7x – 1}{4}\)\(\frac{1}{3}\)(2x – \(\frac{1 – x}{2}\) = \(\frac{10}{3}\)

=> \(\frac{7x – 1}{4}\)\(\frac{2x}{3}\) + \(\frac{1 – x}{3}\) = \(\frac{10}{3}\)

=> \(\frac{21x – 3 – 8x + 2 – 2x}{12}\) = \(\frac{10}{3}\)

=> 11x – 1 = 40

=>  11x = 41

=>      x = \(\frac{41}{11}\)

Verification

L.H.S = \(\frac{75(\frac{41}{11}) – 1}{4} – \frac{1}{3}(2(\frac{41}{11}) – \frac{1 – \frac{41}{11}}{2})\)

= \(\frac{276}{44}\)\(\frac{82}{33}\) + \(\frac{-30}{66}\)

= \(\frac{10}{3}\)

R.H.S = \(\frac{10}{3}\)

Hence, LH.S = R.H.S

Q22 \(\frac{0.5(x – 0.4)}{0.35}\)\(\frac{0.6(x – 2.71)}{0.42}\) = x + 61

Sol:

\(\frac{0.5(x – 0.4)}{0.35}\)\(\frac{0.6(x – 2.71)}{0.42}\) = x + 61

=> \(\frac{x – 0.4}{0.7}\)\(\frac{x – 2.71}{0.7}\) = x + 6.1

=> \(\frac{x – 0.4 – x + 2.71}{0.7}\) = x + 6.1

=> -0.4 + 2.71 = 0.7x + 4.27

=> 0.7x = 2.71 – 0.4 – 4.27

=> 0.7x = -1.96

=>      x = \(\frac{-1.96}{0.7}\)

=>      x = -2.8

Verification

L.H.S = \(\frac{0.5((-2.8) – 0.4)}{0.35}\)\(\frac{0.6((-2.8) – 2.71)}{0.42}\)

= \(\frac{-1.6}{0.35}\) + \(\frac{3.306}{0.42}\)

= -4.571 + 7.871

= 3.3

R.H.S = -2.8 + 6.1

= 3.3

Hence, L.H.S = R.H.S

Q23 6.5x + \(\frac{19.5x – 32.5}{2}\) = 6.5x + 13 + \(\frac{13x – 26}{2}\)

Sol:

6.5x + \(\frac{19.5x – 32.5}{2}\) = 6.5x + 13 + \(\frac{13x – 26}{2}\)

=> \(\frac{19.5x – 32.5}{2}\)\(\frac{13x – 26}{2}\) = 13

=> \(\frac{19.5x – 32.5 – 13x + 26}{2}\) = 13

=> 6.5x – 6.5 = 26

=> 6.5x = 26 + 6.5

=> 6.5x = 32.5

=>      x = \(\frac{32.5}{6.5}\)

= 5

Verification

L.H.S = 6.5(5) + \(\frac{19.5(5) – 32.5}{2}\)

= 65

R.H.S = 6.5(5) + 13 + \(\frac{13(5) – 26}{2}\)

= 65

Hence, L.H.S = R.H.S

Q24 (3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)

Sol:

(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)

=> \(9x^{2} + 6x – 24x – 16 – 8x^{2} – 4x + 22x + 11 = x^{2} + 7x – 3x – 21\)

=> \(x^{2} – 5 = x^{2} + 4x – 21\)

=> 4x = 21 – 5

=>   4x = 16

=>     x = \(\frac{16}{4}\)

= 4

Verification

L.H.S = (3(4) – 8) (3(4) + 2) – (4(4) – 11) (2(4) + 1)

= 4(16) – 5(9)

= 11

R.H.S = (4 – 3) (4 + 7)

= 11

Hence, L.H.S = R.H.S

Q24 \([(2x + 3) + (x + 5)]^{2} + [(2x + 3) – (x + 5)]^{2}\) = \(10x^{2} + 92\)

Sol:

\([(2x + 3) + (x + 5)]^{2} + [(2x + 3) – (x + 5)]^{2}\) = \(10x^{2} + 92\)

=> \((3x + 8)^{2} + (x – 2)^{2} = 10x^{2} + 92\)

=> \(9x^{2} + 48x + 64x + x^{2} – 4x + 4 = 10x^{2} + 92\)

=> \(10x^{2} – 10x^{2} + 44x = 92 – 68\)

=> 44x = 24

=> x = \(\frac{24}{44}\)

=> x = \(\frac{6}{11}\)

Verification

L.H.S = \([(2(\frac{6}{11}) + 3) + (\frac{6}{11} + 5)]^{2}\) + \([(2(\frac{6}{11}) + 3) – (\frac{6}{11} + 5)]^{2}\)

= \([(\frac{45}{11}) + (\frac{61}{11})]^{2}\) + \([(\frac{45}{11}) – (\frac{61}{11})]^{2}\)

= \((\frac{106}{11})^{2} + (\frac{-16}{11})^{2}\)

= \(\frac{11492}{121}\)

R.H.S = \(10(\frac{6}{11})^{2} + 92\)

= \(\frac{360}{121}\) + 92

= \(\frac{11492}{121}\)

Hence, L.H.S = R.H.S


Practise This Question

Determine the line containing the point (2,3)