RD Sharma 2019 Solutions for Class 8 Math Chapter 9 - Linear Equation In One Variable

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RD Sharma 2019 Solutions for Class 8 Math Chapter 9 Linear Equation In One Variable are provided here with simple step-by-step explanations. These solutions for Linear Equation In One Variable are extremely popular among class 8 students for Math Linear Equation In One Variable Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2019 Book of class 8 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2019 Solutions. All RD Sharma 2019 Solutions for class 8 Math are prepared by experts and are 100% accurate.

Page No 9.11:

Question 1:

Solve each of the following equation and also check your result in each case:
$\frac{2 x + 5}{3} = 3 x - 10$

Answer:

$\frac{2 x + 5}{3} = 3 x - 10 or 2 x + 5 = 9 x - 30 or 9 x - 2 x = 5 + 30 or 7 x = 35 or x = \frac{35}{7} or x = 5 Verification : L . H . S . = \frac{10 + 5}{3} = \frac{15}{3} = 5 R . H . S . = 15 - 10 = 5 ∴ L . H . S . = R . H . S . for x = 5 .$

Page No 9.11:

Question 2:

Solve each of the following equation and also check your result in each case:
$\frac{a - 8}{3} = \frac{a - 3}{2}$

Answer:

$\frac{a - 8}{3} = \frac{a - 3}{2} or 2 a - 16 = 3 a - 9 or 3 a - 2 a = - 16 + 9 or a = - 7 Verification : L . H . S . = \frac{- 7 - 8}{3} = \frac{- 15}{3} = - 5 R . H . S . = \frac{- 7 - 3}{2} = \frac{- 10}{2} = - 5$

$∴$ L.H.S. = R.H.S. for a = $-$ 7

Page No 9.11:

Question 3:

Solve each of the following equation and also check your result in each case:
$\frac{7 y + 2}{5} = \frac{6 y - 5}{11}$

Answer:

$\frac{7 y + 2}{5} = \frac{6 y - 5}{11} or 77 y + 22 = 30 y - 25 or 77 y - 30 y = - 25 - 22 or 47 y = - 47 or y = \frac{- 47}{47} = - 1 Verification : L . H . S . = \frac{- 7 + 2}{5} = \frac{- 5}{5} = - 1 R . H . S . = \frac{- 6 - 5}{11} = \frac{- 11}{11} = - 1$

$∴$ L.H.S. = R.H.S. for y = $-$ 1

Page No 9.11:

Question 4:

Solve each of the following equation and also check your result in each case:
$x - 2 x + 2 - \frac{16}{3} x + 5 = 3 - \frac{7}{2} x$

Answer:

$x - 2 x + 2 - \frac{16}{3} x + 5 = 3 - \frac{7}{2} x or \frac{3 x - 6 x + 6 - 16 x + 15}{3} = \frac{6 - 7 x}{2} or \frac{- 19 x + 21}{3} = \frac{6 - 7 x}{2} or - 38 x + 42 = 18 - 21 x or - 21 x + 38 x = 42 - 18 or 17 x = 24 or x = \frac{24}{17} Check : L . H . S . = \frac{24}{17} - 2 \times \frac{24}{17} + 7 - \frac{16}{3} \times \frac{24}{17} = \frac{- 33}{17} R . H . S . = 3 - \frac{7}{2} \times \frac{24}{17} = \frac{- 33}{17}$

$∴$ L.H.S. = R.H.S. for x = $\frac{24}{17}$

Page No 9.11:

Question 5:

Solve each of the following equation and also check your result in each case:
$\frac{1}{2} x + 7 x - 6 = 7 x + \frac{1}{4}$

Answer:

$\frac{1}{2} x + 7 x - 6 = 7 x + \frac{1}{4} or \frac{1}{2} x + 7 x - 7 x = \frac{1}{4} + 6 or \frac{x}{2} = \frac{1 + 24}{4} or \frac{x}{2} = \frac{25}{4} or x = \frac{25}{2} Check : L . H . S . = \frac{1}{2} \times \frac{25}{2} + 7 \times \frac{25}{2} - 6 = \frac{351}{4} R . H . S . = 7 \times \frac{25}{2} + \frac{1}{4} = \frac{351}{4}$

$∴$ L.H.S. = R.H.S. for x = $\frac{25}{2}$

Page No 9.11:

Question 6:

Solve each of the following equation and also check your result in each case:
$\frac{3}{4} x + 4 x = \frac{7}{8} + 6 x - 6$

Answer:

$\frac{3}{4} x + 4 x = \frac{7}{8} + 6 x - 6 or \frac{3}{4} x - 2 x = \frac{7}{8} - 6 or \frac{3 x - 8 x}{4} = \frac{7 - 48}{8} or \frac{- 5 x}{4} = \frac{- 41}{8} or - 40 x = - 164 or x = \frac{- 164}{- 40} = \frac{41}{10} Check : L . H . S . = \frac{3}{4} \times \frac{41}{10} + 4 \times \frac{41}{10} = \frac{123}{40} + \frac{164}{10} = \frac{123 + 656}{40} = \frac{779}{40} R . H . S . = \frac{7}{8} + 6 \times \frac{41}{10} - 6 = \frac{7}{8} + \frac{246}{10} - 6 = \frac{35 + 984 - 240}{40} = \frac{779}{40}$

$∴$ L.H.S. = R.H.S. for x = $\frac{41}{10}$

Page No 9.11:

Question 7:

Solve each of the following equation and also check your result in each case:
$\frac{7}{2} x - \frac{5}{2} x = \frac{20}{3} x + 10$

Answer:

$\frac{7}{2} x - \frac{5}{2} x = \frac{20}{3} x + 10 or \frac{7 x - 5 x}{2} = \frac{20 x + 30}{3} or 40 x + 60 = 6 x or 34 x = - 60 or x = - \frac{60}{34} = - \frac{30}{17} Check : L . H . S . = \frac{7}{2} \times \frac{- 30}{17} - \frac{5}{2} \times - \frac{30}{17} = - \frac{30}{17} R . H . S . = \frac{20}{3} \times \frac{- 30}{17} + 10 = - \frac{30}{17} ∴ L . H . S . = R . H . S . for x = - \frac{30}{17}$

Page No 9.11:

Question 8:

Solve each of the following equation and also check your result in each case:
$\frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Answer:

$\frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3} or \frac{6 x + 1 + 2}{2} = \frac{7 x - 3}{3} or 18 x + 9 = 14 x - 6 or 18 x - 14 x = - 6 - 9 or 4 x = - 15 or x = \frac{- 15}{4} Check : L . H . S . = \frac{6 \times \frac{- 15}{4} + 1}{2} + 1 = \frac{- 45 + 2 + 4}{4} = \frac{- 39}{4} R . H . S . = \frac{7 \times \frac{- 15}{4} - 3}{3} = \frac{- 105 - 12}{12} = \frac{- 39}{4}$

$∴$ L.H.S. = R.H.S. for x = $\frac{- 15}{4}$

Page No 9.11:

Question 9:

Solve each of the following equation and also check your result in each case:
$\frac{3 a - 2}{3} + \frac{2 a + 3}{2} = a + \frac{7}{6}$

Answer:

$\frac{3 a - 2}{3} + \frac{2 a + 3}{2} = a + \frac{7}{6} or \frac{6 a - 4 + 6 a + 9}{6} = a + \frac{7}{6} or 12 a + 5 = 6 a + 7 or 6 a = 7 - 5 or a = \frac{2}{6} = \frac{1}{3} Check : L . H . S . = \frac{3 \times \frac{1}{3} - 2}{3} + \frac{2 \times \frac{1}{3} + 3}{2} = \frac{- 1}{3} + \frac{11}{6} = \frac{9}{6} = \frac{3}{2} R . H . S . = \frac{1}{3} + \frac{7}{6} = \frac{9}{6} = \frac{3}{2}$

$∴$ L.H.S. = R.H.S. for a = $\frac{1}{3}$

Page No 9.11:

Question 10:

Solve each of the following equation and also check your result in each case:
$x - \frac{(x - 1)}{2} = 1 - \frac{(x - 2)}{3}$

Answer:

$x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} or \frac{2 x - x + 1}{2} = \frac{3 - x + 2}{3} or \frac{x + 1}{2} = \frac{5 - x}{3} or 3 x + 3 = 10 - 2 x or 5 x = 10 - 3 or x = \frac{7}{5} Check : L . H . S . = \frac{7}{5} - \frac{\frac{7}{5} - 1}{2} = \frac{7}{5} - \frac{1}{5} = \frac{6}{5} R . H . S . = 1 - \frac{\frac{7}{5} - 2}{3} = 1 - \frac{- 3}{15} = \frac{6}{5}$

$∴$ L.H.S. = R.H.S. for x = $\frac{7}{5}$

Page No 9.11:

Question 11:

Solve each of the following equation and also check your result in each case:
$\frac{3 x}{4} - \frac{(x - 1)}{2} = \frac{(x - 2)}{3}$

Answer:

$\frac{3 x}{4} - \frac{x - 1}{2} = \frac{x - 2}{3} or \frac{3 x - 2 x + 2}{4} = \frac{x - 2}{3} or 4 x - 8 = 3 x + 6 or x = 14 Check : L . H . S . = \frac{3 \times 14}{4} - \frac{14 - 1}{2} = \frac{21}{2} - \frac{13}{2} = \frac{8}{2} = 4 R . H . S . = \frac{14 - 2}{3} = \frac{12}{3} = 4$

$∴$ L.H.S. = R.H.S. for x = 14

Page No 9.11:

Question 12:

Solve each of the following equation and also check your result in each case:
$\frac{5 x}{3} - \frac{(x - 1)}{4} = \frac{(x - 3)}{5}$

Answer:

$\frac{5 x}{3} - \frac{x - 1}{4} = \frac{x - 3}{5} or \frac{20 x - 3 x + 3}{12} = \frac{x - 3}{5} or \frac{17 x + 3}{12} = \frac{x - 3}{5} or 85 x + 15 = 12 x - 36 or 73 x = - 51 or x = \frac{- 51}{73} Check : L . H . S . = \frac{5 \times \frac{- 51}{73}}{3} - \frac{\frac{- 51}{73} - 1}{4} = \frac{- 255}{219} - \frac{- 124}{292} = \frac{- 54}{73} R . H . S . = \frac{\frac{- 51}{73} - 3}{5} = \frac{- 54}{73}$

$∴$ L.H.S. = R.H.S. for x = $\frac{- 51}{73}$

Page No 9.11:

Question 13:

Solve each of the following equation and also check your result in each case:
$\frac{(3 x + 1)}{16} + \frac{(2 x - 3)}{7} = \frac{(x + 3)}{8} + \frac{(3 x - 1)}{14}$

Answer:

$\frac{3 x + 1}{16} + \frac{2 x - 3}{7} = \frac{x + 3}{8} + \frac{3 x - 1}{14} or \frac{3 x + 1}{16} - \frac{x + 3}{8} = \frac{3 x - 1}{14} - \frac{2 x - 3}{7} or \frac{3 x + 1 - 2 x - 6}{16} = \frac{3 x - 1 - 4 x + 6}{14} or \frac{x - 5}{8} = \frac{- x + 5}{7} or 7 x - 35 = - 8 x + 40 or 15 x = 75 or x = \frac{75}{15} = 5 Check : L . H . S . = \frac{3 \times 5 + 1}{16} + \frac{2 \times 5 - 3}{7} = \frac{16}{16} + \frac{7}{7} = 2 R . H . S . = \frac{5 + 3}{8} + \frac{3 \times 5 - 1}{14} = \frac{8}{8} + \frac{14}{14} = 2$

$∴$ L.H.S. = R.H.S. for x = 5

Page No 9.11:

Question 14:

Solve each of the following equation and also check your result in each case:
$\frac{(1 - 2 x)}{7} - \frac{(2 - 3 x)}{8} = \frac{3}{2} + \frac{x}{4}$

Answer:

$\frac{1 - 2 x}{7} - \frac{2 - 3 x}{8} = \frac{3}{2} + \frac{x}{4} or \frac{1 - 2 x}{7} = \frac{3}{2} + \frac{x}{4} + \frac{2 - 3 x}{8} or \frac{1 - 2 x}{7} = \frac{12 + 2 x + 2 - 3 x}{8} or \frac{1 - 2 x}{7} = \frac{14 - x}{8} or 8 - 16 x = 98 - 7 x or - 16 x + 7 x = 98 - 8 or x = \frac{- 90}{9} = - 10 Check : L . H . S . = \frac{1 - 2 \times (- 10)}{7} - \frac{2 - 3 \times (- 10)}{8} = \frac{1 + 20}{7} - \frac{2 + 30}{8} = 3 - 4 = - 1 R . H . S . = \frac{3}{2} + \frac{- 10}{4} = \frac{3}{2} + \frac{- 5}{2} = \frac{3 - 5}{2} = - 1$

$∴$ L.H.S. = R.H.S. for x = $-$ 10

Page No 9.11:

Question 15:

Solve each of the following equation and also check your result in each case:
$\frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36$

Answer:

$\frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 or \frac{63 x + 49 - 14 x + 2 x - 4}{14} = 36 or \frac{51 x + 45}{14} = 36 or 51 x + 45 = 504 or 51 x = 504 - 45 or x = \frac{459}{51} = 9 Thus, x = 9 is the solution of the given equation . Check : Substituting x = 9 in the given equation, we get : L . H . S . = \frac{9 \times 9 + 7}{2} - (9 - \frac{9 - 2}{7}) = \frac{88}{2} - 9 + \frac{7}{7} = 44 - 9 + 1 = 36 R . H . S . = 36 ∴ L . H . S . = R . H . S . for x = 9 .$

Page No 9.11:

Question 16:

Solve each of the following equation and also check your result in each case:
0.18(5x − 4) = 0.5x + 0.8

Answer:

$0.18 (5 x - 4) = 0.5 x + 0.8 or 0.9 x - 0.72 = 0.5 x + 0.8 or 0.9 x - 0.5 x = 0.8 + 0.72 or 0.4 x = 1.52 or x = \frac{1.52}{0.4} or x = 3.8 Thus, x = 3.8 is the solution of the given equation . Check : Substituting x = 3.8 in the given equation, we get : L . H . S . = 0.18 (5 \times 3.8 - 4) = 0.18 \times 15 = 2.7 R . H . S . = 0.5 \times 3.8 + 0.8 = 2.7 ∴ L . H . S . = R . H . S . for x = 3.8 .$

Page No 9.11:

Question 17:

Solve each of the following equation and also check your result in each case:
$\frac{2}{3 x} - \frac{3}{2 x} = \frac{1}{12}$

Answer:

$\frac{2}{3 x} - \frac{3}{2 x} = \frac{1}{12} or \frac{4 - 9}{6 x} = \frac{1}{12} or \frac{- 5}{6 x} = \frac{1}{12} or 6 x = - 60 or x = \frac{- 60}{6} or x = - 10 Thus, x = - 10 is the solution of the given equation . Check : Substituting x = - 10 in the given equation, we get : L . H . S . = \frac{2}{3 \times (- 10)} - \frac{3}{2 \times (- 10)} = \frac{2}{- 30} - \frac{3}{- 20} = \frac{- 4 + 9}{60} = \frac{5}{60} = \frac{1}{12} R . H . S . = \frac{1}{12} ∴ L . H . S . = R . H . S . for x = - 10 .$

Page No 9.11:

Question 18:

Solve each of the following equation and also check your result in each case:
$\frac{4 x}{9} + \frac{1}{3} + \frac{13}{108} x = \frac{8 x + 19}{18}$

Answer:

$\frac{4 x}{9} + \frac{1}{3} + \frac{13}{108} x = \frac{8 x + 19}{18} or \frac{48 x + 36 + 13 x}{108} = \frac{8 x + 19}{18} or \frac{61 x + 36}{108} = \frac{8 x + 19}{18} or 61 x + 36 = 6 (8 x + 19) [Multiply ing both s ides by 108] or 61 x + 36 = 48 x + 114 or 61 x - 48 x = 114 - 36 or 13 x = 78 or x = \frac{78}{13} or x = 6 Thus, x = 6 is the solution of the given equation . Check : Substituting x = 6 in the given equation, we get : L . H . S . = \frac{4 \times 6}{9} + \frac{1}{3} + \frac{13}{108} \times 6 = \frac{24}{9} + \frac{1}{3} + \frac{13}{18} = \frac{48 + 6 + 13}{18} = \frac{67}{18} R . H . S . = \frac{8 \times 6 + 19}{18} = \frac{67}{18} ∴ L . H . S . = R . H . S . for x = 6 .$

Page No 9.12:

Question 19:

Solve each of the following equation and also check your result in each case:
$\frac{(45 - 2 x)}{15} - \frac{(4 x + 10)}{5} = \frac{(15 - 14 x)}{9}$

Answer:

$\frac{45 - 2 x}{15} - \frac{4 x + 10}{5} = \frac{15 - 14 x}{9} or \frac{45 - 2 x - 12 x - 30}{15} = \frac{15 - 14 x}{9} or \frac{15 - 14 x}{5} = \frac{15 - 14 x}{3} [Multiply ing both sides by 3] or 45 - 42 x = 75 - 70 x [After c ross multipl ication] or 70 x - 42 x = 75 - 45 or 28 x = 30 or x = \frac{30}{28} or x = \frac{15}{14} Thus, x = \frac{15}{14} is the solution of the given equation . Check : Substituting x = \frac{15}{14} in the given equation, we get : L . H . S . = \frac{45 - 2 \times \frac{15}{14}}{15} - \frac{4 \times \frac{15}{14} + 10}{5} = \frac{45 \times 7 - 15}{105} - \frac{30 + 70}{35} = \frac{300}{105} - \frac{100}{35} = 0 R . H . S . = \frac{15 - 14 \times \frac{15}{14}}{9} = 0 ∴ L . H . S . = R . H . S . for x = \frac{15}{14}$

Page No 9.12:

Question 20:

Solve each of the following equation and also check your result in each case:
$5 (\frac{7 x + 5}{3}) - \frac{23}{3} = 13 - \frac{4 x - 2}{3}$

Answer:

$5 (\frac{7 x + 5}{3}) - \frac{23}{3} = 13 - \frac{4 x - 2}{3} or \frac{35 x + 25}{3} + \frac{4 x - 2}{3} = 13 + \frac{23}{3} or \frac{35 x + 25 + 4 x - 2}{3} = \frac{39 + 23}{3} or 39 x + 23 = 62 [Multiply ing both sides by 3] or 39 x = 62 - 23 or x = \frac{39}{39} or x = 1 Thus, x = 1 is the solution of the given equation . Check : Substituting x = 1 in the given equation, we get : L . H . S . = 5 (\frac{7 \times 1 + 5}{3}) - \frac{23}{3} = \frac{60}{3} - \frac{23}{3} = \frac{37}{3} R . H . S . = 13 - \frac{4 \times 1 - 2}{3} = \frac{39 - 2}{3} = \frac{37}{3} ∴ L . H . S . = R . H . S . for x = 1 .$

Page No 9.12:

Question 21:

Solve each of the following equation and also check your result in each case:
$\frac{7 x - 1}{4} - \frac{1}{3} (2 x - \frac{1 - x}{2}) = \frac{10}{3}$

Answer:

$\frac{7 x - 1}{4} - \frac{1}{3} (2 x - \frac{1 - x}{2}) = \frac{10}{3} or \frac{7 x - 1}{4} - \frac{2 x}{3} + \frac{1 - x}{6} = \frac{10}{3} or \frac{21 x - 3 - 8 x + 2 - 2 x}{12} = \frac{10}{3} or 11 x - 1 = 40 [Multiply ing both sides by 12] or 11 x = 40 + 1 or x = \frac{41}{11} Thus, x = \frac{41}{11} is the solution of the given equation . Check : Substituting x = \frac{41}{11} in the given equation, we get : L . H . S . = \frac{7 \times \frac{41}{11} - 1}{4} - \frac{1}{3} (2 \times \frac{41}{11} - \frac{1 - \frac{41}{11}}{2}) = \frac{276}{44} - \frac{82}{33} + \frac{- 30}{66} = \frac{10}{3} R . H . S . = \frac{10}{3} ∴ L . H . S . = R . H . S . for x = \frac{41}{11}$

Page No 9.12:

Question 22:

Solve each of the following equation and also check your result in each case:
$\frac{0.5 (x - 0.4)}{0.35} - \frac{0.6 (x - 2.71)}{0.42} = x + 6.1$

Answer:

$\frac{0.5 (x - 0.4)}{0.35} - \frac{0.6 (x - 2.71)}{0.42} = x + 6.1 or \frac{(x - 0.4)}{0.7} - \frac{(x - 2.71)}{0.7} = x + 6.1 or \frac{x - 0.4 - x + 2.71}{0.7} = x + 6.1 or - 0.4 + 2.71 = 0.7 x + 4.27 or 0.7 x = 2.71 - 0.4 - 4.27 or x = \frac{- 1.96}{0.7} = - 2.8 Thus, x = - 2.8 is the solution of the given equation . Check : Substituting x = - 2.8 in the given equation, we get : L . H . S . = \frac{0.5 (- 2.8 - 0.4)}{0.35} - \frac{0.6 (- 2.8 - 2.71)}{0.42} = \frac{- 1.6}{0.35} + \frac{3.306}{0.42} = - 4.571 + 7.871 = 3.3 R . H . S . = - 2.8 + 6.1 = - 3.3 ∴ L . H . S . = R . H . S . for x = - 2.8$

Page No 9.12:

Question 23:

Solve each of the following equation and also check your result in each case:
$6.5 x + \frac{19.5 x - 32.5}{2} = 6.5 x + 13 + (\frac{13 x - 26}{2})$

Answer:

$6.5 x + \frac{19.5 x - 32.5}{2} = 6.5 x + 13 + \frac{13 x - 26}{2} or \frac{19.5 x - 32.5}{2} - \frac{13 x - 26}{2} = 13 or \frac{19.5 x - 32.5 - 13 x + 26}{2} = 13 or 6.5 x - 6.5 = 26 [After c ross multipl ication] or 6.5 x = 26 + 6.5 or x = \frac{32.5}{6.5} = 5 Thus, x = 5 is the solution of the given equation . Check : Substituting x = 5 in the given equation, we get : L . H . S . = 6.5 \times 5 + \frac{19.5 \times 5 - 32.5}{2} = 65 R . H . S . = 6.5 \times 5 + 13 + \frac{13 \times 5 - 26}{2} = 65 ∴ L . H . S . = R . H . S . for x = 5 .$

Page No 9.12:

Question 24:

Solve each of the following equation and also check your result in each case:
(3x − 8)(3x + 2) − (4x − 11)(2x + 1) = (x − 3)(x + 7)

Answer:

$(3 x - 8) (3 x + 2) - (4 x - 11) (2 x + 1) = (x - 3) (x + 7) or 9 x^{2} + 6 x - 24 x - 16 - 8 x^{2} - 4 x + 22 x + 11 = x^{2} + 7 x - 3 x - 21 or x^{2} - 5 = x^{2} + 4 x - 21 or 4 x = - 5 + 21 or x = \frac{16}{4} = 4 Thus, x = 4 is the solution of the given equation . Check : Substituting x = 4 in the given equation, we get : L . H . S . = (3 \times 4 - 8) (3 \times 4 + 2) - (4 \times 4 - 11) (2 \times 4 + 1) = 4 \times 14 - 5 \times 9 = 11 R . H . S . = (4 - 3) (4 + 7) = 11 ∴ L . H . S . = R . H . S . for x = 4 .$

Page No 9.12:

Question 25:

Solve each of the following equation and also check your result in each case:
[(2x + 3) + (x + 5)]² + [(2x + 3) − (x + 5)]² = 10x² + 92

Answer:

$[(2 x + 3) + (x + 5)]^{2} + [(2 x + 3) - (x + 5)]^{2} = 10 x^{2} + 92 or (3 x + 8)^{2} + (x - 2)^{2} = 10 x^{2} + 92 or 9 x^{2} + 48 x + 64 + x^{2} - 4 x + 4 = 10 x^{2} + 92 [(a + b)^{2} = a^{2} + b^{2} + 2 ab and (a - b)^{2} = a^{2} + b^{2} - 2 ab] or 10 x^{2} - 10 x^{2} + 44 x = 92 - 68 or x = \frac{24}{44} or x = \frac{6}{11} Thus, x = \frac{6}{11} is the solution of the given equation . Check : Substituting x = \frac{6}{11} in the given equation, we get : L . H . S . = {[(2 \times \frac{6}{11} + 3) + (\frac{6}{11} + 5)]}^{2} + {[(2 \times \frac{6}{11} + 3) - (\frac{6}{11} + 5)]}^{2} = {[(\frac{45}{11}) + (\frac{61}{11})]}^{2} + {[(\frac{45}{11}) - (\frac{61}{11})]}^{2} = {(\frac{106}{11})}^{2} + {(\frac{- 16}{11})}^{2} = \frac{11492}{121} R . H . S . = 10 \times {(\frac{6}{11})}^{2} + 92 = \frac{360}{121} + 92 = \frac{11492}{121} ∴ L . H . S . = R . H . S . for x = \frac{6}{11}$

Page No 9.17:

Question 1:

Solve the following equation and verify your answer:
$\frac{2 x - 3}{3 x + 2} = - \frac{2}{3}$

Answer:

$\frac{2 x - 3}{3 x + 2} = - \frac{2}{3} or 6 x - 9 = - 6 x - 4 (After c ross multipl ication) or 6 x + 6 x = - 4 + 9 or x = \frac{5}{12} ∴ x = \frac{5}{12} is the solution of the given equation . Check : L . H . S . = \frac{2 \times \frac{5}{12} - 3}{3 \times \frac{5}{12} + 2} = \frac{\frac{5}{6} - 3}{\frac{5}{4} + 2} = \frac{\frac{- 13}{6}}{\frac{13}{4}} = \frac{- 4}{6} = \frac{- 2}{3} R . H . S . = \frac{- 2}{3} ∴ L . H . S . = R . H . S . for x = \frac{5}{12}$

Page No 9.17:

Question 2:

Solve the following equation and verify your answer:
$\frac{2 - y}{y + 7} = \frac{3}{5}$

Answer:

$\frac{2 - y}{y + 7} = \frac{3}{5} or 10 - 5 y = 3 y + 21 (After c ross multipl ication) or 3 y + 5 y = 10 - 21 or 8 y = - 11 or y = \frac{- 11}{8} ∴ y = \frac{- 11}{8} is the solution of the given equation . Check : Substituting y = \frac{- 11}{8} in the given equation, we get : L . H . S . = \frac{2 - \frac{- 11}{8}}{\frac{- 11}{8} + 7} = \frac{16 + 11}{- 11 + 56} = \frac{27}{45} = \frac{3}{5} R . H . S . = \frac{3}{5} ∴ L . H . S . = R . H . S . for y = \frac{- 11}{8}$

Page No 9.17:

Question 3:

Solve the following equation and verify your answer:
$\frac{5 x - 7}{3 x} = 2$

Answer:

$\frac{5 x - 7}{3 x} = 2 or 6 x = 5 x - 7 (After c ross multipl ication) or 6 x - 5 x = - 7 or x = - 7 ∴ x = - 7 is the solution of given equation . Check : Substituting x = - 7 in the given equation, we get : L . H . S = \frac{5 \times (- 7) - 7}{. 3 (- 7)} = \frac{- 35 - 7}{- 21} = \frac{- 42}{- 21} = 2 R . H . S . = 2 ∴ L . H . S . = R . H . S . for x = - 7 .$

Page No 9.17:

Question 4:

Solve the following equation and verify your answer:
$\frac{3 x + 5}{2 x + 7} = 4$

Answer:

$\frac{3 x + 5}{2 x + 7} = 4 or 3 x + 5 = 8 x + 28 or 8 x + 28 = 3 x + 5 (After c ross multipl ication) or 8 x - 3 x = 5 - 28 or 5 x = - 23 or x = \frac{- 23}{5} ∴ x = \frac{- 23}{5} is the solution of given equation . Check : Substituting x = \frac{- 23}{5} in the given equation, we get : L . H . S . = \frac{3 \times \frac{- 23}{5} + 5}{2 \times \frac{- 23}{5} + 7} = \frac{- 69 + 25}{- 46 + 35} = \frac{- 44}{- 11} = 4 R . H . S . = 4 ∴ L . H . S . = R . H . S . for x = \frac{- 23}{5}$

Page No 9.17:

Question 5:

Solve the following equation and verify your answer:
$\frac{2 y + 5}{y + 4} = 1$

Answer:

$\frac{2 y + 5}{y + 4} = 1 or 2 y + 5 = y + 4 or 2 y - y = 4 - 5 or y = - 1 Thus, y = - 1 is the solution of the given equation . Check : Substituting y = - 1 in the given equation, we get : L . H . S . = \frac{2 (- 1) + 5}{- 1 + 4} = \frac{- 2 + 5}{3} = \frac{3}{3} = 1 R . H . S . = 1 ∴ L . H . S . = R . H . S . for y = - 1 .$

Page No 9.17:

Question 6:

Solve the following equation and verify your answer:
$\frac{2 x + 1}{3 x - 2} = \frac{5}{9}$

Answer:

$\frac{2 x + 1}{3 x - 2} = \frac{5}{9} or 18 x + 9 = 15 x - 10 [After c ross multipl ication] or 18 x - 15 x = - 10 - 9 or 3 x = - 19 or x = \frac{- 19}{3} Thus, x = \frac{- 19}{3} is the solution of the given equation . Check : Substituting x = \frac{- 19}{3} in the given equation, we get : L . H . S . = \frac{2 (\frac{- 19}{3}) + 1}{3 (\frac{- 19}{3}) - 2} = \frac{- 38 + 3}{- 57 - 6} = \frac{- 35}{- 63} = \frac{5}{9} R . H . S . = \frac{5}{9} ∴ L . H . S . = R . H . S . for x = \frac{- 19}{3}$

Page No 9.17:

Question 7:

Solve the following equation and verify your answer:
$\frac{1 - 9 y}{19 - 3 y} = \frac{5}{8}$

Answer:

$\frac{1 - 9 y}{19 - 3 y} = \frac{5}{8} or 8 - 72 y = 95 - 15 y [After c ross multipl ication] or 95 - 15 y = 8 - 72 y or 72 y - 15 y = 8 - 95 or 57 y = - 87 or y = \frac{- 87}{57} or y = \frac{- 29}{19} Thus y = \frac{- 29}{19} is the solution of the given equation . Check : Substituting y = \frac{- 29}{19} in the given equation, we get : L . H . S . = \frac{1 - 9 (\frac{- 29}{19})}{19 - 3 (\frac{- 29}{19})} = \frac{19 + 261}{361 + 87} = \frac{280}{448} = \frac{5}{8} R . H . S . = \frac{5}{8} ∴ L . H . S . = R . H . S . for y = \frac{- 29}{19}$

Page No 9.17:

Question 8:

Solve the following equation and verify your answer:
$\frac{2 x}{3 x + 1} = - 3$

Answer:

$\frac{2 x}{3 x + 1} = - 3 or 2 x = - 9 x - 3 [After c ross multipl ication] or 2 x + 9 x = - 3 or 11 x = - 3 or x = \frac{- 3}{11} Thus, x = \frac{- 3}{11} is the solution of the given equation . Check : Substituting x = \frac{- 3}{11} in the given equation, we get : L . H . S . = \frac{2 (\frac{- 3}{11})}{3 (\frac{- 3}{11}) + 1} = \frac{- 6}{- 9 + 11} = \frac{- 6}{2} = - 3 R . H . S . = - 3 ∴ L . H . S . = R . H . S . for x = \frac{- 3}{11}$

Page No 9.17:

Question 9:

Solve the following equation and verify your answer:
$\frac{y - (7 - 8 y)}{9 y - (3 + 4 y)} = \frac{2}{3}$

Answer:

$\frac{y - (7 - 8 y)}{9 y - (3 + 4 y)} = \frac{2}{3} or \frac{9 y - 7}{5 y - 3} = \frac{2}{3} or 27 y - 21 = 10 y - 6 [After c ross multipl ication] or 27 y - 10 y = - 6 + 21 or 17 y = 15 or y = \frac{15}{17} Thus, y = \frac{15}{17} is the solution of the given equation . Check : Substituting y = \frac{15}{17} in the given equation, we get : L . H . S . = \frac{9 (\frac{15}{17}) - 7}{5 (\frac{15}{17}) - 3} = \frac{135 - 119}{75 - 51} = \frac{16}{24} = \frac{2}{3} R . H . S . = \frac{2}{3} ∴ L . H . S . = R . H . S . for y = \frac{15}{17}$

Page No 9.17:

Question 10:

Solve the following equation and verify your answer:
$\frac{6}{2 x - (3 - 4 x)} = \frac{2}{3}$

Answer:

$\frac{6}{2 x - (3 - 4 x)} = \frac{2}{3} or \frac{6}{6 x - 3} = \frac{2}{3} or 12 x - 6 = 18 [After c ross multipl ication] or 12 x = 18 + 6 or x = \frac{24}{12} or x = 2 Thus, x = 2 is the solution of the given equation . Check : Substituting x = 2 in the given equation, we get : L . H . S . = \frac{6}{2 \times 2 - (3 - 4 \times 2)} = \frac{6}{9} = \frac{2}{3} R . H . S . = \frac{2}{3} ∴ L . H . S . = R . H . S . for x = 2 .$

Page No 9.17:

Question 11:

Solve the following equation and verify your answer:
$\frac{2}{3 x} - \frac{3}{2 x} = \frac{1}{12}$

Answer:

$\frac{2}{3 x} - \frac{3}{2 x} = \frac{1}{12} or, \frac{4 - 9}{6 x} = \frac{1}{12} or, \frac{- 5}{x} = \frac{1}{2} or, x = - 10 [After c ross multipl ication] Thus, x = - 10 is the solution of the given equation . Check : Substituting x = - 10 in the given equation, we get : L . H . S . = \frac{2}{3 (- 10)} - \frac{3}{2 (- 10)} = \frac{2}{- 30} - \frac{3}{- 20} = \frac{4 - 9}{- 60} = \frac{- 5}{- 60} = \frac{1}{12} R . H . S . = \frac{1}{12} ∴ L . H . S . = R . H . S . for x = - 10 .$

Page No 9.17:

Question 12:

Solve the following equation and verify your answer:
$\frac{3 x + 5}{4 x + 2} = \frac{3 x + 4}{4 x + 7}$

Answer:

$\frac{3 x + 5}{4 x + 2} = \frac{3 x + 4}{4 x + 7} or, 12 x^{2} + 20 x + 21 x + 35 = 12 x^{2} + 16 x + 6 x + 8 [Cross multiply] or, 12 x^{2} - 12 x^{2} + 41 x - 22 x = 8 - 35 or, 19 x = - 27 or, x = \frac{- 27}{19} Thus, x = \frac{- 27}{19} is the solution of the given equation Check : Substituting x = \frac{- 27}{19} in the given equation, we get : L . H . S . = \frac{3 (\frac{- 27}{19}) + 5}{4 (\frac{- 27}{19}) + 2} = \frac{- 81 + 95}{- 108 + 38} = \frac{14}{- 70} = \frac{- 1}{5} R . H . S . = \frac{3 (\frac{- 27}{19}) + 4}{4 (\frac{- 27}{19}) + 7} = \frac{- 81 + 76}{- 108 + 133} = \frac{- 5}{25} = \frac{- 1}{5} ∴ L . H . S . = R . H . S . for x = \frac{- 27}{19}$

Page No 9.17:

Question 13:

Solve the following equation and verify your answer:
$\frac{7 x - 2}{5 x - 1} = \frac{7 x + 3}{5 x + 4}$

Answer:

$\frac{7 x - 2}{5 x - 1} = \frac{7 x + 3}{5 x + 4} or 35 x^{2} + 28 x - 10 x - 8 = 35 x^{2} + 15 x - 7 x - 3 [After c ross multipl ication] or 35 x^{2} - 35 x^{2} + 18 x - 8 x = - 3 + 8 or 10 x = 5 or x = \frac{5}{10} or x = \frac{1}{2} Thus, x = \frac{1}{2} is the solution of the given equation . Check : Substituting x = \frac{1}{2} in the given equation, we get : L . H . S . = \frac{7 (\frac{1}{2}) - 2}{5 (\frac{1}{2}) - 1} = \frac{7 - 4}{5 - 2} = \frac{3}{3} = 1 R . H . S . = \frac{7 (\frac{1}{2}) + 3}{5 (\frac{1}{2}) + 4} = \frac{7 + 6}{5 + 8} = \frac{13}{13} = 1 ∴ L . H . S . = R . H . S . for x = \frac{1}{2}$

Page No 9.17:

Question 14:

Solve the following equation and verify your answer:
${(\frac{x + 1}{x + 2})}^{2} = \frac{x + 2}{x + 4}$

Answer:

${(\frac{x + 1}{x + 2})}^{2} = \frac{x + 2}{x + 4} or \frac{x^{2} + 2 x + 1}{x^{2} + 4 x + 4} = \frac{x + 2}{x + 4} or x^{3} + 2 x^{2} + x + 4 x^{2} + 8 x + 4 = x^{3} + 4 x^{2} + 4 x + 2 x^{2} + 8 x + 8 [After c ross multipl ication] or x^{3} - x^{3} + 6 x^{2} - 6 x^{2} + 9 x - 12 x = 8 - 4 or - 3 x = 4 or x = \frac{4}{- 3} = \frac{- 4}{3} Thus, x = \frac{- 4}{3} is the solution of the given equation . Check : Substituting x = \frac{- 4}{3} in the given equation, we get : L . H . S . = {(\frac{\frac{- 4}{3} + 1}{\frac{- 4}{3} + 2})}^{2} = {(\frac{- 4 + 3}{- 4 + 6})}^{2} = \frac{1}{4} R . H . S . = \frac{\frac{- 4}{3} + 2}{\frac{- 4}{3} + 4} = \frac{- 4 + 6}{- 4 + 12} = \frac{2}{8} = \frac{1}{4} ∴ L . H . S . = R . H . S . for x = \frac{- 4}{3}$

Page No 9.17:

Question 15:

Solve the following equation and verify your answer:
${(\frac{x + 1}{x - 4})}^{2} = \frac{x + 8}{x - 2}$

Answer:

${(\frac{x + 1}{x - 4})}^{2} = \frac{x + 8}{x - 2} or \frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 16} = \frac{x + 8}{x - 2} [(a + b)^{2} = a^{2} + b^{2} + 2 a b a n d (a - b)^{2} = a^{2} + b^{2} - 2 a b] or x^{3} + 2 x^{2} + x - 2 x^{2} - 4 x - 2 = x^{3} - 8 x^{2} + 16 x + 8 x^{2} - 64 x + 128 [After c ross multipl ication] or x^{3} - x^{3} - 3 x + 48 x = 128 + 2 or 45 x = 130 or x = \frac{130}{45} = \frac{26}{9} Thus x = \frac{26}{9} is the solution of the given equation . Check : Substituting x = \frac{26}{9} in the given equation, we get : L . H . S . = {(\frac{\frac{26}{9} + 1}{\frac{26}{9} - 4})}^{2} = {(\frac{26 + 9}{26 - 36})}^{2} = \frac{1225}{100} = \frac{49}{4} R . H . S . = (\frac{\frac{26}{9} + 8}{\frac{26}{9} - 2}) = (\frac{26 + 72}{26 - 18}) = \frac{98}{8} = \frac{49}{4} ∴ L . H . S . = R . H . S . for x = \frac{26}{9}$

Page No 9.17:

Question 16:

Solve the following equation and verify your answer:
$\frac{9 x - 7}{3 x + 5} = \frac{3 x - 4}{x + 6}$

Answer:

$\frac{9 x - 7}{3 x + 5} = \frac{3 x - 4}{x + 6} or 9 x^{2} - 7 x + 54 x - 42 = 9 x^{2} - 12 x + 15 x - 20 [After c ross multipl ication] or 9 x^{2} - 9 x^{2} + 47 x - 3 x = - 20 + 42 or 44 x = 22 or x = \frac{22}{44} or x = \frac{1}{2} Thus, x = \frac{1}{2} is the solution of the given equation . Check : Substituting x = \frac{1}{2} in the given equation, we get : L . H . S . = \frac{9 (\frac{1}{2}) - 7}{3 (\frac{1}{2}) + 5} = \frac{9 - 14}{3 + 10} = \frac{- 5}{13} R . H . S . = \frac{3 (\frac{1}{2}) - 4}{\frac{1}{2} + 6} = \frac{3 - 8}{1 + 12} = \frac{- 5}{13} ∴ L . H . S . = R . H . S . for x = \frac{1}{2}$

Page No 9.17:

Question 17:

Solve the following equation and verify your answer:
$\frac{x + 2}{x + 5} = \frac{x}{x + 6}$

Answer:

$\frac{x + 2}{x + 5} = \frac{x}{x + 6} or x^{2} + 2 x + 6 x + 12 = x^{2} + 5 x [After c ross multipl ication] or x^{2} - x^{2} + 8 x - 5 x = - 12 or 3 x = - 12 or x = \frac{- 12}{3} or x = - 4 Thus, x = - 4 is the solution of given equation . Check : Substituting x = - 4 in the given equation, we get : L . H . S . = \frac{- 4 + 2}{- 4 + 5} = - 2 R . H . S . = \frac{- 4}{- 4 + 6} = - 2 ∴ L . H . S . = R . H . S . for x = - 4 .$

Page No 9.17:

Question 18:

Solve the following equation and verify your answer:
$\frac{2 x - (7 - 5 x)}{9 x - (3 + 4 x)} = \frac{7}{6}$

Answer:

$\frac{2 x - (7 - 5 x)}{9 x - (3 + 4 x)} = \frac{7}{6} or \frac{7 x - 7}{5 x - 3} = \frac{7}{6} or 42 x - 42 = 35 x - 21 [After c ross multipl ication] or 42 x - 35 x = - 21 + 42 or 7 x = 21 or x = \frac{21}{7} or x = 3 Thus, x = 3 is the solution of the given equation . Check : Substituting x = 3 in the given equation, we get : L . H . S . = \frac{2 \times 3 - (7 - 5 \times 3)}{9 \times 3 - (3 + 4 \times 3)} = \frac{6 - (7 - 15)}{27 - (3 + 12)} = \frac{6 + 8}{27 - 15} = \frac{14}{12} = \frac{7}{6} R . H . S . = \frac{7}{6} ∴ L . H . S . = R . H . S . for x = 3 .$

Page No 9.17:

Question 19:

Solve the following equation and verify your answer:
$\frac{15 (2 - x) - 5 (x + 6)}{1 - 3 x} = 10$

Answer:

$\frac{15 (2 - x) - 5 (x + 6)}{1 - 3 x} = 10 or \frac{30 - 15 x - 5 x - 30}{1 - 3 x} = 10 or \frac{- 20 x}{1 - 3 x} = 10 or 10 - 30 x = - 20 x [After c ross multipl ication] or - 20 x + 30 x = 10 or 10 x = 10 or x = 1 Thus, x = 1 is the solution of the given equation . Check : Substituting x = 1 in the given equation, we get : L . H . S . = \frac{15 (2 - 1) - 5 (1 + 6)}{1 - 3 (1)} = \frac{15 - 35}{- 2} = \frac{- 20}{- 2} = 10 R . H . S . = 10 ∴ L . H . S . = R . H . S . for x = 1 .$

Page No 9.17:

Question 20:

Solve the following equation and verify your answer:
$\frac{x + 3}{x - 3} + \frac{x + 2}{x - 2} = 2$

Answer:

$\frac{x + 3}{x - 3} + \frac{x + 2}{x - 2} = 2 or \frac{x + 3}{x - 3} = 2 - \frac{x + 2}{x - 2} or \frac{x + 3}{x - 3} = \frac{2 x - 4 - x - 2}{x - 2} or \frac{x + 3}{x - 3} = \frac{x - 6}{x - 2} or x^{2} - 2 x + 3 x - 6 = x^{2} - 3 x - 6 x + 18 [After c ross multipl ication] or x^{2} - x^{2} + x + 9 x = 18 + 6 or 10 x = 24 or x = \frac{24}{10} or x = \frac{12}{5} Thus, x = \frac{12}{5} is the solution of the given equation . Check : Substituting x = \frac{12}{5} in the given equation, we get : L . H . S . = \frac{\frac{12}{5} + 3}{\frac{12}{5} - 3} + \frac{\frac{12}{5} + 2}{\frac{12}{5} - 2} = \frac{12 + 15}{12 - 15} + \frac{12 + 10}{12 - 10} = \frac{27}{- 3} + \frac{22}{2} = \frac{54 - 66}{- 6} = \frac{- 12}{- 6} = 2 R . H . S . = 2 ∴ L . H . S . = R . H . S . for x = \frac{12}{5}$

Page No 9.17:

Question 21:

Solve the following equation and verify your answer:
$\frac{(x + 2) (2 x - 3) - 2 x^{2} + 6}{x - 5} = 2$

Answer:

$\frac{(x + 2) (2 x - 3) - 2 x^{2} + 6}{x - 5} = 2 or \frac{2 x^{2} + x - 6 - 2 x^{2} + 6}{x - 5} = 2 or \frac{x}{x - 5} = 2 or 2 x - 10 = x [After c ross multipl ication] or 2 x - x = 10 or x = 10 Thus, x = 10 is the solution of the given equation . Check : Substituting x = 10 in the given equation, we get : L . H . S . = \frac{(10 + 2) (2 \times 10 - 3) - 2 \times 10^{2} + 6}{10 - 5} = \frac{12 \times 17 - 200 + 6}{5} = \frac{10}{5} = 2 R . H . S . = 2 ∴ L . H . S . = R . H . S . for x = 10 .$

Page No 9.17:

Question 22:

Solve the following equation and verify your answer:
$\frac{x^{2} - (x + 1) (x + 2)}{5 x + 1} = 6$

Answer:

$\frac{x^{2} - (x + 1) (x + 2)}{5 x + 1} = 6 or \frac{x^{2} - x^{2} - 2 x - x - 2}{5 x + 1} = 6 or \frac{- 3 x - 2}{5 x + 1} = 6 or 30 x + 6 = - 3 x - 2 [After c ross multipl ication] or 30 x + 3 x = - 2 - 6 or 33 x = - 8 or x = \frac{- 8}{33} Thus, x = \frac{- 8}{33} is the solution of the given equation . Check : Substituting x = \frac{- 8}{33} in the given equation, we get : L . H . S . = \frac{(\frac{- 8}{33})^{2} - (\frac{- 8}{33} + 1) (\frac{- 8}{33} + 2)}{5 (\frac{- 8}{33}) + 1} = \frac{\frac{64}{1089} - \frac{25}{33} \times \frac{58}{33}}{\frac{- 40}{33} + 1} = \frac{\frac{64}{1089} - \frac{1450}{1089}}{\frac{- 7}{33}} = \frac{\frac{- 1386}{1089}}{\frac{- 7}{33}} = \frac{42}{7} = R . H . S . = 6 ∴ L . H . S . = R . H . S . for x = \frac{- 8}{33}$

Page No 9.17:

Question 23:

Solve the following equation and verify your answer:
$\frac{(2 x + 3) - (5 x - 7)}{6 x + 11} = - \frac{8}{3}$

Answer:

$\frac{(2 x + 3) - (5 x - 7)}{6 x + 11} = \frac{- 8}{3} or \frac{- 3 x + 10}{6 x + 11} = \frac{- 8}{3} or - 9 x + 30 = - 48 x - 88 [After c ross multipl ication] or - 9 x + 48 x = - 88 - 30 or 39 x = - 118 or x = \frac{- 118}{39} Thus, x = \frac{- 118}{39} is the solution of the given equation . Check : Substituting x = \frac{- 118}{39} in the given equation, we get : L . H . S . = \frac{- 3 (\frac{- 118}{39}) + 10}{6 (\frac{- 118}{39}) + 11} = \frac{354 + 390}{- 708 + 429} = \frac{744}{- 279} = \frac{- 8}{- 3} R . H . S . = \frac{- 8}{3} ∴ L . H . S . = R . H . S . for x = \frac{- 118}{39}$

Page No 9.17:

Question 24:

Find a positive value of x for which the given equation is satisfied:
(i) $\frac{x^{2} - 9}{5 + x^{2}} = - \frac{5}{9}$
(ii) $\frac{y^{2} + 4}{3 y^{2} + 7} = \frac{1}{2}$

Answer:

$(i) \frac{x^{2} - 9}{5 + x^{2}} = \frac{- 5}{9} or 9 x^{2} - 81 = - 25 - 5 x^{2} [After c ross multipl ication] or 9 x^{2} + 5 x^{2} = - 25 + 81 or 14 x^{2} = 56 or x^{2} = \frac{56}{14} or x^{2} = 4 = 2^{2} or x = 2 Thus, x = 2 is the solution of the given equation . Check : Substituting x = 2 in the given equation, we get : L . H . S . = \frac{2^{2} - 9}{5 + 2^{2}} = \frac{4 - 9}{5 + 4} = \frac{- 5}{9} R . H . S . = \frac{- 5}{9} ∴ L . H . S . = R . H . S . for x = 2 . (ii) \frac{y^{2} + 4}{3 y^{2} + 7} = \frac{1}{2} or 3 y^{2} + 7 = 2 y^{2} + 8 [After c ross multipl ication] or 3 y^{2} - 2 y^{2} = 8 - 7 or y^{2} = 1 or y = 1 Thus, y = 1 is the solution of the given equation . Check : Substituting y = 1 in the given equation, we get : L . H . S . = \frac{1^{2} + 4}{3 (1)^{2} + 7} = \frac{5}{10} = \frac{1}{2} R . H . S . = \frac{1}{2} ∴ L . H . S . = R . H . S . for y = 1 .$

Page No 9.29:

Question 1:

Four-fifth of a number is more than three-fourth of the number by 4. Find the number.

Answer:

$Let the number be x . According to the question, \frac{4}{5} x - \frac{3}{4} x = 4 or \frac{16 x - 15 x}{20} = 4 or x = 80 [After c ross multipl ication] Thus, the required number is 80 .$

Page No 9.29:

Question 2:

The difference between the squares of two consecutive numbers is 31. Find the numbers.

Answer:

$Let the numbers be x and x + 1 . According to the question, (x + 1)^{2} - x^{2} = 31 or x^{2} + 2 x + 1 - x^{2} = 31 or 2 x = 31 - 1 or x = \frac{30}{2} or x = 15 Thus, the numbers are 15 and 16 .$

Page No 9.29:

Question 3:

Find a number whose double is 45 greater than its half.

Answer:

$Let the number be x . According to the question, 2 x = \frac{1}{2} x + 45 o r 2 x - \frac{1}{2} x = 45 or \frac{4 x - x}{2} = 45 or 3 x = 90 [After c ross multipl ication] or x = \frac{90}{3} or x = 30 Thus, the number is 30 .$

Page No 9.29:

Question 4:

Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.

Answer:

$Let the number be x . According to the question, 5 x - 5 = 2 x + 4 or 5 x - 2 x = 4 + 5 or 3 x = 9 or x = \frac{9}{3} or x = 3 Thus, the number is 3 .$

Page No 9.29:

Question 5:

A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.

Answer:

$Let the number be x . According to the question, \frac{x}{5} + 5 = \frac{x}{4} - 5 or \frac{x}{5} - \frac{x}{4} = - 5 - 5 or \frac{4 x - 5 x}{20} = - 10 or - x = - 200 [After c ross multipl ication] or x = 200 Thus, the number is 200 .$

Page No 9.29:

Question 6:

A number consists of two digits whose sum is 9. If 27 is subtracted from the number, its digits are reversed. Find the number.

Answer:

$Let the units digit be x . ∵ Sum of two digits = 9 ∴ Tens digit = (9 - x) ∴ Original number = 10 \times (9 - x) + x Reversed number = 10 x + (9 - x) According to the question, 10 \times (9 - x) + x - 27 = 10 x + (9 - x) or 90 - 10 x + x - 27 = 10 x + 9 - x or 9 x + 9 x = 90 - 27 - 9 or 18 x = 54 or x = \frac{54}{18} = 3 ∴ The n umber = 10 \times (9 - 3) + 3 = 63$

Page No 9.29:

Question 7:

Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.

Answer:

$Let the first part of 184 be x . Therefore, the other part will be (184 - x) . According to the question, \frac{1}{3} x - \frac{1}{7} (184 - x) = 8 or \frac{7 x - 552 + 3 x}{21} = 8 or 10 x - 552 = 168 [After c ross multipl ication] or 10 x = 168 + 552 or x = \frac{720}{10} = 72 Thus, the parts of 184 are 72 and 112 (184 - 72 = 112) .$

Page No 9.29:

Question 8:

The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to $\frac{2}{3}$ . What is the original fraction equal to?

Answer:

$Let the denominator of the fraction be x . Therefore, the numerator will be (x - 6) . ∴ Fraction = \frac{x - 6}{x} According to the question, \frac{x - 6 + 3}{x} = \frac{2}{3} or \frac{x - 3}{x} = \frac{2}{3} or 3 x - 9 = 2 x [After cross multiplication] or 3 x - 2 x = 9 or x = 9 Thus, the original fraction = \frac{9 - 6}{9} = \frac{1}{3}$

Page No 9.29:

Question 9:

A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50, find the number of notes of each type.

Answer:

$Let the number of Rs . 10 notes be x . Therefore, the number of Rs . 20 notes will be (50 - x) . Value of Rs . 10 notes = 10 x Value of Rs . 20 notes = 20 (50 - x) According to the question, 10 x + 20 (50 - x) = 800 or 10 x + 1000 - 20 x = 800 or 10 x = 1000 - 800 or x = \frac{200}{10} = 20 ∴ Number of Rs . 10 notes = 20 Number of Rs . 20 notes = (50 - 20) = 30 .$

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Question 10:

Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty-five paise coins as she has fifty-paise coins. How many coins of each kind does she have?

Answer:

$Let the number of 50 paise coins be x . Therefore, the number of 25 paise coins will be 2 x . Value of 50 paise coins = Rs . 0.5 x Value of 25 paise coins = Rs . 0.25 \times 2 x According to the question, 0.5 x + 0.25 \times 2 x = 9 or x = 9 ∴ Number of fifty paise coins = 9 Number of twenty five paise coins = 2 \times 9 = 18 Total number of coins = 9 + 18 = 27 .$

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Question 11:

Sunita is twice as old as Ashima. If six years is subtracted from Ashima's age and four years added to Sunita's age, then Sunita will be four times Ashima's age. How old were they two years ago?

Answer:

$Let the age of Ashima be x years . Therefore, the age of Sunita will be 2 x years . According to the question, 4 (x - 6) = 2 x + 4 or 4 x - 24 = 2 x + 4 or 4 x - 2 x = 4 + 24 or 2 x = 28 or x = 14 ∴ Age of Ashima = 14 years . Age of Sunita = 2 \times 14 = 28 years .$

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Question 12:

The ages of sonu and Monu are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer:

$It is given that the ratio of the ages of Sonu and Monu is 7 : 5 . Let the present ages of Sonu and Monu be 7 x and 5 x years . After ten years : Age of Sonu = 7 x + 10 years Age of Monu = 5 x + 10 years According to the question, \frac{7 x + 10}{5 x + 10} = \frac{9}{7} or 49 x + 70 = 45 x + 90 or 49 x - 45 x = 90 - 70 o r 4 x = 20 or x = 5 ∴ Present age of Sonu = 7 \times 5 = 35 years . Present age of Monu = 5 \times 5 = 25 years .$

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Question 13:

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer:

$Five years ago : Let the age of the son be x years . Therefore, the age of the father will be 7 x years . ∴ Present age of the son = (x + 5) years Present age of the father = (7 x + 5) years After five years : Age of the son = (x + 5 + 5) = (x + 10) years Age of the father = (7 x + 5 + 5) = (7 x + 10) years According to the question, 7 x + 10 = 3 (x + 10) or 7 x - 3 x = 30 - 10 or 4 x = 20 or x = 5 ∴ Present age of the son = (5 + 5) = 10 years . Present age of the father = (7 \times 5 + 5) = 40 years .$

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Question 14:

I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now?

Answer:

$Let the age of my son be x years . Therefore, my age will be 5 x years . After 6 years : Age of my son = (x + 6) years My age = (5 x + 6) years According to the question, 5 x + 6 = 3 (x + 6) or 5 x - 3 x = 18 - 6 or 2 x = 12 or x = 6 ∴ Age of my son = 6 years . My age = 5 \times 6 = 30 years .$

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Question 15:

I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

Answer:

$Let the number of five - rupee notes be x . Therefore, the number of ten - rupee notes will be (x + 10) . Now, Value of five - rupee notes = Rs . 5 x Value of ten - rupee notes = Rs . 10 (x + 10) According to the question, 5 x + 10 (x + 10) = 1000 or 15 x = 1000 - 100 or x = \frac{900}{15} = 60 ∴ Number of five - rupee notes = 60 . Number of ten - rupee notes = 60 + 10 = 70 .$

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Question 16:

At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice and just three did not drink any thing. How many guests were in all?

Answer:

$Let the total number of guests be x . Therefore, the number of guests, who drank colas, would be \frac{1}{4} x . The number of guests, who drank squash, would be \frac{1}{3} x . The number of guests, who drank fruit juice, would be \frac{2}{5} x . The number of guests, who did not drink, would be 3 . According to the question, x - (\frac{x}{4} + \frac{x}{3} + \frac{2 x}{5}) = 3 or \frac{60 x - 15 x - 20 x - 24 x}{60} = 3 or x = 180 Thus, total number of guests = 180 .$

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Question 17:

There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?

Answer:

$Let the number of correctly answered questions be x . Therefore, the number of unattempted or wrongly answered questions will be (180 - x) . According to the question, 4 x - 1 (180 - x) = 450 or 5 x = 450 + 180 or x = \frac{630}{5} = 126 Thus, number of correctly answered questions = 126 . Number of unattempted or wrongly answered questions = 180 - 126 = 54 .$

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Question 18:

A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day, he works and he will be fined Rs 5 for each day, he is absent. If he receives Rs 745 in all, for how many days he remained absent?

Answer:

$Let the number of days for which the labour er is absent be x . Therefore, the number of days for which he is present will be (20 - x) . ∴ Earnings = Rs . 60 (20 - x) Fine = Rs . 5 x According to the question, 60 (20 - x) - 5 x = 745 or 1200 - 60 x - 5 x = 745 or 65 x = 1200 - 745 or x = \frac{455}{65} = 7 Thus, the labourer was absent for 7 days .$

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Question 19:

Ravish has three boxes whose total weight is $60 \frac{1}{2}$ kg. Box B weighs $3 \frac{1}{2}$ kg more than box A and box C weighs $5 \frac{1}{3}$ kg more than box B. Find the weight of box A.

Answer:

$Let the weight of box A be x kg . Therefore, the weights of box B and box C will be (x + 3 \frac{1}{2}) kg and (x + 3 \frac{1}{2} + 5 \frac{1}{3}) kg, respectively . According to the question, x + (x + 3 \frac{1}{2}) + (x + 3 \frac{1}{2} + 5 \frac{1}{3}) = 60 \frac{1}{2} or 3 x = \frac{121}{2} - \frac{7}{2} - \frac{7}{2} - \frac{16}{3} or 3 x = \frac{363 - 21 - 21 - 32}{6} or 3 x = \frac{289}{6} or x = \frac{289}{18} Thus, weight of box A = \frac{289}{18} kg$

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Question 20:

The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.

Answer:

$Let, the denominator of the rational number be x . ∴ T he numerator of the rational number will be x - 3 . ∴ T he rational number = \frac{x - 3}{x} According to the question, \frac{x - 3 + 2}{x + 5} = \frac{1}{2} or \frac{x - 1}{x + 5} = \frac{1}{2} or 2 x - 2 = x + 5 or 2 x - x = 5 + 2 or x = 7 ∴ T he rational number = \frac{7 - 3}{7} = \frac{4}{7}$

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Question 21:

In a rational number, twice the numerator is 2 more than the denominator. If 3 is added to each, the numerator and the denominator, the new fraction is 2/3. Find the original number.

Answer:

$Let the denominator be x . ∴ T he numerator = \frac{x + 2}{2} ∴ T he rational number = \frac{x + 2}{2 x} According to the question, \frac{\frac{x + 2}{2} + 3}{x + 3} = \frac{2}{3} o r \frac{x + 2 + 6}{2 (x + 3)} = \frac{2}{3} or \frac{x + 8}{2 x + 6} = \frac{2}{3} or 3 x + 24 = 4 x + 12 or x = 24 - 12 or x = 12 ∴ T he rational number = \frac{12 + 2}{2 \times 12} = \frac{14}{24} = \frac{7}{12}$

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Question 22:

The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.

Answer:

$Let, the speed of the first train be x km / h . Then, the speed of the other train will be (x + 5) km / h . 2 hours after they start ed: Distance of the first train from the starting point = 2 x km Distance of the other train from the starting point = 2 (x + 5) km Now, 2 (x + 5) + 2 x + 30 = 340 or 4 x + 10 + 30 = 340 or 4 x = 340 - 40 or x = \frac{300}{4} = 75 ∴ Speed of the first train = 75 km / h . Speed of the other train = (75 + 5) = 80 km / h .$

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Question 23:

A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/hr, find the speed of the steamer in still water and the distance between the ports.

Answer:

$It is given that the speed of the stream is 1 km / h . Let the speed of the steamer in still water be x km / h . ∴ Downstream speed = (x + 1) km / h Upstream speed = (x - 1) km / h The downstream and upstream distances are same; therefore, we have : 9 (x + 1) = 10 (x - 1) or 9 x + 9 = 10 x - 10 or x = 19 ∴ Speed of the steamer in still water = 19 km / h . Distance between the ports = 9 (19 + 1) = 180 km .$

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Question 24:

Bhagwanti inherited Rs 12000.00. She invested part of it as 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00. How much did she invest at each rate?

Answer:

$At the rate of 10 %, let the investment by Bhagwanti be Rs . x . Therefore, at the rate of 12 %, the investment will be Rs . (12000 - x) . At the rate of 10 %, her annual income = x \times 10 % At the rate of 12 %, her annual income = (12000 - x) \times 12 % So, x \times 0.1 + 0.12 (12000 - x) = 1280 or 0.1 x - 0.12 x = 1280 - 1440 or 0.02 x = 160 or x = 8000 Thus, at the rate of 10 %, she invested Rs . 8000 and at the rate of 12 %, she invested Rs . 4000 (12000 - 8000) .$

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Question 25:

The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm² more than that of the given rectangle. Find the length and breath of the given rectangle.

Answer:

$Let the breadth of the rectangle be x cm . Therefore, the length of the rectangle will be (x + 9) cm . ∴ Area of the rectangle = x (x + 9) {cm}^{2} . If the length and breadth are increased by 3 cm each, a rea = (x + 3) (x + 9 + 3) {cm}^{2} . Now, (x + 3) (x + 12) - x (x + 9) = 84 or x^{2} + 15 x + 36 - x^{2} - 9 x = 84 or 6 x = 84 - 36 or x = \frac{48}{6} = 8 . Thus, brea d th of the rectangle = 8 cm . Length of the rectangle = (8 + 9) = 17 cm .$

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Question 26:

The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?

Answer:

$Let Anup' s age be x years . Therefore, his father' s age will be (100 - x) years . When Anup i s as old as his father after (100 - 2 x) years, Anuj' s age = (\frac{100 - x}{5} + 100 - 2 x) years = \frac{600 - 11 x}{5} years . Again, when Anup is as old as his father, Anuj' s age = x + 8 . Now, \frac{600 - 11 x}{5} = x + 8 or 600 - 11 x = 5 x + 40 or 16 x = 560 or x = 35 . Thus, Anup' s age = 35 years Anup' s f ather' s age = 100 - x = 100 - 35 = 65 years Anuj' s age = x + 8 = 35 + 8 = 43 years$

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Question 27:

A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?

Answer:

$Suppose, the lady start ed with x rupee s . Money spent on shopping = \frac{x}{2} rupee s Remaining amount = x - \frac{x}{2} = \frac{x}{2} rupee s A fter giving a rupee she had = (\frac{x}{2} - 1) rupee s Money spent on lunch = \frac{1}{2} (\frac{x}{2} - 1) rupee s A fter giving a two - rup e e tip she had = \frac{1}{2} (\frac{x}{2} - 1) - 2 = \frac{x - 2 - 8}{4} = \frac{x - 10}{4} rupee s Money spent on a book = \frac{1}{2} (\frac{x - 10}{4}) rupee s A fter spending three rupees on bus fare she had = \frac{1}{2} (\frac{x - 10}{4}) - 3 = \frac{x - 10 - 24}{8} = \frac{x - 34}{8} rupee s Now, \frac{x - 34}{8} = 1 or x - 34 = 8 or x = 42 Therefore, she start ed with 42 rupees .$

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Question 1:

Solve each of the following equation and also verify your solution:
$9 \frac{1}{4} = y - 1 \frac{1}{3}$

Answer:

$9 \frac{1}{4} = y - 1 \frac{1}{3} o r \frac{37}{4} + \frac{4}{3} = y or y = \frac{127}{12} ∴ y = \frac{127}{12} for the given equation . C heck : L . H . S = 9 \frac{1}{4} R . H . S = \frac{127}{12} - 1 \frac{1}{3} = \frac{127}{12} - \frac{4}{3} = \frac{127 - 16}{12} = \frac{111}{12} = 9 \frac{1}{4} So, L . H . S = R . H . S for y = \frac{127}{12}$

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Question 2:

Solve each of the following equation and also verify your solution:
$\frac{5 x}{3} + \frac{2}{5} = 1$

Answer:

$\frac{5 x}{3} + \frac{2}{5} = 1 \Rightarrow \frac{5 x}{3} = 1 - \frac{2}{5} \Rightarrow \frac{5 x}{3} = \frac{3}{5} \Rightarrow x = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} Verification : L . H . S . = \frac{5}{3} \times \frac{9}{25} + \frac{2}{5} = \frac{3}{5} + \frac{2}{5} = 1 R . H . S . = 1 ∴ L . H . S . = R . H . S . for x = \frac{9}{25}$

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Question 3:

Solve each of the following equation and also verify your solution:
$\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 13$

Answer:

$\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 13 \Rightarrow \frac{x \times 6 + x \times 4 + x \times 3}{12} = 13 \Rightarrow \frac{13 x}{12} = 13 \Rightarrow x = 13 \times \frac{12}{13} = 12 Verification : L . H . S . = \frac{12}{2} + \frac{12}{3} + \frac{12}{4} = 6 + 4 + 3 = 13 = R . H . S .$

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Question 4:

Solve each of the following equation and also verify your solution:
$\frac{x}{2} + \frac{x}{8} = \frac{1}{8}$

Answer:

$\frac{x}{2} + \frac{x}{8} = \frac{1}{8} or \frac{4 x + x}{8} = \frac{1}{8} or \frac{5 x}{8} = \frac{1}{8} or x = \frac{1}{8} \times \frac{8}{5} = \frac{1}{5} Verification : L . H . S . = \frac{1}{2} \times \frac{1}{5} + \frac{1}{8} \times \frac{1}{5} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40} = \frac{1}{8} = R . H . S .$

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Question 5:

Solve each of the following equation and also verify your solution:
$\frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12}$

Answer:

$\frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} or \frac{16 x - 9 x}{24} = \frac{7}{12} or \frac{7 x}{24} = \frac{7}{12} or x = \frac{7}{12} \times \frac{24}{7} = 2 Verification : L . H . S . = \frac{4}{3} - \frac{6}{8} = \frac{32 - 18}{24} = \frac{7}{12} R . H . S . = \frac{7}{12} ∴ R . H . S . = L . H . S . for x = 2$

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Question 6:

Solve each of the following equation and also verify your solution:
(x + 2)(x + 3) + (x − 3)(x − 2) − 2x(x + 1) = 0

Answer:

$(x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 or x^{2} + 5 x + 6 + x^{2} - 5 x + 6 - 2 x^{2} - 2 x = 0 or 12 - 2 x = 0 or x = \frac{12}{2} = 6 Verification : L . H . S . = (6 + 2) (6 + 3) + (6 - 3) (6 - 2) - 2 \times 6 (6 + 1) = 72 + 12 - 84 = 0 = R . H . S .$

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Question 7:

Solve each of the following equation and also verify your solution:
$\frac{x}{2} - \frac{4}{5} + \frac{x}{5} + \frac{3 x}{10} = \frac{1}{5}$

Answer:

$\frac{x}{2} - \frac{4}{5} + \frac{x}{5} + \frac{3 x}{10} = \frac{1}{5} or \frac{x}{2} + \frac{x}{5} + \frac{3 x}{10} = \frac{1}{5} + \frac{4}{5} or \frac{5 x + 2 x + 3 x}{10} = \frac{5}{5} or \frac{10 x}{10} = 1 or x = 1 Verification : L . H . S . = \frac{1}{2} - \frac{4}{5} + \frac{1}{5} + \frac{3}{10} = \frac{5 - 8 + 2 + 3}{10} = \frac{1}{5} = R . H . S .$

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Question 8:

Solve each of the following equation and also verify your solution:
$\frac{7}{x} + 35 = \frac{1}{10}$

Answer:

$\frac{7}{x} + 35 = \frac{1}{10} or \frac{7}{x} = \frac{1}{10} - 35 or \frac{7}{x} = \frac{1 - 350}{10} or \frac{x}{7} = \frac{10}{- 349} or x = \frac{- 10 \times 7}{349} = \frac{- 70}{349} Verification : L . H . S . = \frac{7}{\frac{- 70}{349}} + 35 = 7 \times \frac{349}{- 70} + 35 = \frac{349}{- 10} + 35 = \frac{1}{10} = R . H . S .$

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Question 9:

Solve each of the following equation and also verify your solution:
$\frac{2 x - 1}{3} - \frac{6 x - 2}{5} = \frac{1}{3}$

Answer:

$\frac{2 x - 1}{3} - \frac{6 x - 2}{5} = \frac{1}{3} or \frac{10 x - 5 - 18 x + 6}{15} = \frac{1}{3} or \frac{- 8 x + 1}{15} = \frac{1}{3} or - 24 x + 3 = 15 or 24 x = 3 - 15 or x = \frac{- 12}{24} = \frac{- 1}{2} Verification : L . H . S . = \frac{2 \times \frac{- 1}{2} - 1}{3} - \frac{6 \times \frac{- 1}{2} - 2}{5} = \frac{- 2}{3} - \frac{- 5}{5} = \frac{- 2 + 3}{3} = \frac{1}{3} = R . H . S .$

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Question 10:

Solve each of the following equation and also verify your solution:
13(y − 4) − 3(y − 9) − 5(y + 4) = 0

Answer:

$13 (y - 4) - 3 (y - 9) - 5 (y + 4) = 0 or 13 y - 52 - 3 y + 27 - 5 y - 20 = 0 or 5 y = 45 or y = \frac{45}{5} = 9 Verification : L . H . S . = 13 (9 - 4) - 3 (9 - 9) - 5 (9 + 4) = 13 \times 5 - 3 \times 0 - 5 \times 13 = 0 = R . H . S .$

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Question 11:

Solve each of the following equation and also verify your solution:
$\frac{2}{3} (x - 5) - \frac{1}{4} (x - 2) = \frac{9}{2}$

Answer:

$\frac{2}{3} (x - 5) - \frac{1}{4} (x - 2) = \frac{9}{2} or \frac{2 x - 10}{3} - \frac{x - 2}{4} = \frac{9}{2} or \frac{8 x - 40 - 3 x + 6}{12} = \frac{9}{2} or \frac{5 x - 34}{12} = \frac{9}{2} or 10 x - 68 = 108 or 10 x = 108 + 68 or x = \frac{176}{10} = \frac{88}{5} Verification : L . H . S . = \frac{2}{3} (\frac{88}{5} - 5) - \frac{1}{4} (\frac{88}{5} - 2) = \frac{2}{3} \times \frac{63}{5} - \frac{1}{4} \times \frac{78}{5} = \frac{9}{2} = R . H . S .$

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