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Linear Equation in One Variable Questions and Answers



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Q31. $(\text{x} - 2)^{2} = (\text{x} - 1)(\text{x} - 1)$


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$\Rightarrow (\text{x} - 2)^{2} = (\text{x} - 1)(\text{x} - 1)$

$\Rightarrow (\text{x} - 2)(\text{x} - 2) = \text{x}^{2} - \text{x} + \text{x} - 1$

$\Rightarrow \text{x}^{2} - 2\text{x} - 2\text{x} + 4 = \text{x}^{2} - 1$

$\Rightarrow \text{x}^{2} - 4\text{x} + 4 = \text{x}^{2} - 1$

$\Rightarrow \text{x}^{2} - 4\text{x} - \text{x}^{2} = -1 - 4$

$\Rightarrow -4\text{x} = -5$

$\therefore \text{x} = \dfrac{5}{4} = 1.25$ Ans


Q32. $3(3\text{x} - 4) - 2(4\text{x} - 5) = 6$


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$\Rightarrow 3(3\text{x} - 4) - 2(4\text{x} - 5) = 6$

$\Rightarrow 9\text{x} - 12 - 8\text{x} + 10 = 6$

$\Rightarrow \text{x} - 2 = 6$

$\Rightarrow \text{x} = 6 + 2$

$\therefore \text{x} = 8$ Ans


Q33. $\dfrac{\text{x} - 1}{2} - \dfrac{\text{x} + 1}{3} = 5 - \text{x}$


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$\Rightarrow \dfrac{\text{x} - 1}{2} - \dfrac{\text{x} + 1}{3} = 5 - \text{x}$

L.C.M. of denominators = 6

$\Rightarrow 6(\dfrac{\text{x} - 1}{2}) - 6(\dfrac{\text{x} + 1}{3}) = 6(5 - \text{x})$

$\Rightarrow 3(\text{x} - 1) - 2(\text{x} + 1) = 30 - 6\text{x}$

$\Rightarrow 3\text{x} - 3 - 2\text{x} - 2 = 30 - 6\text{x}$

$\Rightarrow \text{x} - 5 = 30 - 6\text{x}$

$\Rightarrow \text{x} + 6\text{x} = 30 + 5$

$\Rightarrow 7\text{x} = 35$

$\therefore \text{x} = \dfrac{35}{7} = 5$ Ans


Q34. $\dfrac{3\text{x} - 1}{5} - \dfrac{\text{x}}{7} = 3$


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$\Rightarrow \dfrac{3\text{x} - 1}{5} - \dfrac{\text{x}}{7} = 3$

L.C.M. of denominators = 35

$\Rightarrow 35(\dfrac{3\text{x} - 1}{5}) - 35(\dfrac{\text{x}}{7}) = 35(3)$

$\Rightarrow 7(3\text{x} - 1) - 5(\text{x}) = 105$

$\Rightarrow 21\text{x} - 7 - 5\text{x} = 105$

$\Rightarrow 16\text{x} - 7 = 105$

$\Rightarrow 16\text{x} = 105 + 7$

$\Rightarrow 16\text{x} = 112$

$\therefore \text{x} = \dfrac{112}{16} = 7$ Ans


Q35. $\dfrac{1}{3}(\text{x} - 3) - \text{x} - 3 = 2(\text{x} - 1)$


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$\Rightarrow \dfrac{1}{3}(\text{x} - 3) - \text{x} - 3 = 2(\text{x} - 1)$

$\Rightarrow \dfrac{1}{3}(\text{x}) - \dfrac{1}{3}(3) - \text{x} - 3 = 2\text{x} - 2$

$\Rightarrow \dfrac{\text{x}}{3} - 1 - \text{x} - 3 = 2\text{x} - 2$

$\Rightarrow \dfrac{\text{x}}{3} - \text{x} - 4 = 2\text{x} - 2$

$\Rightarrow (\dfrac{\text{x - 3\text{x}}}{3}) - 4 = 2\text{x} - 2$

$\Rightarrow -\dfrac{\text{2\text{x}}}{3} - 4 = 2\text{x} - 2$

$\Rightarrow -\dfrac{\text{2\text{x}}}{3} - 2\text{x} = 4 - 2$

$\Rightarrow \dfrac{\text{-2\text{x} - 6\text{x}}}{3} = 2$

$\Rightarrow -\dfrac{\text{8\text{x}}}{3} = 2$

$\Rightarrow -8\text{x} = 3(2)$

$\Rightarrow -8\text{x} = 6$

$\therefore \text{x} = -\dfrac{6}{8} = -\dfrac{3}{4}$ Ans


Q36. $\dfrac{5\text{y} - 4}{8} - \dfrac{\text{y} - 3}{5} = \dfrac{\text{y} + 6}{4}$


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$\Rightarrow \dfrac{5\text{y} - 4}{8} - \dfrac{\text{y} - 3}{5} = \dfrac{\text{y} + 6}{4}$

L.C.M. of denominators = 40

$\Rightarrow 40(\dfrac{5\text{y} - 4}{8}) - 40(\dfrac{\text{y} - 3}{5}) = 40(\dfrac{\text{y} + 6}{4})$

$\Rightarrow 5(5\text{y} - 4) - 8(\text{y} - 3) = 10(\text{y} + 6)$

$\Rightarrow 25\text{y} - 20 - 8\text{y} + 24 = 10\text{y} + 60$

$\Rightarrow 17\text{y} + 4 = 10\text{y} + 60$

$\Rightarrow 17\text{y} - 10\text{y} = 60 - 4$

$\Rightarrow 7\text{y} = 56$

$\therefore \text{y} = \dfrac{56}{7} = 8$ Ans


Q37. $\dfrac{\text{x} - 2}{4} - \dfrac{1}{3} = \text{x} - \dfrac{2\text{x} - 1}{3}$


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$\Rightarrow \dfrac{\text{x} - 2}{4} - \dfrac{1}{3} = \text{x} - \dfrac{2\text{x} - 1}{3}$

L.C.M. of denominators = 12

$\Rightarrow 12(\dfrac{\text{x} - 2}{4}) - 12(\dfrac{1}{3}) = 12(\text{x}) - 12(\dfrac{2\text{x} - 1}{3})$

$\Rightarrow 3(\text{x} - 2) - 4(1) = 12\text{x} - 4(2\text{x} - 1)$

$\Rightarrow 3\text{x} - 6 - 4 = 12\text{x} - 8\text{x} - 4$

$\Rightarrow 3\text{x} - 10 = 4\text{x} - 4$

$\Rightarrow 3\text{x} - 4\text{x} = 10 - 4$

$\Rightarrow -\text{x} = 6$

$\therefore \text{x} = \dfrac{1}{6} = -6$ Ans


Q38. $\dfrac{5\text{x} - 4}{7} - \dfrac{1}{2} = \dfrac{3\text{x} - 1}{14}$


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$\Rightarrow \dfrac{5\text{x} - 4}{7} - \dfrac{1}{2} = \dfrac{3\text{x} - 1}{14}$

L.C.M. of denominators = 14

$\Rightarrow 14(\dfrac{5\text{x} - 4}{7}) - 14(\dfrac{1}{2}) = 14(\dfrac{3\text{x} - 1}{14})$

$\Rightarrow 2(5\text{x} - 4) - 7(1) = 3\text{x} - 1$

$\Rightarrow 10\text{x} - 8 - 7 = 3\text{x} - 1$

$\Rightarrow 10\text{x} - 15 = 3\text{x} - 1$

$\Rightarrow 10\text{x} - 3\text{x} = 15 - 1$

$\Rightarrow 7\text{x} = 14$

$\therefore \text{x} = \dfrac{14}{7} = 2$ Ans


Q39. $\dfrac{2\text{x} - 3}{2\text{x} - 1} = \dfrac{3\text{x} - 1}{3\text{x} + 5}$


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$\Rightarrow \dfrac{2\text{x} - 3}{2\text{x} - 1} = \dfrac{3\text{x} - 1}{3\text{x} + 5}$

$\Rightarrow (3\text{x} + 5)(2\text{x} - 3) = (2\text{x} - 1)(3\text{x} - 1)$

$\Rightarrow 6\text{x}^{2} + 10\text{x} - 9\text{x} - 15 = 6\text{x}^{2} - 2\text{x} - 3\text{x} + 1$

$\Rightarrow 6\text{x}^{2} + \text{x} - 15 = 6\text{x}^{2} - 5\text{x} + 1$

$\Rightarrow 6\text{x}^{2} + \text{x} - (6\text{x}^{2} - 5\text{x}) = 15 + 1$

$\Rightarrow 6\text{x}^{2} + \text{x} - 6\text{x}^{2} + 5\text{x} = 15 + 1$

$\Rightarrow 6\text{x} = 16$

$\therefore \text{x} = \dfrac{16}{6} = \dfrac{8}{3} = 2\dfrac{2}{3}$ Ans


Q40. $\dfrac{4}{5}(\text{x} + \dfrac{5}{6}) + \dfrac{2}{3}(\text{x} - \dfrac{1}{4}) = 1\dfrac{1}{9}$


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$\Rightarrow \dfrac{4}{5}(\text{x} + \dfrac{5}{6}) + \dfrac{2}{3}(\text{x} - \dfrac{1}{4}) = 1\dfrac{1}{9}$

$\Rightarrow \dfrac{4}{5}(\text{x}) + \dfrac{4}{5}(\dfrac{5}{6}) + \dfrac{2}{3}(\text{x}) - \dfrac{2}{3}(\dfrac{1}{4}) = \dfrac{10}{9}$

$\Rightarrow \dfrac{4\text{x}}{5} + \dfrac{4}{6} + \dfrac{2\text{x}}{3} - \dfrac{1}{6} = \dfrac{10}{9}$

$\Rightarrow (\dfrac{4\text{x}}{5} + \dfrac{2\text{x}}{3}) + (\dfrac{4}{6} - \dfrac{1}{6}) = \dfrac{10}{9}$

$\Rightarrow (\dfrac{12\text{x} + 10\text{x}}{15}) + (\dfrac{4 - 1}{6}) = \dfrac{10}{9}$

$\Rightarrow \dfrac{22\text{x}}{15} + \dfrac{3}{6} = \dfrac{10}{9}$

$\Rightarrow \dfrac{22\text{x}}{15} = \dfrac{10}{9} - \dfrac{3}{6}$

$\Rightarrow \dfrac{22\text{x}}{15} = \dfrac{20 - 9}{18}$

$\Rightarrow \dfrac{22\text{x}}{15} = \dfrac{11}{18}$

$\Rightarrow 18(22\text{x}) = 15(11)$

$\Rightarrow 396\text{x} = 165$

$\therefore \text{x} = \dfrac{165}{396} = \dfrac{55}{132} = \dfrac{5}{12}$ Ans


Q41. $4 - \dfrac{\text{x} - 3}{2} = 3$


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$\Rightarrow 4 - \dfrac{\text{x} - 3}{2} = 3$

Simplify by taking L.C.M. = 2

$\Rightarrow 2(4) - 2(\dfrac{\text{x} - 3}{2}) = 2(3)$

$\Rightarrow 8 - (\text{x} - 3) = 6$

$\Rightarrow 8 - \text{x} + 3 = 6$

$\Rightarrow -\text{x} + 11 = 6$

$\Rightarrow -\text{x} = 6 - 11$

$\Rightarrow -\text{x} = -5$

$\therefore \text{x} = 5$ Ans


Q42. $\dfrac{3\text{x} - 1}{4\dfrac{1}{2}} - \dfrac{\text{x} - 1}{2} = 0$


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$\Rightarrow \dfrac{3\text{x} - 1}{4\dfrac{1}{2}} - \dfrac{\text{x} - 1}{2} = 0$

$\Rightarrow \dfrac{3\text{x} - 1}{\dfrac{9}{2}} - \dfrac{\text{x} - 1}{2} = 0$

$\Rightarrow \dfrac{2(3\text{x} - 1)}{9} - \dfrac{\text{x} - 1}{2} = 0$

$\Rightarrow \dfrac{6\text{x} - 2}{9} - \dfrac{\text{x} - 1}{2} = 0$

Simplify by taking L.C.M. = 18

$\Rightarrow 18(\dfrac{6\text{x} - 2}{9}) - 18(\dfrac{\text{x} - 1}{2}) = 18(0)$

$\Rightarrow 2(6\text{x} - 2) - 9(\text{x} - 1) = 0$

$\Rightarrow 12\text{x} - 4 - 9\text{x} + 9 = 0$

$\Rightarrow 3\text{x} + 5 = 0$

$\Rightarrow 3\text{x} = -5$

$\therefore \text{x} = -\dfrac{5}{3} = -1\dfrac{2}{3}$ Ans


Q43. $\dfrac{2}{5}(\text{x} - 1) = 1 - \dfrac{3}{5}(3\text{x} - 5)$


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$\Rightarrow \dfrac{2}{5}(\text{x} - 1) = 1 - \dfrac{3}{5}(3\text{x} - 5)$

$\Rightarrow \dfrac{2}{5}(\text{x}) - \dfrac{2}{5}(1) = 1 - \dfrac{3}{5}(3\text{x}) + \dfrac{3}{5}(5)$

$\Rightarrow \dfrac{2\text{x}}{5} - \dfrac{2}{5} = 1 - \dfrac{9\text{x}}{5} + 3$

$\Rightarrow \dfrac{2\text{x}}{5} - \dfrac{2}{5} = 4 - \dfrac{9\text{x}}{5}$

$\Rightarrow \dfrac{2\text{x}}{5} + \dfrac{9\text{x}}{5} = 4 + \dfrac{2}{5}$

$\Rightarrow \dfrac{2\text{x} + 9\text{x}}{5} = \dfrac{20 + 2}{5}$

$\Rightarrow \dfrac{11\text{x}}{5} = \dfrac{22}{5}$

Upon simplifying by L.C.M. = 5, we have:

$\Rightarrow 11\text{x} = 22$

$\therefore \text{x} = \dfrac{22}{11} = 2$ Ans


Q44. $0.5 + \dfrac{\text{x}}{3} = \text{x} - 3$


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$\Rightarrow 0.5 + \dfrac{\text{x}}{3} = \text{x} - 3$

Simplify by taking L.C.M. = 3

$\Rightarrow 3(0.5) + 3(\dfrac{\text{x}}{3}) = 3(\text{x} - 3)$

$\Rightarrow 1.5 + \text{x} = 3\text{x} - 9$

$\Rightarrow \text{x} - 3\text{x} = -9 - 1.5$

$\Rightarrow -2\text{x} = -10.5$

$\Rightarrow 2\text{x} = 10.5$

$\therefore \text{x} = \dfrac{10.5}{2} = 5.25$ Ans


Q45. $\dfrac{1}{2} - \dfrac{9 - \text{x}}{11} = \dfrac{3\text{x}}{22}$


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$\Rightarrow \dfrac{1}{2} - \dfrac{9 - \text{x}}{11} = \dfrac{3\text{x}}{22}$

Simplify by taking L.C.M. = 22

$\Rightarrow 22(\dfrac{1}{2}) - 22(\dfrac{9 - \text{x}}{11}) = 22(\dfrac{3\text{x}}{22})$

$\Rightarrow 11(1) - 2(9 - \text{x}) = 3\text{x}$

$\Rightarrow 11 - 18 + 2\text{x} = 3\text{x}$

$\Rightarrow -7 = 3\text{x} - 2\text{x}$

$\Rightarrow -7 = \text{x}$

$\therefore \text{x} = -7$ Ans


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