Q1. $20 = 6 + 2\text{x}$
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$\Rightarrow 20 = 6 + 2\text{x}$
$\Rightarrow 20 - 6 = 2\text{x}$
$\Rightarrow 14 = 2\text{x}$
$\therefore \text{x} = \dfrac{14}{2} = 7$ Ans
$\Rightarrow 20 - 6 = 2\text{x}$
$\Rightarrow 14 = 2\text{x}$
$\therefore \text{x} = \dfrac{14}{2} = 7$ Ans
Q2. $15 + \text{x} = 5\text{x} + 3$
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$\Rightarrow 15 + \text{x} = 5\text{x} + 3$
$\Rightarrow 15 - 3 = 5\text{x} - \text{x}$
$\Rightarrow 12 = 4\text{x}$
$\therefore \text{x} = \dfrac{12}{4} = 3$ Ans
$\Rightarrow 15 - 3 = 5\text{x} - \text{x}$
$\Rightarrow 12 = 4\text{x}$
$\therefore \text{x} = \dfrac{12}{4} = 3$ Ans
Q3. $\dfrac{3\text{x} + 2}{\text{x} - 6} = -7$
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$\Rightarrow \dfrac{3\text{x} + 2}{\text{x} - 6} = -7$
$\Rightarrow 3\text{x} + 2 = -7(\text{x} - 6)$
$\Rightarrow 3\text{x} + 2 = -7\text{x} + 42$
$\Rightarrow 3\text{x} + 7\text{x} = 42 - 2$
$\Rightarrow 10\text{x} = 40$
$\therefore \text{x} = \dfrac{40}{10} = 4$ Ans
$\Rightarrow 3\text{x} + 2 = -7(\text{x} - 6)$
$\Rightarrow 3\text{x} + 2 = -7\text{x} + 42$
$\Rightarrow 3\text{x} + 7\text{x} = 42 - 2$
$\Rightarrow 10\text{x} = 40$
$\therefore \text{x} = \dfrac{40}{10} = 4$ Ans
Q4. $3\text{a} - 4 = 2(4 - \text{a})$
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$\Rightarrow 3\text{a} - 4 = 2(4 - \text{a})$
$\Rightarrow 3\text{a} - 4 = 8 - 2\text{a}$
$\Rightarrow 3\text{a} + 2\text{a} = 8 + 4$
$\Rightarrow 5\text{a} = 12$
$\therefore \text{a} = \dfrac{12}{5} = 2.4$ Ans
$\Rightarrow 3\text{a} - 4 = 8 - 2\text{a}$
$\Rightarrow 3\text{a} + 2\text{a} = 8 + 4$
$\Rightarrow 5\text{a} = 12$
$\therefore \text{a} = \dfrac{12}{5} = 2.4$ Ans
Q5. $3(\text{b} - 4) = 2(4 - \text{b})$
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$\Rightarrow 3(\text{b} - 4) = 2(4 - \text{b})$
$\Rightarrow 3\text{b} - 12 = 8 - 2\text{b}$
$\Rightarrow 3\text{b} + 2\text{b} = 8 + 12$
$\Rightarrow 5\text{b} = 20$
$\therefore \text{b} = \dfrac{20}{5} = 4$ Ans
$\Rightarrow 3\text{b} - 12 = 8 - 2\text{b}$
$\Rightarrow 3\text{b} + 2\text{b} = 8 + 12$
$\Rightarrow 5\text{b} = 20$
$\therefore \text{b} = \dfrac{20}{5} = 4$ Ans
Q6. $\dfrac{\text{x} + 2}{9} = \dfrac{\text{x} + 4}{11}$
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$\Rightarrow \dfrac{\text{x} + 2}{9} = \dfrac{\text{x} + 4}{11}$
$\Rightarrow 11(\text{x} + 2) = 9(\text{x} + 4)$
$\Rightarrow 11\text{x} - 9\text{x} = 36 - 22$
$\Rightarrow 2\text{x} = 14$
$\therefore \text{x} = \dfrac{14}{2} = 7$ Ans
$\Rightarrow 11(\text{x} + 2) = 9(\text{x} + 4)$
$\Rightarrow 11\text{x} - 9\text{x} = 36 - 22$
$\Rightarrow 2\text{x} = 14$
$\therefore \text{x} = \dfrac{14}{2} = 7$ Ans
Q7. $\dfrac{\text{x} - 8}{5} = \dfrac{\text{x} - 12}{9}$
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$\Rightarrow \dfrac{\text{x} - 8}{5} = \dfrac{\text{x} - 12}{9}$
$\Rightarrow 9(\text{x} - 8) = 5(\text{x} - 12)$
$\Rightarrow 9\text{x} - 72 = 5\text{x} - 60$
$\Rightarrow 9\text{x} - 5\text{x} = 72 - 60$
$\Rightarrow 4\text{x} = 12$
$\therefore \text{x} = \dfrac{12}{4} = 3$ Ans
$\Rightarrow 9(\text{x} - 8) = 5(\text{x} - 12)$
$\Rightarrow 9\text{x} - 72 = 5\text{x} - 60$
$\Rightarrow 9\text{x} - 5\text{x} = 72 - 60$
$\Rightarrow 4\text{x} = 12$
$\therefore \text{x} = \dfrac{12}{4} = 3$ Ans
Q8. $5(8\text{x} + 3) = 9(4\text{x} + 7)$
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$\Rightarrow 5(8\text{x} + 3) = 9(4\text{x} + 7)$
$\Rightarrow 40\text{x} + 15 = 36\text{x} + 63$
$\Rightarrow 40\text{x} - 36\text{x} = 63 - 15$
$\Rightarrow 4\text{x} = 48$
$\therefore \text{x} = \dfrac{48}{4} = 12$ Ans
$\Rightarrow 40\text{x} + 15 = 36\text{x} + 63$
$\Rightarrow 40\text{x} - 36\text{x} = 63 - 15$
$\Rightarrow 4\text{x} = 48$
$\therefore \text{x} = \dfrac{48}{4} = 12$ Ans
Q9. $3(\text{x} + 1) = 12 + 4(\text{x} - 1)$
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$\Rightarrow 3(\text{x} + 1) = 12 + 4(\text{x} - 1)$
$\Rightarrow 3\text{x} + 3 = 12 + 4\text{x} - 4$
$\Rightarrow 3\text{x} + 3 = 4\text{x} + 8$
$\Rightarrow 3 - 8 = 4\text{x} - 3\text{x}$
$\therefore \text{x} = -5$ Ans
$\Rightarrow 3\text{x} + 3 = 12 + 4\text{x} - 4$
$\Rightarrow 3\text{x} + 3 = 4\text{x} + 8$
$\Rightarrow 3 - 8 = 4\text{x} - 3\text{x}$
$\therefore \text{x} = -5$ Ans
Q10. $\dfrac{3\text{x}}{4} - \dfrac{\text{1}}{4}(\text{x} - 20) = \dfrac{\text{x}}{4} + 32$
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$\Rightarrow \dfrac{3\text{x}}{4} - \dfrac{\text{1}}{4}(\text{x} - 20) = \dfrac{\text{x}}{4} + 32$
$\Rightarrow \dfrac{3\text{x}}{4} - \dfrac{\text{x} - 20}{4} = \dfrac{\text{x}}{4} + 32$
Simplify by taking L.C.M. = 4
$\Rightarrow 4(\dfrac{3\text{x}}{4}) - 4(\dfrac{\text{x} - 20}{4}) = 4(\dfrac{\text{x}}{4}) + 4(32)$
$\Rightarrow 3\text{x} - \text{x} - 20 = \text{x} + 128$
$\Rightarrow 2\text{x} - 20 = \text{x} + 128$
$\Rightarrow 2\text{x} - \text{x} = 128 - 20$
$\therefore \text{x} = 108$ Ans
$\Rightarrow \dfrac{3\text{x}}{4} - \dfrac{\text{x} - 20}{4} = \dfrac{\text{x}}{4} + 32$
Simplify by taking L.C.M. = 4
$\Rightarrow 4(\dfrac{3\text{x}}{4}) - 4(\dfrac{\text{x} - 20}{4}) = 4(\dfrac{\text{x}}{4}) + 4(32)$
$\Rightarrow 3\text{x} - \text{x} - 20 = \text{x} + 128$
$\Rightarrow 2\text{x} - 20 = \text{x} + 128$
$\Rightarrow 2\text{x} - \text{x} = 128 - 20$
$\therefore \text{x} = 108$ Ans
Q11. $3\text{a} - \dfrac{1}{5} = \dfrac{\text{a}}{5} + 5\dfrac{2}{5}$
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$\Rightarrow 3\text{a} - \dfrac{1}{5} = \dfrac{\text{a}}{5} + 5\dfrac{2}{5}$
Simplify by taking L.C.M. = 5
$\Rightarrow 5(3\text{a}) - 5(\dfrac{1}{5}) = 5(\dfrac{\text{a}}{5}) + 5(\dfrac{27}{5})$
$\Rightarrow 15\text{a} - 1 = \text{a} + 27$
$\Rightarrow 15\text{a} - \text{a} = 27 + 1$
$\Rightarrow 14\text{a} = 28$
$\therefore \text{a} = \dfrac{28}{14} = 2$ Ans
Simplify by taking L.C.M. = 5
$\Rightarrow 5(3\text{a}) - 5(\dfrac{1}{5}) = 5(\dfrac{\text{a}}{5}) + 5(\dfrac{27}{5})$
$\Rightarrow 15\text{a} - 1 = \text{a} + 27$
$\Rightarrow 15\text{a} - \text{a} = 27 + 1$
$\Rightarrow 14\text{a} = 28$
$\therefore \text{a} = \dfrac{28}{14} = 2$ Ans
Q12. $\dfrac{\text{x}}{3} - 2\dfrac{1}{2} = \dfrac{4\text{x}}{9} - \dfrac{2\text{x}}{3}$
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$\Rightarrow \dfrac{\text{x}}{3} - 2\dfrac{1}{2} = \dfrac{4\text{x}}{9} - \dfrac{2\text{x}}{3}$
$\Rightarrow \dfrac{\text{x}}{3} - \dfrac{5}{2} = \dfrac{4\text{x}}{9} - \dfrac{2\text{x}}{3}$
L.C.M. of denominators = 18
$\Rightarrow 18(\dfrac{\text{x}}{3}) - 18(\dfrac{5}{2}) = 18(\dfrac{4\text{x}}{9}) - 18(\dfrac{2\text{x}}{3})$
$\Rightarrow 6\text{x} - 9(5) = 2(4\text{x}) - 6(2\text{x})$
$\Rightarrow 6\text{x} - 45 = 8\text{x} - 12\text{x}$
$\Rightarrow 6\text{x} - 45 = -4\text{x}$
$\Rightarrow 6\text{x} + 4\text{x} = 45$
$\Rightarrow 10\text{x} = 45$
$\therefore \text{x} = \dfrac{45}{10} = 4.5$ Ans
$\Rightarrow \dfrac{\text{x}}{3} - \dfrac{5}{2} = \dfrac{4\text{x}}{9} - \dfrac{2\text{x}}{3}$
L.C.M. of denominators = 18
$\Rightarrow 18(\dfrac{\text{x}}{3}) - 18(\dfrac{5}{2}) = 18(\dfrac{4\text{x}}{9}) - 18(\dfrac{2\text{x}}{3})$
$\Rightarrow 6\text{x} - 9(5) = 2(4\text{x}) - 6(2\text{x})$
$\Rightarrow 6\text{x} - 45 = 8\text{x} - 12\text{x}$
$\Rightarrow 6\text{x} - 45 = -4\text{x}$
$\Rightarrow 6\text{x} + 4\text{x} = 45$
$\Rightarrow 10\text{x} = 45$
$\therefore \text{x} = \dfrac{45}{10} = 4.5$ Ans
Q13. $\dfrac{4(\text{y} + 2)}{5} = 7 + \dfrac{5\text{y}}{13}$
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$\Rightarrow \dfrac{4(\text{y} + 2)}{5} = 7 + \dfrac{5\text{y}}{13}$
$\Rightarrow \dfrac{4\text{y} + 8}{5} = 7 + \dfrac{5\text{y}}{13}$
L.C.M. of denominators = 65
$\Rightarrow 65(\dfrac{4\text{y} + 8}{5}) = 65(7) + 65(\dfrac{5\text{y}}{13})$
$\Rightarrow 13(4\text{y} + 8) = 455 + 5(5\text{y})$
$\Rightarrow 52\text{y} + 104 = 455 + 25\text{y}$
$\Rightarrow 52\text{y} - + 25\text{y} = 455 -104$
$\Rightarrow 27\text{y} = 351$
$\therefore \text{y} = \dfrac{351}{27} = 13$ Ans
$\Rightarrow \dfrac{4\text{y} + 8}{5} = 7 + \dfrac{5\text{y}}{13}$
L.C.M. of denominators = 65
$\Rightarrow 65(\dfrac{4\text{y} + 8}{5}) = 65(7) + 65(\dfrac{5\text{y}}{13})$
$\Rightarrow 13(4\text{y} + 8) = 455 + 5(5\text{y})$
$\Rightarrow 52\text{y} + 104 = 455 + 25\text{y}$
$\Rightarrow 52\text{y} - + 25\text{y} = 455 -104$
$\Rightarrow 27\text{y} = 351$
$\therefore \text{y} = \dfrac{351}{27} = 13$ Ans
Q14. $\dfrac{\text{a} + 5}{6} - \dfrac{\text{a} + 1}{9} = \dfrac{\text{a} + 3}{4}$
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$\Rightarrow \dfrac{\text{a} + 5}{6} - \dfrac{\text{a} + 1}{9} = \dfrac{\text{a} + 3}{4}$
L.C.M. of denominators = 36
$\Rightarrow 36(\dfrac{\text{a} + 5}{6}) - 36(\dfrac{\text{a} + 1}{9}) = 36(\dfrac{\text{a} + 3}{4})$
$\Rightarrow 6(\text{a} + 5) - 4(\text{a} + 1) = 9(\text{a} + 3)$
$\Rightarrow 6\text{a} + 30 - 4\text{a} - 4 = 9\text{a} + 27$
$\Rightarrow 2\text{a} + 26 = 9\text{a} + 27$
$\Rightarrow 26 - 27 = 9\text{a} - 2\text{a}$
$\Rightarrow -1 = 7\text{a}$
$\therefore \text{a} = -\dfrac{1}{7}$ Ans
L.C.M. of denominators = 36
$\Rightarrow 36(\dfrac{\text{a} + 5}{6}) - 36(\dfrac{\text{a} + 1}{9}) = 36(\dfrac{\text{a} + 3}{4})$
$\Rightarrow 6(\text{a} + 5) - 4(\text{a} + 1) = 9(\text{a} + 3)$
$\Rightarrow 6\text{a} + 30 - 4\text{a} - 4 = 9\text{a} + 27$
$\Rightarrow 2\text{a} + 26 = 9\text{a} + 27$
$\Rightarrow 26 - 27 = 9\text{a} - 2\text{a}$
$\Rightarrow -1 = 7\text{a}$
$\therefore \text{a} = -\dfrac{1}{7}$ Ans
Q15. $\dfrac{2\text{x} - 13}{5} - \dfrac{\text{x} - 3}{11} = \dfrac{\text{x} - 9}{5} + 1$
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$\Rightarrow \dfrac{2\text{x} - 13}{5} - \dfrac{\text{x} - 3}{11} = \dfrac{\text{x} - 9}{5} + 1$
L.C.M. of denominators = 55
$\Rightarrow 55(\dfrac{2\text{x} - 13}{5}) - 55(\dfrac{\text{x} - 3}{11}) = 55(\dfrac{\text{x} - 9}{5}) + 55(1)$
$\Rightarrow 11(2\text{x} - 13) - 5(\text{x} - 3) = 11(\text{x} - 9) + 55$
$\Rightarrow 22\text{x} - 143 - 5\text{x} + 15 = 11\text{x} - 99 + 55$
$\Rightarrow 17\text{x} - 128 = 11\text{x} - 44$
$\Rightarrow 17\text{x} - 11\text{x} = 128 - 44$
$\Rightarrow 6\text{x} = 128 - 44$
$\Rightarrow 6\text{x} = 84$
$\therefore \text{x} = \dfrac{84}{6} = 14$ Ans
L.C.M. of denominators = 55
$\Rightarrow 55(\dfrac{2\text{x} - 13}{5}) - 55(\dfrac{\text{x} - 3}{11}) = 55(\dfrac{\text{x} - 9}{5}) + 55(1)$
$\Rightarrow 11(2\text{x} - 13) - 5(\text{x} - 3) = 11(\text{x} - 9) + 55$
$\Rightarrow 22\text{x} - 143 - 5\text{x} + 15 = 11\text{x} - 99 + 55$
$\Rightarrow 17\text{x} - 128 = 11\text{x} - 44$
$\Rightarrow 17\text{x} - 11\text{x} = 128 - 44$
$\Rightarrow 6\text{x} = 128 - 44$
$\Rightarrow 6\text{x} = 84$
$\therefore \text{x} = \dfrac{84}{6} = 14$ Ans