Q16. $6(6\text{x} - 5) - 5(7\text{x} - 8) = 12(4 - \text{x}) + 1$
Show Answer
$\Rightarrow 6(6\text{x} - 5) - 5(7\text{x} - 8) = 12(4 - \text{x}) + 1$
$\Rightarrow 36\text{x} - 30 - 35\text{x} + 40 = 48 - 12\text{x} + 1$
$\Rightarrow \text{x} + 10 = 49 - 12\text{x}$
$\Rightarrow \text{x} + 12\text{x} = 49 - 10$
$\Rightarrow 13\text{x} = 39$
$\Rightarrow \text{x} = \dfrac{39}{13} = 3$ Ans
$\Rightarrow 36\text{x} - 30 - 35\text{x} + 40 = 48 - 12\text{x} + 1$
$\Rightarrow \text{x} + 10 = 49 - 12\text{x}$
$\Rightarrow \text{x} + 12\text{x} = 49 - 10$
$\Rightarrow 13\text{x} = 39$
$\Rightarrow \text{x} = \dfrac{39}{13} = 3$ Ans
Q17. $(\text{x} - 5)(\text{x} + 3) = (\text{x} - 7)(\text{x} + 4)$
Show Answer
$\Rightarrow (\text{x} - 5)(\text{x} + 3) = (\text{x} - 7)(\text{x} + 4)$
$\Rightarrow \text{x}^{2} + 3\text{x} - 5\text{x} - 15 = \text{x}^{2} + 4\text{x} - 7\text{x} - 28$
$\Rightarrow \text{x}^{2} - 2\text{x} - 15 = \text{x}^{2} - 3\text{x} - 28$
$\Rightarrow \text{x}^{2} - 2\text{x} - \text{x}^{2} + 3\text{x} = 15 - 28$
$\Rightarrow \text{x} = -13$ Ans
$\Rightarrow \text{x}^{2} + 3\text{x} - 5\text{x} - 15 = \text{x}^{2} + 4\text{x} - 7\text{x} - 28$
$\Rightarrow \text{x}^{2} - 2\text{x} - 15 = \text{x}^{2} - 3\text{x} - 28$
$\Rightarrow \text{x}^{2} - 2\text{x} - \text{x}^{2} + 3\text{x} = 15 - 28$
$\Rightarrow \text{x} = -13$ Ans
Q18. $(\text{x} - 5)^{2} - (\text{x} + 2)^{2} = -2$
Show Answer
$\Rightarrow (\text{x} - 5)^{2} - (\text{x} + 2)^{2} = -2$
We have the formula:
(i) $(\text{a} - \text{b})^{2} = \text{a}^{2} - 2\text{ab} + \text{b}^{2}$
(ii) $(\text{a} + \text{b})^{2} = \text{a}^{2} + 2\text{ab} + \text{b}^{2}$
Upon applying the formula, we have:
$\therefore (\text{x} - 5)^{2} = \text{x}^{2} - 2(\text{x})(5) + (5)^{2} \Rightarrow \text{x}^{2} - 10\text{x} + 25$
$\therefore (\text{x} + 2)^{2} = \text{x}^{2} + 2(\text{x})(2) + (2)^{2} \Rightarrow \text{x}^{2} + 4\text{x} + 4$
Now:
$\Rightarrow (\text{x}^{2} - 10\text{x} + 25) - (\text{x}^{2} + 4\text{x} + 4) = -2$
$\Rightarrow \text{x}^{2} - 10\text{x} + 25 - \text{x}^{2} - 4\text{x} - 4 = -2$
$\Rightarrow -14\text{x} + 21 = -2$
$\Rightarrow -14\text{x} = -21 - 2$
$\Rightarrow -14\text{x} = -23$
$\therefore \text{x} = \dfrac{23}{14} = 1\dfrac{3}{14}$ Ans
We have the formula:
(i) $(\text{a} - \text{b})^{2} = \text{a}^{2} - 2\text{ab} + \text{b}^{2}$
(ii) $(\text{a} + \text{b})^{2} = \text{a}^{2} + 2\text{ab} + \text{b}^{2}$
Upon applying the formula, we have:
$\therefore (\text{x} - 5)^{2} = \text{x}^{2} - 2(\text{x})(5) + (5)^{2} \Rightarrow \text{x}^{2} - 10\text{x} + 25$
$\therefore (\text{x} + 2)^{2} = \text{x}^{2} + 2(\text{x})(2) + (2)^{2} \Rightarrow \text{x}^{2} + 4\text{x} + 4$
Now:
$\Rightarrow (\text{x}^{2} - 10\text{x} + 25) - (\text{x}^{2} + 4\text{x} + 4) = -2$
$\Rightarrow \text{x}^{2} - 10\text{x} + 25 - \text{x}^{2} - 4\text{x} - 4 = -2$
$\Rightarrow -14\text{x} + 21 = -2$
$\Rightarrow -14\text{x} = -21 - 2$
$\Rightarrow -14\text{x} = -23$
$\therefore \text{x} = \dfrac{23}{14} = 1\dfrac{3}{14}$ Ans
Q19. $(\text{x} - 1)(\text{x} + 6) - (\text{x} - 2)(\text{x} - 3) = 3$
Show Answer
$\Rightarrow (\text{x} - 1)(\text{x} + 6) - (\text{x} - 2)(\text{x} - 3) = 3$
$\Rightarrow (\text{x}^{2} + 6\text{x} - \text{x} - 6) - (\text{x}^{2} - 3\text{x} - 2\text{x} + 6) = 3$
$\Rightarrow (\text{x}^{2} + 5\text{x} - 6) - (\text{x}^{2} - 5\text{x} + 6) = 3$
$\Rightarrow \text{x}^{2} + 5\text{x} - 6 - \text{x}^{2} + 5\text{x} - 6 = 3$
$\Rightarrow 10\text{x} - 12 = 3$
$\Rightarrow 10\text{x} = 3 + 12$
$\Rightarrow 10\text{x} = 15$
$\therefore \text{x} = \dfrac{15}{10} = \dfrac{3}{2} = 1\dfrac{1}{2}$ Ans
$\Rightarrow (\text{x}^{2} + 6\text{x} - \text{x} - 6) - (\text{x}^{2} - 3\text{x} - 2\text{x} + 6) = 3$
$\Rightarrow (\text{x}^{2} + 5\text{x} - 6) - (\text{x}^{2} - 5\text{x} + 6) = 3$
$\Rightarrow \text{x}^{2} + 5\text{x} - 6 - \text{x}^{2} + 5\text{x} - 6 = 3$
$\Rightarrow 10\text{x} - 12 = 3$
$\Rightarrow 10\text{x} = 3 + 12$
$\Rightarrow 10\text{x} = 15$
$\therefore \text{x} = \dfrac{15}{10} = \dfrac{3}{2} = 1\dfrac{1}{2}$ Ans
Q20. $\dfrac{3\text{x}}{\text{x} + 6} - \dfrac{\text{x}}{\text{x} + 5} = 2$
Show Answer
$\Rightarrow \dfrac{3\text{x}}{\text{x} + 6} - \dfrac{\text{x}}{\text{x} + 5} = 2$
L.C.M. of denominators = $\text{x}^{2} + 11\text{x} + 30$
$\Rightarrow 3\text{x}(\text{x} + 5) - \text{x}(\text{x} + 6) = 2(\text{x}^{2} + 11\text{x} + 30)$
$\Rightarrow 3\text{x}^{2} + 15\text{x} - \text{x}^{2} - 6\text{x} = 2\text{x}^{2} + 22\text{x} + 60$
$\Rightarrow 2\text{x}^{2} + 9\text{x} = 2\text{x}^{2} + 22\text{x} + 60$
$\Rightarrow 2\text{x}^{2} + 9\text{x} - 2\text{x}^{2} - 22\text{x} = 60$
$\Rightarrow -13\text{x} = 60$
$\therefore \text{x} = -\dfrac{60}{13} = -4\dfrac{8}{13}$ Ans
L.C.M. of denominators = $\text{x}^{2} + 11\text{x} + 30$
$\Rightarrow 3\text{x}(\text{x} + 5) - \text{x}(\text{x} + 6) = 2(\text{x}^{2} + 11\text{x} + 30)$
$\Rightarrow 3\text{x}^{2} + 15\text{x} - \text{x}^{2} - 6\text{x} = 2\text{x}^{2} + 22\text{x} + 60$
$\Rightarrow 2\text{x}^{2} + 9\text{x} = 2\text{x}^{2} + 22\text{x} + 60$
$\Rightarrow 2\text{x}^{2} + 9\text{x} - 2\text{x}^{2} - 22\text{x} = 60$
$\Rightarrow -13\text{x} = 60$
$\therefore \text{x} = -\dfrac{60}{13} = -4\dfrac{8}{13}$ Ans
Q21. $\dfrac{1}{\text{x} - 1} + \dfrac{2}{\text{x} - 2} = \dfrac{3}{\text{x} - 3}$
Show Answer
$\Rightarrow \dfrac{1}{\text{x} - 1} + \dfrac{2}{\text{x} - 2} = \dfrac{3}{\text{x} - 3}$
L.C.M. of denominators = $\text{x}^{3} - 4\text{x}^{2} + 11\text{x} - 6$
$\Rightarrow (\text{x}^{2} - 5\text{x} + 6) + 2(\text{x}^{2} - 4\text{x} + 3) = 3(\text{x}^{2} - 3\text{x} + 2)$
$\Rightarrow \text{x}^{2} - 5\text{x} + 6 + 2\text{x}^{2} - 8\text{x} + 6 = 3\text{x}^{2} - 9\text{x} + 6$
$\Rightarrow 3\text{x}^{2} - 13\text{x} + 12 = 3\text{x}^{2} - 9\text{x} + 6$
$\Rightarrow 3\text{x}^{2} - 13\text{x} - 3\text{x}^{2} + 9\text{x} = 6 - 12$
$\Rightarrow -4\text{x} = -6$
$\therefore \text{x} = \dfrac{6}{4} = \dfrac{3}{2} = 1\dfrac{1}{2}$ Ans
L.C.M. of denominators = $\text{x}^{3} - 4\text{x}^{2} + 11\text{x} - 6$
$\Rightarrow (\text{x}^{2} - 5\text{x} + 6) + 2(\text{x}^{2} - 4\text{x} + 3) = 3(\text{x}^{2} - 3\text{x} + 2)$
$\Rightarrow \text{x}^{2} - 5\text{x} + 6 + 2\text{x}^{2} - 8\text{x} + 6 = 3\text{x}^{2} - 9\text{x} + 6$
$\Rightarrow 3\text{x}^{2} - 13\text{x} + 12 = 3\text{x}^{2} - 9\text{x} + 6$
$\Rightarrow 3\text{x}^{2} - 13\text{x} - 3\text{x}^{2} + 9\text{x} = 6 - 12$
$\Rightarrow -4\text{x} = -6$
$\therefore \text{x} = \dfrac{6}{4} = \dfrac{3}{2} = 1\dfrac{1}{2}$ Ans
Q22. $\dfrac{\text{x} - 1}{7\text{x} - 14} = \dfrac{\text{x} - 3}{7\text{x} - 26}$
Show Answer
$\Rightarrow \dfrac{\text{x} - 1}{7\text{x} - 14} = \dfrac{\text{x} - 3}{7\text{x} - 26}$
$\Rightarrow 7\text{x} - 26(\text{x} - 1) = 7\text{x} - 14(\text{x} - 3)$
$\Rightarrow 7\text{x}^{2} - 26\text{x} - 7\text{x} + 26 = 7\text{x}^{2} - 14\text{x} - 21\text{x} + 42$
$\Rightarrow 7\text{x}^{2} - 33\text{x} + 26 = 7\text{x}^{2} - 35\text{x} + 42$
$\Rightarrow 7\text{x}^{2} - 33\text{x} - 7\text{x}^{2} + 35\text{x} = 42 - 26$
$\Rightarrow 2\text{x} = 16$
$\therefore \text{x} = \dfrac{16}{2} = 8$ Ans
$\Rightarrow 7\text{x} - 26(\text{x} - 1) = 7\text{x} - 14(\text{x} - 3)$
$\Rightarrow 7\text{x}^{2} - 26\text{x} - 7\text{x} + 26 = 7\text{x}^{2} - 14\text{x} - 21\text{x} + 42$
$\Rightarrow 7\text{x}^{2} - 33\text{x} + 26 = 7\text{x}^{2} - 35\text{x} + 42$
$\Rightarrow 7\text{x}^{2} - 33\text{x} - 7\text{x}^{2} + 35\text{x} = 42 - 26$
$\Rightarrow 2\text{x} = 16$
$\therefore \text{x} = \dfrac{16}{2} = 8$ Ans
Q23. $\dfrac{1}{\text{x} - 1} - \dfrac{1}{\text{x}} = \dfrac{1}{\text{x} + 3} - \dfrac{1}{\text{x} + 4}$
Show Answer
$\Rightarrow \dfrac{1}{\text{x} - 1} - \dfrac{1}{\text{x}} = \dfrac{1}{\text{x} + 3} - \dfrac{1}{\text{x} + 4}$
(i) L.C.M. of denominator $\text{x}(\text{x} - 1) = \text{x}^{2} - \text{x}$
(ii) L.C.M. of denominator $(\text{x} + 3)(\text{x} + 4) = \text{x}^{2} + 7\text{x} + 12$
$\Rightarrow \dfrac{\text{x} - (\text{x} - 1)}{\text{x}^{2} - \text{x}} = \dfrac{(\text{x} + 4) - (\text{x} + 3)}{\text{x}^{2} + 7\text{x} + 12}$
$\Rightarrow \dfrac{\text{x} - \text{x} + 1}{\text{x}^{2} - \text{x}} = \dfrac{\text{x} + 4 - \text{x} - 3}{\text{x}^{2} + 7\text{x} + 12}$
$\Rightarrow \dfrac{1}{\text{x}^{2} - \text{x}} = \dfrac{1}{\text{x}^{2} + 7\text{x} + 12}$
$\Rightarrow \text{x}^{2} + 7\text{x} + 12 = \text{x}^{2} - \text{x}$
$\Rightarrow \text{x}^{2} + 7\text{x} - \text{x}^{2} + \text{x} = 12$
$\Rightarrow 8\text{x} = 12$
$\therefore \text{x} = \dfrac{12}{8} = \dfrac{3}{2} = 1\dfrac{1}{2}$ Ans
(i) L.C.M. of denominator $\text{x}(\text{x} - 1) = \text{x}^{2} - \text{x}$
(ii) L.C.M. of denominator $(\text{x} + 3)(\text{x} + 4) = \text{x}^{2} + 7\text{x} + 12$
$\Rightarrow \dfrac{\text{x} - (\text{x} - 1)}{\text{x}^{2} - \text{x}} = \dfrac{(\text{x} + 4) - (\text{x} + 3)}{\text{x}^{2} + 7\text{x} + 12}$
$\Rightarrow \dfrac{\text{x} - \text{x} + 1}{\text{x}^{2} - \text{x}} = \dfrac{\text{x} + 4 - \text{x} - 3}{\text{x}^{2} + 7\text{x} + 12}$
$\Rightarrow \dfrac{1}{\text{x}^{2} - \text{x}} = \dfrac{1}{\text{x}^{2} + 7\text{x} + 12}$
$\Rightarrow \text{x}^{2} + 7\text{x} + 12 = \text{x}^{2} - \text{x}$
$\Rightarrow \text{x}^{2} + 7\text{x} - \text{x}^{2} + \text{x} = 12$
$\Rightarrow 8\text{x} = 12$
$\therefore \text{x} = \dfrac{12}{8} = \dfrac{3}{2} = 1\dfrac{1}{2}$ Ans
Q24. Solve: $\dfrac{2\text{x}}{3} - \dfrac{\text{x} - 1}{6} + \dfrac{7\text{x} - 1}{4} = 2\dfrac{1}{6}$
Hence, find the value of 'a', if $\dfrac{1}{\text{a}} + 5\text{x} = 8$
Show Answer
$\Rightarrow \dfrac{2\text{x}}{3} - \dfrac{\text{x} - 1}{6} + \dfrac{7\text{x} - 1}{4} = 2\dfrac{1}{6}$
L.C.M. of denominators = 12
$\Rightarrow 12(\dfrac{\text{2x}}{3}) - 12(\dfrac{\text{x} - 1}{6}) + 12(\dfrac{7\text{x} - 1}{4}) = 12(\dfrac{13}{6})$
$\Rightarrow 4(2\text{x}) - 2(\text{x} - 1) + 3(7\text{x} - 1) = 2(13)$
$\Rightarrow 8\text{x} - 2\text{x} + 2 + 21\text{x} - 3 = 26$
$\Rightarrow 27\text{x} - 1 = 26$
$\Rightarrow 27\text{x} = 26 + 1$
$\Rightarrow 27\text{x} = 27$
$\therefore \text{x} = \dfrac{27}{27} = 1$ Ans
Now $\dfrac{1}{\text{a}} + 5\text{x} = 8$, where $\text{x} = 1$
$\Rightarrow \dfrac{1}{\text{a}} + 5(1) = 8$
$\Rightarrow \dfrac{1}{\text{a}} + 5 = 8$
$\Rightarrow \dfrac{1}{\text{a}} = 8 - 5$
$\Rightarrow \dfrac{1}{\text{a}} = 3$
$\therefore \text{a} = \dfrac{1}{3}$ Ans
L.C.M. of denominators = 12
$\Rightarrow 12(\dfrac{\text{2x}}{3}) - 12(\dfrac{\text{x} - 1}{6}) + 12(\dfrac{7\text{x} - 1}{4}) = 12(\dfrac{13}{6})$
$\Rightarrow 4(2\text{x}) - 2(\text{x} - 1) + 3(7\text{x} - 1) = 2(13)$
$\Rightarrow 8\text{x} - 2\text{x} + 2 + 21\text{x} - 3 = 26$
$\Rightarrow 27\text{x} - 1 = 26$
$\Rightarrow 27\text{x} = 26 + 1$
$\Rightarrow 27\text{x} = 27$
$\therefore \text{x} = \dfrac{27}{27} = 1$ Ans
Now $\dfrac{1}{\text{a}} + 5\text{x} = 8$, where $\text{x} = 1$
$\Rightarrow \dfrac{1}{\text{a}} + 5(1) = 8$
$\Rightarrow \dfrac{1}{\text{a}} + 5 = 8$
$\Rightarrow \dfrac{1}{\text{a}} = 8 - 5$
$\Rightarrow \dfrac{1}{\text{a}} = 3$
$\therefore \text{a} = \dfrac{1}{3}$ Ans
Q25. Solve: $\dfrac{4 - 3\text{x}}{5} + \dfrac{7 - \text{x}}{3} + 4\dfrac{1}{3} = 0$
Hence, find the value of 'p', if $3\text{p} - 2\text{x} + 1 = 0$
Show Answer
$\Rightarrow \dfrac{4 - 3\text{x}}{5} + \dfrac{7 - \text{x}}{3} + 4\dfrac{1}{3} = 0$
$\Rightarrow \dfrac{4 - 3\text{x}}{5} + \dfrac{7 - \text{x}}{3} + \dfrac{13}{3} = 0$
$\Rightarrow \dfrac{4 - 3\text{x}}{5} + \dfrac{7 - \text{x}}{3} = -\dfrac{13}{3}$
L.C.M. of denominators = 15
$\Rightarrow 15(\dfrac{4 - 3\text{x}}{5}) + 15(\dfrac{7 - \text{x}}{3}) = 15(-\dfrac{13}{3})$
$\Rightarrow 3(4 - 3\text{x}) + 5(7 - \text{x}) = 5(-13)$
$\Rightarrow 12 - 9\text{x} + 35 - 5\text{x} = -65$
$\Rightarrow 47 - 14\text{x} = -65$
$\Rightarrow -14\text{x} = -65 - 47$
$\Rightarrow -14\text{x} = -112$
$\Rightarrow 14\text{x} = \dfrac{112}{14} = 8$
Now $3\text{p} - 2\text{x} + 1 = 0$, where $\text{x} = 8$
$\Rightarrow 3\text{p} - 2\text{x} + 1 = 0$
$\Rightarrow 3\text{p} - 2(8) + 1 = 0$
$\Rightarrow 3\text{p} - 16 + 1 = 0$
$\Rightarrow 3\text{p} - 15 = 0$
$\Rightarrow 3\text{p} = 15$
$\therefore \text{p} = \dfrac{15}{3} = 5$ Ans
$\Rightarrow \dfrac{4 - 3\text{x}}{5} + \dfrac{7 - \text{x}}{3} + \dfrac{13}{3} = 0$
$\Rightarrow \dfrac{4 - 3\text{x}}{5} + \dfrac{7 - \text{x}}{3} = -\dfrac{13}{3}$
L.C.M. of denominators = 15
$\Rightarrow 15(\dfrac{4 - 3\text{x}}{5}) + 15(\dfrac{7 - \text{x}}{3}) = 15(-\dfrac{13}{3})$
$\Rightarrow 3(4 - 3\text{x}) + 5(7 - \text{x}) = 5(-13)$
$\Rightarrow 12 - 9\text{x} + 35 - 5\text{x} = -65$
$\Rightarrow 47 - 14\text{x} = -65$
$\Rightarrow -14\text{x} = -65 - 47$
$\Rightarrow -14\text{x} = -112$
$\Rightarrow 14\text{x} = \dfrac{112}{14} = 8$
Now $3\text{p} - 2\text{x} + 1 = 0$, where $\text{x} = 8$
$\Rightarrow 3\text{p} - 2\text{x} + 1 = 0$
$\Rightarrow 3\text{p} - 2(8) + 1 = 0$
$\Rightarrow 3\text{p} - 16 + 1 = 0$
$\Rightarrow 3\text{p} - 15 = 0$
$\Rightarrow 3\text{p} = 15$
$\therefore \text{p} = \dfrac{15}{3} = 5$ Ans
Q26. $3\text{x} - \dfrac{1}{2}\text{x} = 2\dfrac{1}{2}$
Show Answer
$\Rightarrow 3\text{x} - \dfrac{1}{2}\text{x} = 2\dfrac{1}{2}$
$\Rightarrow \dfrac{6\text{x} - \text{x}}{2} = \dfrac{5}{2}$
$\Rightarrow \dfrac{5\text{x}}{2} = \dfrac{5}{2}$
$\Rightarrow 2(5\text{x}) = 2(5)$
$\Rightarrow 10\text{x} = 10$
$\therefore \text{x} = \dfrac{10}{10} = 1$ Ans
$\Rightarrow \dfrac{6\text{x} - \text{x}}{2} = \dfrac{5}{2}$
$\Rightarrow \dfrac{5\text{x}}{2} = \dfrac{5}{2}$
$\Rightarrow 2(5\text{x}) = 2(5)$
$\Rightarrow 10\text{x} = 10$
$\therefore \text{x} = \dfrac{10}{10} = 1$ Ans
Q27. $4\text{x} - 6 = \dfrac{3\text{x}}{4} + 20$
Show Answer
$\Rightarrow 4\text{x} - 6 = \dfrac{3\text{x}}{4} + 20$
$\Rightarrow 4\text{x} - 6 = \dfrac{3\text{x} + 80}{4}$
$\Rightarrow 4(4\text{x} - 6) = 3\text{x} + 80$
$\Rightarrow 16\text{x} - 24 = 3\text{x} + 80$
$\Rightarrow 16\text{x} - 3\text{x} = 80 + 24$
$\Rightarrow 13\text{x} = 104$
$\therefore \text{x} = \dfrac{104}{13} = 8$ Ans
$\Rightarrow 4\text{x} - 6 = \dfrac{3\text{x} + 80}{4}$
$\Rightarrow 4(4\text{x} - 6) = 3\text{x} + 80$
$\Rightarrow 16\text{x} - 24 = 3\text{x} + 80$
$\Rightarrow 16\text{x} - 3\text{x} = 80 + 24$
$\Rightarrow 13\text{x} = 104$
$\therefore \text{x} = \dfrac{104}{13} = 8$ Ans
Q28. $7 - 2(\text{x} - 3) = 9$
Show Answer
$\Rightarrow 7 - 2(\text{x} - 3) = 9$
$\Rightarrow 7 - 2\text{x} + 6 = 9$
$\Rightarrow 13 - 2\text{x}= 9$
$\Rightarrow -2\text{x}= 9 - 13$
$\Rightarrow -2\text{x}= -4$
$\therefore \text{x}= \dfrac{4}{2} = 2$ Ans
$\Rightarrow 7 - 2\text{x} + 6 = 9$
$\Rightarrow 13 - 2\text{x}= 9$
$\Rightarrow -2\text{x}= 9 - 13$
$\Rightarrow -2\text{x}= -4$
$\therefore \text{x}= \dfrac{4}{2} = 2$ Ans
Q29. $3\dfrac{3}{4}\text{x} = 2\text{x} - 1\dfrac{1}{2}$
Show Answer
$\Rightarrow 3\dfrac{3}{4}\text{x} = 2\text{x} - 1\dfrac{1}{2}$
$\Rightarrow \dfrac{15}{4}\text{x} = 2\text{x} - \dfrac{3}{2}$
$\Rightarrow \dfrac{15}{4}\text{x} = \dfrac{4\text{x} - 3}{2}$
$\Rightarrow 2(15\text{x}) = 4(4\text{x} - 3)$
$\Rightarrow 30\text{x} = 16\text{x} - 12$
$\Rightarrow 30\text{x} - 16\text{x} = -12$
$\Rightarrow 14\text{x} = -12$
$\therefore \text{x} = -\dfrac{12}{14} = -\dfrac{6}{7}$ Ans
$\Rightarrow \dfrac{15}{4}\text{x} = 2\text{x} - \dfrac{3}{2}$
$\Rightarrow \dfrac{15}{4}\text{x} = \dfrac{4\text{x} - 3}{2}$
$\Rightarrow 2(15\text{x}) = 4(4\text{x} - 3)$
$\Rightarrow 30\text{x} = 16\text{x} - 12$
$\Rightarrow 30\text{x} - 16\text{x} = -12$
$\Rightarrow 14\text{x} = -12$
$\therefore \text{x} = -\dfrac{12}{14} = -\dfrac{6}{7}$ Ans
Q30. $0.06 + 2\text{x} = 0.19$
Show Answer
$\Rightarrow 0.06 + 2\text{x} = 0.19$
$\Rightarrow 2\text{x} = 0.19 - 0.06$
$\Rightarrow 2\text{x} = 0.13$
$\therefore \text{x} = \dfrac{0.13}{2} = 0.065$ Ans
$\Rightarrow 2\text{x} = 0.19 - 0.06$
$\Rightarrow 2\text{x} = 0.13$
$\therefore \text{x} = \dfrac{0.13}{2} = 0.065$ Ans