Question
$\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)-\frac{9}{2}$

Answer

$\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)-\frac{9}{2}$
$\Rightarrow\frac{2}{3}\text{x}-\frac{10}{3}-\frac{1}{4}\text{x}+\frac{1}{2}=\frac{9}{2}$
$\Rightarrow\frac{2}{3}\text{x}-\frac{1}{4}\text{x}-\frac{10}{3}+\frac{1}{2}=\frac{9}{2}$
$\frac{8\text{x} - 3\text{x}-40+6=54}{12}=\frac{9}{2}$ (L.C.M. of 3, 4, 2 = 12)
$5\text{x}=-34=\frac{9\times12}{2}=9\times6=54$

$5\text{x}-34=54$ (Adding 34 to both sides)

$5\text{x}-34+34=54+34$

$\Rightarrow5\text{x}=88$

$\Rightarrow\text{x}=\frac{88}{5}$

Verification:
$\text{L.H.S.}=\frac{2}{3}(\text{x}-5)-\frac{1}{4}(\text{x}-2)$
$=\frac{2}{3}\Big(\frac{88}{5}-5\Big)-\frac{1}{4}\Big(\frac{88}{5}-2\Big)$
$=\frac{2}{3}\Big(\frac{88 - 25}{5}\Big)-\frac{1}{4}\Big(\frac{88-10}{5}\Big)$
$=\frac{2}{3}\times\frac{63}{5}-\frac{1}{4}\times\frac{78}{5}$
$=\frac{42}{5}-\frac{39}{10}$
$=\frac{84-39}{10}=\frac{45}{10}=\frac{9}{2}=\text{R.H.S.}$

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