MCQ
If $\frac{\text{x}}{6}+\frac{\text{x}}{4}=\frac{\text{x}}{2}+\frac{3}{4},$ then $x =$
- A$9$
- B$-6$
- ✓$-9$
- D$4$
As, $\frac{\text{x}}{6}+\frac{\text{x}}{4}=\frac{\text{x}}{2}+\frac34$
$\Rightarrow\frac{\text{x}}{6}+\frac{\text{x}}{4}-\frac{\text{x}}{2}=\frac{3}{4} ($By transposing $\frac{\text{x}}{2}$ to $L.H.S.)$
$\Rightarrow\frac{2\text{x}}{12}+\frac{3\text{x}}{12}-\frac{6\text{x}}{12}=\frac{3}{4}$
$\Rightarrow\frac{2\text{x}+3\text{x}-6\text{x}}{12}=\frac34$
$\Rightarrow\frac{-\text{x}}{12}=\frac34$
$\Rightarrow-\text{x}\times4=3\times12$ (By cross multiplication)
$\Rightarrow-4\text{x}=36$
$\Rightarrow\text{x}=\frac{36}{-4}$
$\therefore\text{x}=-9$
Hence, the correct alternative is option $(c).$
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