MCQ
If $2\text{x}+\frac53=\frac14\text{x}+4,$ then $x =$
- A$3$
- B$4$
- C$\frac34$
- ✓$\frac43$
✓
Answer
Correct option: D.
$\frac43$
As, $2\text{x}+\frac53=\frac14\text{x}+4$
$\Rightarrow2\text{x}-\frac14\text{x}=4-\frac53$ (By transposing $\frac53$ to $R.H.S.$ and $\frac14\text{x}$ to $L.H.S.)$
$\Rightarrow\frac{\text{2x}}{1}-\frac{\text{x}}{4}=\frac41-\frac53$
$\Rightarrow \frac{\text{8x}}{4}-\frac{\text{x}}{4}=\frac{12}{3}-\frac53$
$\Rightarrow \frac{8\text{x}-\text{x}}{4}=\frac{12-5}{3}$
$\Rightarrow\frac{7\text{x}}{4}=\frac{7}{3}$
$\Rightarrow\text{7x}\times3=4\times7$ (By cross multiplication)
$\Rightarrow21\text{x}=28$
$\Rightarrow\text{x}=\frac{28}{21}$
$\therefore\text{x}=\frac{4}{3}$
Hence, the correct alternative is option $(d).$
$\Rightarrow2\text{x}-\frac14\text{x}=4-\frac53$ (By transposing $\frac53$ to $R.H.S.$ and $\frac14\text{x}$ to $L.H.S.)$
$\Rightarrow\frac{\text{2x}}{1}-\frac{\text{x}}{4}=\frac41-\frac53$
$\Rightarrow \frac{\text{8x}}{4}-\frac{\text{x}}{4}=\frac{12}{3}-\frac53$
$\Rightarrow \frac{8\text{x}-\text{x}}{4}=\frac{12-5}{3}$
$\Rightarrow\frac{7\text{x}}{4}=\frac{7}{3}$
$\Rightarrow\text{7x}\times3=4\times7$ (By cross multiplication)
$\Rightarrow21\text{x}=28$
$\Rightarrow\text{x}=\frac{28}{21}$
$\therefore\text{x}=\frac{4}{3}$
Hence, the correct alternative is option $(d).$
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