MCQ
If $4\text{x}+3<6\text{x}+7$ then x belongs to the interval:
- A$(2,\infty)$
- B$(-2,\infty)$
- C$(-\infty, 2)$
- D$(-4,\infty)$
Solution:
$4\text{x}+3<6\text{x}+7$
Subtracting 3 from both sides,
$4\text{x}+3<6\text{x}+7-3$
$\Rightarrow4\text{x}<6\text{x}+4$
Subtracting 6x from both sides,
$4\text{x} – 6\text{x} <6\text{x} + 4 – 6\text{x}$
$\Rightarrow– 2\text{x}<4$ or
$\Rightarrow\text{x}>-2\text{ i.e..,}$ all the real numbers greater than –2, are the solutions of the given inequality.
Hence, the solution set is $(–2,\infty), \text{i}.\text{e}.\text{x}\in(-2,\infty)$
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