MCQ
If − 3x + 17 < -13, then:
- A$\text{x}\in(10,\infty)$
- B$\text{x}\in[10,\infty)$
- C$\text{x}\in(-\infty,10]$
- D$\text{x}\in[-10,10)$
Solution:
− 3x + 17 < −13
Subtracting 17 on both sides, we get
⇒ −3x + 17 − 17 < −13 − 17
⇒ −3x < − 30
Dividing −3 on both sides, we get
$\Rightarrow\frac{-3\text{x}}{-3}>\frac{-30}{-3}$
$\Rightarrow\text{x}>10$
$\Rightarrow\text{x}\in(10,\infty)$
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