Question
Solve the following linear inequations in R:
$\frac{3}{\text{x}-2}<1$
$\frac{3}{\text{x}-2}<1$
✓
Answer
$\frac{3}{\text{x}-2}<1$
$\frac{3}{\text{x}-2}-1<0$
$\frac{3-(\text{x}-2)}{\text{x}-2}<0$
$\frac{3-\text{x}+2}{\text{x}-2}<0$
$\frac{5-\text{x}}{\text{x}-2}<0$
$\frac{\text{x}-5}{\text{x}-2}>0$
Case1: x - 5 > 0 and x - 2 > 0
⇒ x > 5 and x > 2
⇒ x > 5
Case 2: x - 5 < 0 and x - 2 < 0
⇒ x < 5 and x < 2
⇒ x < 2
$\therefore$ solution set is $(-\infty,2)\cup(5,\infty)$
$\frac{3}{\text{x}-2}-1<0$
$\frac{3-(\text{x}-2)}{\text{x}-2}<0$
$\frac{3-\text{x}+2}{\text{x}-2}<0$
$\frac{5-\text{x}}{\text{x}-2}<0$
$\frac{\text{x}-5}{\text{x}-2}>0$
Case1: x - 5 > 0 and x - 2 > 0
⇒ x > 5 and x > 2
⇒ x > 5
Case 2: x - 5 < 0 and x - 2 < 0
⇒ x < 5 and x < 2
⇒ x < 2
$\therefore$ solution set is $(-\infty,2)\cup(5,\infty)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate the following limits: $\lim _{x \rightarrow \sqrt{2}}\left[\frac{x^2+x \sqrt{2}-4}{x^2-3 x \sqrt{2}+4}\right]$
→Find n, if
→$\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-7) !}=1: 6$
Evaluate the following limits: $\lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]$
→If $\text{y}=\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{bx}-\text{a}},$ show that x = f(y).
→$\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
→If $S$ denotes the sum of an infinite G.P. and $S_1$ denotes the sum of the squares of its terms, then prove that the first terms and common ratio are respectively $\frac{2\text{SS}_1}{\text{S}^2+\text{S}_1}\text{ and }\frac{\text{S}^2-\text{S}_1}{\text{S}^2+\text{S}_1}.$
→∆PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle within the triangle.
Compute the variance and S.D.
The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.
Find the general solutions of the following equations:
$\sin2\text{x}=\cos3\text{x}$
→$\sin2\text{x}=\cos3\text{x}$