Solve for $x$ : $\dfrac{x+3}{x-2} \leq 2$.

> [!TLDR] > $\dfrac{x+3}{x-2}\le 2$ holds for $x [!FORMULA] > Rearrange to $\dfrac{x+3}{x-2}-2\le 0$ and combine into a single fraction, then use a sign chart. Never multiply across by $x-2$ while its sign is unknown. 1. Combine: $\dfrac{x+3}{x-2}-2=\dfrac{(x+3)-2(x-2)}{x-2}=\dfrac{-x+7}{x-2}=\dfrac{7-x}{x-2}.$ 2. The inequality becomes $\dfrac{7-x}{x-2}\le 0$, i.e. (multiplying numerator and denominator by $-1$) $\dfrac{x-7}{x-2}\ge 0.$ 3. Critical points: $x=7$ (numerator $0$) and $x=2$ (denominator $0$, so $x=2$ is excluded). …

Solve for x : (x+3)/(x-2) ≤ 2.

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$\dfrac{x+3}{x-2}\le 2$ holds for $x<2$ or $x\ge 7$ .

\dfrac{x+3}{x-2}-2\le 0

Combine: $\dfrac{x+3}{x-2}-2=\dfrac{(x+3)-2(x-2)}{x-2}=\dfrac{-x+7}{x-2}=\dfrac{7-x}{x-2}.$
The inequality becomes $\dfrac{7-x}{x-2}\le 0$ , i.e. (multiplying numerator and denominator by $-1$ ) $\dfrac{x-7}{x-2}\ge 0.$
Critical points: $x=7$ (numerator $0$ ) and $x=2$ (denominator $0$ , so $x=2$ is excluded). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/S/WXYZ/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0, x \in R$
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$\dfrac{|x+1|}{x+1}>0$ holds precisely where $x+1>0$ , i.e. $x>-1$ , excluding $x=-1$ where it is undefined.

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Let $t = x+1$ ; the expression is $\dfrac{|t|}{t}$ .

If $t>0$ , $|t| = t$ , so $\dfrac{|t|}{t} = 1 > 0$ . ✓

If $t<0$ , $|t| = -t$ , so $\dfrac{|t|}{t} = -1 < 0$ . ✗ …
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CBSE 2025Set 465/S/WXYZ/41 markMCQ
$|3x - 2| \leq \dfrac{1}{2}$
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The Assertion's solution set $[\tfrac12,\tfrac56]$ is correct, but the Reason misstates the modulus rule (it describes $\ge$ , not $\le$ ), so (C).

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Solve the Assertion: $|3x-2|\le \tfrac12 \iff -\tfrac12 \le 3x-2 \le \tfrac12$ .

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Divide by 3: $\tfrac12 \le x \le \tfrac56$ , i.e. solution set $\left[\tfrac12,\tfrac56\right]$ — Assertion is true. …
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CBSE 2024Set 465/RQPS/41 markMCQ
$x > y$
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Dividing both sides of $x>y$ by the negative number $z$ reverses the inequality, so $\dfrac{x}{z}<\dfrac{y}{z}$ .

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Given $x>y$ and $z<0$ . …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0$
(a) $x \in [-1, \infty)$
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$\dfrac{|x+1|}{x+1}>0$ forces $x+1>0$ , i.e. $x\in(-1,\infty)$ .

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Let $t=x+1$ . The expression $\dfrac{|t|}{t}>0$ holds only when $t>0$ .

So $x+1>0 \Rightarrow x>-1$ . …
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Q.Solve for $x$ : $\dfrac{x+3}{x-2} \leq 2$ .

Concept understanding — Linear and Modulus Inequalities

Q.Solve for xxx : x+3x−2≤2\dfrac{x+3}{x-2} \leq 2x−2x+3​≤2.

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Q.Solve for $x$ : $\dfrac{x+3}{x-2} \leq 2$ .

Unlock everything free for 14 days