If (3x - 5)/6 + 8 ≥ 4 + 2x/3, then Option 1: x (-∞, 19] Option 2: x (-∞, -19] Option 3: x [19, ∞) Option 4: x [-19, ∞)

Answer: x (-∞, 19]. Solution: Given: (3x - 5)/6 + 8 ≥ 4 + 2x/3 Combining constants: (3x - 5)/6 + 4 ≥ 2x/3 Multiplying all terms by 6 (the LCM of 6 and 3): 6 x (3x - 5)/6 + 6 x 4 ≥ 6 x 2x/3 (3x - 5) + 24 ≥ 4x Simplifying: 3x - 5 + 24 ≥ 4x 3x + 19 ≥ 4x Rearranging terms: 19 ≥ 4x - 3x 19 ≥ x Therefore x ≤ 19 In interval notation: x (-∞, 19]

Algebra

Inequalities

Medium

applied

If $\frac{3x - 5}{6} + 8 \geq 4 + \frac{2x}{3}$ , then

$x \in (-\infty, 19]$

$x \in (-\infty, -19]$

$x \in [19, \infty)$

$x \in [-19, \infty)$

✅ Correct Option: 1

Given: $\frac{3x - 5}{6} + 8 \geq 4 + \frac{2x}{3}$

Combining constants:

$\frac{3x - 5}{6} + 4 \geq \frac{2x}{3}$

Multiplying all terms by 6 (the LCM of 6 and 3):

$6 \times \frac{3x - 5}{6} + 6 \times 4 \geq 6 \times \frac{2x}{3}$

$(3x - 5) + 24 \geq 4x$

Simplifying:

$3x - 5 + 24 \geq 4x$

$3x + 19 \geq 4x$

Rearranging terms:

$19 \geq 4x - 3x$

$19 \geq x$

Therefore $x \leq 19$

In interval notation: $x \in (-\infty, 19]$

2025: 13 May Shift 1

If $a, b$ and $c$ are positive real numbers, then

Match List-I with List-II

List-I	List-II
(Expression)	(The Least value of the expression)
(A) $(a + b)(b + c)(c + a)$	(I) $8abc$
(B) $(a + b + c)(ab + bc + ca)$	(II) $9a^2b^2c^2$
(C) $(a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2)$	(III) $9abc$
(D) $(a + b)^2(b + c)^2(c + a)^2$	(IV) $64a^2b^2c^2$

Choose the correct answer from the options given below:

2025: 30 May Shift 2

The solution set of the inequality $\frac{2x+3}{x-1} < 0$ is:

2026: 21 May Shift 2

The solution set for the inequality $|5x-3| \le \dfrac{1}{5}$ is:

2025: 16 May Shift 1

If, in a pair of consecutive positive integers, both numbers are greater than 5 and their sum is less than 23, then the number of such pairs are:

2025: 15 May Shift 1

The inequality $\frac{5x - 2}{3} - \frac{7x - 3}{5} > \frac{x}{4}$ holds when

2026: 18 May Shift 2

Match the LIST-I with LIST-II

	LIST-I (Inequality)		LIST-II (Solution set)
A.	$\vert x-3 \vert < 1$	I.	$x \in \left(-\dfrac{2}{5}, \dfrac{8}{5}\right)$
B.	$\vert 3-5x \vert < 5$	II.	$x \in \left(-\infty, -\dfrac{1}{2}\right)$
C.	$2x+1 < \vert 2x+1 \vert$	III.	$x \in \left(-\dfrac{1}{3}, 5\right)$
D.	$\vert 7-3x \vert < 8$	IV.	$x \in (2,4)$

Choose the correct answer from the options given below:

If 3x−56+8≥4+2x3\frac{3x - 5}{6} + 8 \geq 4 + \frac{2x}{3}63x−5​+8≥4+32x​, then

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $\frac{3x - 5}{6} + 8 \geq 4 + \frac{2x}{3}$ , then