CUET UG 2025 — General Test Quantitative Reasoning

A system of two linear equations has no solution when the lines are parallel. This occurs when the lines have equal slopes but different y-intercepts.

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Converting the first equation $3x - ky - 3 = 0$3x−ky−3=0 to slope-intercept form:

$$ky = 3x - 3$$ky=3x−3

$$y = \frac{3}{k}x - \frac{3}{k}$$y=k3​x−k3​

The slope is $\frac{3}{k}$k3​ and the y-intercept is $-\frac{3}{k}$−k3​.

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Converting the second equation $2x - 3y - 4 = 0$2x−3y−4=0 to slope-intercept form:

$$3y = 2x - 4$$3y=2x−4

$$y = \frac{2}{3}x - \frac{4}{3}$$y=32​x−34​

The slope is $\frac{2}{3}$32​ and the y-intercept is $-\frac{4}{3}$−34​.

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For the lines to be parallel, the slopes must be equal:

$$\frac{3}{k} = \frac{2}{3}$$k3​=32​

$$3 \times 3 = k \times 2$$3×3=k×2

$$9 = 2k$$9=2k

$$k = \frac{9}{2}$$k=29​

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When $k = \frac{9}{2}$k=29​, the y-intercepts are:

First equation: $-\frac{3}{9/2} = -\frac{3 \times 2}{9} = -\frac{2}{3}$−9/23​=−93×2​=−32​

Second equation: $-\frac{4}{3}$−34​

Since $-\frac{2}{3} \neq -\frac{4}{3}$−32​=−34​, the lines are parallel with different y-intercepts, confirming no solution exists.

Therefore, $k = \frac{9}{2}$k=29​.