CUET UG 2025 — Mathematics Algebra

The four constraints are:

1. $x - 2y \geq 0$x−2y≥0
2. $2x - y \leq -4$2x−y≤−4
3. $x \geq 0$x≥0
4. $y \geq 0$y≥0

The solution set contains points that satisfy all four constraints simultaneously.

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From constraint 1: $x - 2y \geq 0$x−2y≥0

$x \geq 2y$x≥2y ... (A)

From constraint 2: $2x - y \leq -4$2x−y≤−4

$$-y \leq -4 - 2x$$−y≤−4−2x

$y \geq 2x + 4$y≥2x+4 ... (B)

Constraints 3 and 4 indicate the first quadrant.

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From (B): $y \geq 2x + 4$y≥2x+4

Substituting into (A):

$$x \geq 2y$$x≥2y

$$x \geq 2(2x + 4)$$x≥2(2x+4)

$$x \geq 4x + 8$$x≥4x+8

$$-3x \geq 8$$−3x≥8

$$x \leq -\dfrac{8}{3}$$x≤−38​

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The constraints require:

$x \leq -\dfrac{8}{3}$x≤−38​ (approximately $-2.67$−2.67)

AND

$x \geq 0$x≥0 (from constraint 3)

These cannot be satisfied simultaneously. No point can have $x \leq -\dfrac{8}{3}$x≤−38​ and $x \geq 0$x≥0 at the same time.

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Since no point satisfies all four constraints, the solution set is empty.

Therefore, the solution set is $\phi$ϕ.