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Match List - I with List - II.

List - IList - II
(1) 14−110(x+3)=56x14 - \frac{1}{10}(x+3) = \frac{5}{6}x(I) 4
(2) (x−5)2−(x+3)2=48(x-5)^2 - (x+3)^2 = 48(II) −232-\frac{23}{2}
(3) 6(x−4)=4(x−3)−(3x−8)6(x-4) = 4(x-3) - (3x-8)(III) 61
(4) (2x−1)(2x+3)=(2x−7)(2x+7)(2x-1)(2x+3) = (2x-7)(2x+7)(IV) −2-2

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

Eq (3): 6(x−4)=4(x−3)−(3x−8)⇒6x−24=x−4⇒x=46(x-4) = 4(x-3) - (3x-8) \Rightarrow 6x-24 = x - 4 \Rightarrow x = 4 (I). Eq (2): (x−5)2−(x+3)2=−16x+16=48⇒x=−2(x-5)^2 - (x+3)^2 = -16x + 16 = 48 \Rightarrow x = -2 (IV). Eq (4): 4x2+4x−3=4x2−49⇒x=−23/24x^2+4x-3 = 4x^2 - 49 \Rightarrow x = -23/2 (II). Eq (1) therefore maps to III (61). Match: (1)-(III), (2)-(IV), (3)-(I), (4)-(II).

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