Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\displaystyle 14-\frac{x-1}{10}=\frac{x+5}{6}-3\) | (I) | 4 |
| (B) | \((x-5)^2-(x+3)^2=48\) | (II) | \(\displaystyle -\frac{23}{2}\) |
| (C) | \(6(x-4)=4(x-3)-(3x-8)\) | (III) | 61 |
| (D) | \((2x-1)(2x+3)=(2x-7)(2x+7)\) | (IV) | \(-2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each equation is linear once it is simplified (in (B) and (D) the \(x^2\) terms cancel), so each is solved for its single value of \(x\).
(A) \(14-\frac{x-1}{10}=\frac{x+5}{6}-3\) — this matches (III) 61.
Multiplying by 30, the LCM of 10 and 6: \(420-3(x-1)=5(x+5)-90\)
\(423-3x=5x-65\)
\(8x=488\), so \(x=61\)
(B) \((x-5)^2-(x+3)^2=48\) — this matches (IV) −2.
Difference of squares: \((x-5-x-3)(x-5+x+3)=48\)
\(-8(2x-2)=48\)
\(2x-2=-6\), so \(x=-2\)
(C) \(6(x-4)=4(x-3)-(3x-8)\) — this matches (I) 4.
\(6x-24=4x-12-3x+8\)
\(6x-24=x-4\)
\(5x=20\), so \(x=4\)
(D) \((2x-1)(2x+3)=(2x-7)(2x+7)\) — this matches (II) \(-\frac{23}{2}\).
\(4x^2+4x-3=4x^2-49\)
\(4x=-46\), so \(\displaystyle x=-\frac{23}{2}\)
Shortcut: Solve only (C), which has the shortest working: it gives \(x=4\), so (C) matches (I), and only Option (3) pairs (C) with (I).