This question asks us to match equations from List I with values from List II by solving each equation. Let's solve each equation step by step.

Solving Equation (A)

The equation is: \( \frac{14 - (x - 1)}{10} = \frac{x + 5}{6} - 3 \)

First, simplify both sides of the equation:

Left side: \( \frac{14 - x + 1}{10} = \frac{15 - x}{10} \)

Right side: \( \frac{x + 5}{6} - 3 = \frac{x + 5}{6} - \frac{18}{6} = \frac{x + 5 - 18}{6} = \frac{x - 13}{6} \)

So the equation becomes: \( \frac{15 - x}{10} = \frac{x - 13}{6} \)

To eliminate the denominators, multiply both sides by the least common multiple (LCM) of 10 and 6, which is 30.

\( 30 \times \left( \frac{15 - x}{10} \right) = 30 \times \left( \frac{x - 13}{6} \right) \)

\( 3(15 - x) = 5(x - 13) \)

Distribute the numbers on both sides:

\( 45 - 3x = 5x - 65 \)

Now, we want to gather the terms with \( x \) on one side and the constant terms on the other side. Add \( 3x \) to both sides:

\( 45 = 5x + 3x - 65 \)

\( 45 = 8x - 65 \)

Add 65 to both sides:

\( 45 + 65 = 8x \)

\( 110 = 8x \)

Divide both sides by 8:

\( x = \frac{110}{8} \)

Simplify the fraction:

\( x = \frac{55}{4} \)

The solution for equation (A) is \( x = \frac{55}{4} \).

Solving Equation (B)

The equation is: \( (x - 5)^2 - (x + 3)^2 = 48 \)

This equation is in the form of a difference of squares, \( a^2 - b^2 \), which can be factored as \( (a - b)(a + b) \).

Here, \( a = x - 5 \) and \( b = x + 3 \).

Substitute \( a \) and \( b \) into the factored form:

\( ( (x - 5) - (x + 3) ) ( (x - 5) + (x + 3) ) = 48 \)

Simplify the expressions inside the parentheses:

First parenthesis: \( (x - 5 - x - 3) = -8 \)

Second parenthesis: \( (x - 5 + x + 3) = 2x - 2 \)

So the equation becomes:

\( (-8)(2x - 2) = 48 \)

Divide both sides by \( -8 \):

\( 2x - 2 = \frac{48}{-8} \)

\( 2x - 2 = -6 \)

Add 2 to both sides:

\( 2x = -6 + 2 \)

\( 2x = -4 \)

Divide both sides by 2:

\( x = \frac{-4}{2} \)

\( x = -2 \)

The solution for equation (B) is \( x = -2 \).

Solving Equation (C)

The equation is: \( 6(x - 4) = 4(x - 3) - 3(x - 8) \)

First, expand both sides of the equation by distributing the numbers:

Left side: \( 6 \times x - 6 \times 4 = 6x - 24 \)

Right side: \( (4 \times x - 4 \times 3) - (3 \times x - 3 \times 8) = (4x - 12) - (3x - 24) \)

Right side continued: \( 4x - 12 - 3x + 24 \)

Combine like terms on the right side:

\( (4x - 3x) + (-12 + 24) = x + 12 \)

So the equation becomes:

\( 6x - 24 = x + 12 \)

Subtract \( x \) from both sides:

\( 6x - x - 24 = 12 \)

\( 5x - 24 = 12 \)

Add 24 to both sides:

\( 5x = 12 + 24 \)

\( 5x = 36 \)

Divide both sides by 5:

\( x = \frac{36}{5} \)

The solution for equation (C) is \( x = \frac{36}{5} \).

Solving Equation (D)

The equation is: \( (2x - 1)(2x + 3) = (2x - 7)(2x + 7) \)

Expand both sides of the equation.

Left side: \( (2x - 1)(2x + 3) \)

Using FOIL (First, Outer, Inner, Last): \( (2x)(2x) + (2x)(3) + (-1)(2x) + (-1)(3) = 4x^2 + 6x - 2x - 3 = 4x^2 + 4x - 3 \)

Right side: \( (2x - 7)(2x + 7) \)

This is a difference of squares form \( (a - b)(a + b) = a^2 - b^2 \). Here \( a = 2x \) and \( b = 7 \).

\( (2x)^2 - 7^2 = 4x^2 - 49 \)

So the equation becomes:

\( 4x^2 + 4x - 3 = 4x^2 - 49 \)

Subtract \( 4x^2 \) from both sides:

\( 4x - 3 = -49 \)

Add 3 to both sides:

\( 4x = -49 + 3 \)

\( 4x = -46 \)

Divide both sides by 4:

\( x = \frac{-46}{4} \)

Simplify the fraction:

\( x = \frac{-23}{2} \)

The solution for equation (D) is \( x = \frac{-23}{2} \).

Matching Lists Based on Provided Answer

List I contains the equations and List II contains potential values. We have solved each equation. Now we present the lists and the matching as provided in the correct answer option text.

Based on the provided correct answer option, the matching is:

• (A) matches with (I)
• (B) matches with (II)
• (C) matches with (III)
• (D) matches with (IV)

Revision Table for Equation Solving

Reviewing the techniques used to solve these types of equations is helpful for exam preparation.

Additional Information on Algebraic Equations

Algebraic equations involve variables, constants, and mathematical operations. Solving an equation means finding the value(s) of the variable(s) that make the equation true.

• Linear Equations: These are equations where the highest power of the variable is 1. They typically have one unique solution. The goal is to isolate the variable on one side of the equation using inverse operations (addition/subtraction, multiplication/division).
• Difference of Squares: The algebraic identity \( a^2 - b^2 = (a - b)(a + b) \) is very useful for simplifying or factoring expressions, as seen in Equation (B). It helps convert a difference of squares into a product of binomials.
• Expanding Binomials: Multiplying two binomials, like \( (ax + b)(cx + d) \), can be done using the FOIL method (First, Outer, Inner, Last) or simply by distributing each term from the first binomial to the second.
• LCM in Equations: When dealing with fractions in an equation, multiplying all terms by the least common multiple (LCM) of the denominators is a standard method to clear the fractions and simplify the equation into a form that is easier to solve.

Practice solving various types of algebraic equations helps build proficiency and confidence in algebraic manipulation skills required for exams.