Subtracting 3 from both sides gives 2x = 4 and dividing by 2 yields x = 2. This method is a fundamental approach to isolating the variable.

The slope-intercept form of a line is y = mx + b where m represents the slope. In this equation, m equals 4, which is the slope.

The expression combines like terms, as both terms have the same variable x. Adding the coefficients (3 + 2) gives 5, resulting in 5x.

Multiplying both sides by 4 isolates x, yielding x = 12. This is a direct application of solving simple algebraic equations.

The distributive property states that multiplying a number by a sum is the same as doing each multiplication separately. Option A correctly applies this principle by distributing a over the sum inside the parentheses.

Expanding the left side gives 2x - 6, and then isolating x reveals that x equals -4 after collecting like terms and dividing by the coefficient. This is a typical linear equation solution method.

Perpendicular lines have slopes that are negative reciprocals of each other. Since the given line has a slope of ½, the perpendicular line must have a slope of -2 as shown in Option A.

To factor the quadratic, we need two numbers that multiply to 6 and add to -5. The numbers -2 and -3 fit these conditions, so the expression factors to (x - 2)(x - 3).

Subtracting 2x from both sides gives x - 4 < 5, and adding 4 to both sides yields x < 9. This process appropriately isolates the variable in the inequality.

Evaluating f(x) at x = 2 gives 2² - 4 = 0, and evaluating g(x) at x = 2 gives 2(2) + 3 = 7. Their sum is 0 + 7, which equals 7.

When the numbers are arranged in order, the median is the middle value. For this set of five numbers, the median is 9.

The area of a triangle is calculated using the formula ½ × base × height. Substituting the given values, the area is ½ × 8 × 5 = 20.

First, distribute 2 across (3x + 4) to get 6x + 8, then subtract 5x to combine like terms. The simplified expression is x + 8.

The absolute value equation |x - 3| = 5 splits into two cases: x - 3 = 5 and x - 3 = -5. Solving these gives x = 8 or x = -2.

The numerator factors as a difference of squares: (x - 3)(x + 3). Canceling the common factor (x + 3) with the denominator leaves x - 3.

Using the substitution method, express x from the second equation as x = y + 1 and substitute into the first equation to solve for y. The solution is y = 2 and then x = 3.

The vertex of a quadratic in the form y = ax² + bx + c is found using -b/(2a) for the x-coordinate. Substituting x = 2 into the equation gives the y-coordinate of -5, so the vertex is (2, -5).

Since the square root function requires a nonnegative argument, the expression inside the square root, x - 1, must be greater than or equal to 0. This condition leads to the domain x ≥ 1.

Using the area formula A = πr², setting πr² = 25π gives r² = 25, so r = 5. The circumference is then calculated using 2πr, resulting in 10π.

Substituting x = -2 into the expression gives a numerator of (-8) - 2(4) + 4, which simplifies to -12, and a denominator of -3. Dividing -12 by -3 yields 4.