Multiplication is performed before addition. Thus, 8 × 2 equals 16 and adding 7 gives 23.

To add the fractions, convert them to a common denominator (6): 1/2 becomes 3/6 and 1/3 becomes 2/6. Their sum is 5/6.

Divide both sides of the equation by 2 to isolate x, resulting in x = 5. This tests basic algebraic manipulation.

The sequence increases by 2 each time; adding 2 to 7 gives 9. This tests basic pattern recognition.

The area of a rectangle is found by multiplying the length by the width: 6 × 4 equals 24. This reinforces basic geometric calculations.

Divide both sides by 3 to obtain x - 2 = 4, then add 2 to both sides to find x = 6. This tests the distributive property and linear equation solving.

Both the numerator and denominator are divisible by 5; dividing each by 5 simplifies 15/35 to 3/7. This reinforces fraction simplification skills.

Subtract 2x from both sides to get 3 = x - 4, and then add 4 to both sides to obtain x = 7. This demonstrates balancing operations across the equation.

Cross-multiply to get 4 × 15 = 5x, which simplifies to 60 = 5x and then x = 12. This tests the method of solving proportions.

2^3 equals 8 and adding 4 results in 12. This reinforces understanding of exponents and the order of operations.

Factoring the quadratic gives (x - 2)(x - 3) = 0, so the solutions are x = 2 and x = 3. This tests factoring and solving quadratic equations.

The slope is calculated by (change in y)/(change in x): (8 - 2)/(3 - 1) equals 6/2, which simplifies to 3. This reinforces the concept of slope in coordinate geometry.

A square has four equal sides. Multiplying 5 by 4 gives a perimeter of 20 units.

Combining the like terms 2x and 3x results in 5x. This question reinforces the concept of combining coefficients.

Multiply both sides of the equation by 3 to isolate x, which gives x = 12. This is a fundamental exercise in solving simple linear equations.

First, solve x - y = 1 to get x = y + 1, then substitute into 2x + 3y = 12 to solve for y, giving y = 2 and x = 3. This uses substitution to solve a system of linear equations.

50 can be factored into 25 × 2, so √50 equals √25 × √2, which simplifies to 5√2. This tests the ability to simplify radical expressions by extracting square factors.

Cross-multiply to obtain 3(x - 1) = (x + 2), then simplify to get 2x = 5, so x = 5/2. This problem assesses skills in solving rational equations.

Using the formula A = πr², substitute the given values to solve for r², which simplifies to 49, and thus r = 7. This question applies algebra to geometric formulas.

Apply the Pythagorean theorem: the hypotenuse is the square root of (9² + 12²), which equals √(81 + 144) or √225, resulting in 15. This reinforces the application of the Pythagorean theorem in right triangles.