This question tests translating a word problem into a linear equation. The total charge consists of a flat fee F plus a per-mile charge of $0.25 times the number of miles m. The equation representing the total charge is F + 0.25m = 37.50, where F is the flat fee and 0.25m is the mileage charge. This matches choice B exactly. Choice A incorrectly multiplies the flat fee by 0.25, while choice C incorrectly applies the rate to the sum of F and m rather than just m.

This question tests solving linear inequalities involving fractions. Starting with (x - 5)/3 > 2, we multiply both sides by 3 to get x - 5 > 6. Adding 5 to both sides yields x > 11. Since this is a strict inequality (>) rather than ≥, the solution is x > 11, not x ≥ 11. A common error would be to incorrectly simplify the right side after multiplying, getting x - 5 > 2 instead of x - 5 > 6.

This question tests recognizing dependent equations in a system. Notice that the second equation 2x + 2y = 18 is exactly twice the first equation x + y = 9. When we multiply the first equation by 2, we get 2x + 2y = 18, which is identical to the second equation. This means the two equations represent the same line, so the system has infinitely many solutions - any point (x, y) satisfying x + y = 9 is a solution. The system is not inconsistent (which would have no solutions) nor does it have a unique solution.

This question tests finding values that satisfy a system of linear inequalities. The correct approach is to solve each inequality separately and find the intersection of the solution sets. For 2x + 1 > 7, subtract 1 and divide by 2 to get x > 3; combined with x ≤ 5, the range is 3 < x ≤ 5. Among the options, x = 5 satisfies both as 2(5) + 1 = 11 > 7 and 5 ≤ 5. This is justified because it fits the intersection precisely. A common incorrect option like x = 6 fails as it exceeds x ≤ 5. Another error might be selecting x = 3, which does not satisfy the strict inequality >7.

This question tests solving linear equations in a cost model. The correct approach is to isolate the variable for pages after subtracting the fixed cost. Given 50 + 0.2p = 86, subtract 50 to get 0.2p = 36, then p = 180. This is justified as 50 + 0.2(180) = 86. A common incorrect option like 170 might come from dividing 86 by 0.2 without subtracting. Another error could be using 0.2 incorrectly, leading to 150 or 200.

This question tests solving systems of linear equations. The correct approach is to use elimination by subtracting equations. From 2x + y = 13 and x + y = 9, subtract to get x = 4. This is justified by substituting back: for x = 4, y = 5, and 2(4) + 5 = 13. A common incorrect option like 3 might come from adding instead of subtracting. Another error could be solving for y first, leading to 9 or 13.

This question tests solving linear equations in one variable. The correct approach is to isolate the variable x by moving terms involving x to one side and constants to the other. Starting with 5x - 7 = 3x + 9, subtract 3x from both sides to get 2x - 7 = 9, then add 7 to both sides yielding 2x = 16. Dividing both sides by 2 gives x = 8. This solution is justified because substituting x = 8 back into the original equation gives 5(8) - 7 = 33 and 3(8) + 9 = 33, which are equal. A common incorrect option like 16 might result from forgetting to divide by 2 after isolating the term. Another error could be mishandling signs, leading to negative values like -8 or -1.

This question tests linear equations combined with inequalities. Given x + y = 10 and x ≥ y, we need to determine what must be true. Since x ≥ y and x + y = 10, we can substitute y = 10 - x into the inequality: x ≥ 10 - x. Adding x to both sides gives 2x ≥ 10, so x ≥ 5. This means x must be at least 5, making choice A correct. Choice B (y ≥ 5) is false because if x = 6 and y = 4, the conditions are satisfied but y < 5.

This question tests solving linear inequalities with attention to inequality direction. Starting with 3 - 2x ≤ 11, we first subtract 3 from both sides to get -2x ≤ 8. When dividing both sides by -2, we must reverse the inequality sign, yielding x ≥ -4. This means x can be any value greater than or equal to -4. A common error is forgetting to flip the inequality sign when dividing by a negative number, which would incorrectly give x ≤ -4.

This question tests solving linear inequalities with variables on both sides. Starting with 2(t - 3) ≤ 4t + 6, we first expand the left side to get 2t - 6 ≤ 4t + 6. Subtracting 2t from both sides gives -6 ≤ 2t + 6. Subtracting 6 from both sides yields -12 ≤ 2t. Dividing by 2 gives -6 ≤ t, which is equivalent to t ≥ -6. A common error would be to incorrectly move terms, potentially reversing the inequality sign when it shouldn't be reversed.