Linear growth and decay

The governing concept is solving a linear equation by setting two linear expressions equal. Since p(x)=q(x), write 260−15x = 68+3x. Subtracting 68 gives 192−15x=3x. Adding 15x to both sides gives 192=18x, and division by 18 gives x=192/18=32/3, approximately 10.67. Therefore option D is correct. A quick check confirms the result: at x=32/3, both expressions have the same value. The integer choices 8, 9, and 10 do not satisfy the equality; for example, at x=10, p(x)=110 while q(x)=98. The negative coefficient affects the algebra but does not change the method.

The governing concept is substitution into a linear function followed by solving a one-variable equation. Evaluate each required function value: f(3) = a + 9(3) = a + 27, f(7) = a + 63, and f(11) = a + 99. Adding them gives (a + 27) + (a + 63) + (a + 99) = 3a + 189. The problem states that this sum is 279, so 3a + 189 = 279. Subtracting 189 gives 3a = 90, and dividing by 3 yields a = 30. Hence option A is correct. The coefficient of a is 3 because a appears in all three function values; overlooking that repetition or adding the numerical terms incorrectly can produce the distractors.

The governing concept is substitution into a linear expression and collection of like terms. Calculate the three values: g(5) = b - 8(5) = b - 40, g(9) = b - 72, and g(13) = b - 104. Their sum is (b - 40) + (b - 72) + (b - 104) = 3b - (40 + 72 + 104) = 3b - 216. The given condition therefore becomes 3b - 216 = 288. Adding 216 to both sides gives 3b = 504, and dividing by 3 gives b = 168. Thus option D is correct. The distractors can arise from adding only some subtracted terms, treating b as appearing once, or making an arithmetic error when solving the final equation.