These can be solved independently. Now x will always equal x, so only the second equation is
used to define the points that satisfy the original equation. Solving for y, you haveThis is the solution to the original equation, and it is the answer to the problem.28 . B
When x is subtracted from each of the numbers 8, 16, and 40, the resulting three numbers
are 8 − x, 16 − x, and 40 − x. In a geometric sequence a 1 , a 2 , a 3 , ..., an,...for all integers m ≥ 1,
, where r is a constant. Here, 8 − x, 16 − x, and 40 − x form a geometric progression.
So . Solve this equation for x. Cross multiplying, you have (16 − x)(16 − x) = (8 −
x)(40 − x). Multiplying out each side and solving for x, you have
You can check that this is correct. The original numbers were 8, 16, and 40. When x = 4 is
subtracted from each of these numbers, the resulting numbers are 8 − 4, 16 − 4, and 40 − 4,
which are the numbers 4, 12, and 36. Then and .You have the same ratio, so you know that the value of x is 4 and (B) is correct.29 . C
Substitute −1 for x and 1 for x into f(x) = ax^2 + bx + c and then set the resulting expressions
equal to the values of the function f at x = −1 and x = 1.
When x = −1, f(x) = f(−1) = ax^2 + bx + c = a((−1)^2 ) + b(−1) + c = a − b + c. You know that f(−1) =
−18, so a − b + c = −18.When x = 1, f(x) = f(1) = ax^2 + bx + c = a(1^2 ) + b(1) + c = a + b + c. You know that f(1) = 10, so a +
b + c = 10.