Solve the problem.
-An electronics company kept comparative statistics on two products, A and B. For the years 2003 to 2011, the total number of Product A ever sold (in thousands) is given by the equation y = 77x + 200, where x is the
Number of years since 2003. For that same period, the total number of Product B ever sold (in thousands) is
Given by the equation y = -30x + 434, where x is the number of years since 2003. Use the substitution method to
Solve the system and choose the statement that most accurately describes the solution.
A) At about 2.2 years (to the nearest tenth) since 2003, the company sold the same number of Product A as Product B.
B) At some point between 2003 and 2011, the company had sold 2200 of each product.
C) When 369,000 of Product A had been sold, 2.2 times as many of Product B had been sold.
D) The company sold about 2.2 times as many of Product B as Product A.

Question

Quizplus · Accepted Answer

To solve the system using the substitution method, set the two equations equal to each other because we're looking for when the total number of products sold is the same. So, $77x + 200 = -30x + 434$. Solving for $x$ gives $107x = 234$, so $x \approx 2.2$. This means that approximately 2.2 years after 2003, the company sold the same number of Product A as Product B.

Solve the Problem. -An Electronics Company Kept Comparative Statistics on Two Products, a Two