Whenever the question includes variables and the answers are numbers, think Plugging In the Answers. In (A), x = 3 and y = -3. Plug these numbers into the equation to get -3 = 5(3 – 3)² – 3. Solve the right side of the equation to get -3 = 5(0)² – 3 or -3 = 0 – 3. The correct answer is (A). None of the other points work when plugged into the equation, so eliminate (B), (C), and (D).

Plugging In would not be straightforward for this problem, given the fractions and negative numbers. A better approach would be to simplify the expressions first and then plug in or solve. Distribute the (c + 2) term to both sides of the equation. On the left side, this will cancel out with the (c + 2) term in the denominator. On the right side, make sure to distribute the (c + 2) to both terms inside the parentheses. The equation becomes ​\( \frac{3(c+2)}{c+2} \)​= 5(c + 2)​\( \frac{c(c+2)}{c+2} \)​ – or 3 = 5c + 10 – c. Combine the c terms and subtract 10 from both sides to get -7 = 4c. Divide both sides by 4 to find that c = ​\( -\frac{7}{4} \)​. The answer is (B).

First, rewrite the equations so that they are in the slope-intercept form of a line, y = mx + b, where m = slope. The first equation becomes 3y = -x + 42 or y = ​\( -\frac{1}{3}x \)​ + 14. The slope of this first line is therefore ​\( -\frac{1}{3} \)​. The second equation becomes -y = -3x + 8 or y = 3x – 8. The slope of this line is therefore 3. The slopes of the two lines are negative reciprocals of each other, which means that the two lines are perpendicular to each other. The correct answer is (D).

The graph crosses the x-axis at three distinct points. When the function is set to 0, there should be three real solutions for x. Use Process of Elimination to solve this question. Set the equation in (A) to 0 to get 0 = (x – q)². In this equation, the root is at x = q, thereby providing only one real value for x. Eliminate (A). Set the equation in (B) to 0 to get 0 = (x – q)(x + s). The solutions for this equation are x = q or x = -s. Therefore, there are only two real solutions for x. Eliminate (B). Set the equation in (C) to 0 to get 0 = (x – r)(x + s)(x + t). The solutions for this equation are x = r, x = -s, and x = -t. Therefore, there are three real solutions for x. The correct answer is (C).