Explanation for Question 17 From the Math (No Calc) Section on the 2019 March Sat
Question 17 asks what is the solution to the equation above? 2 So I'm going to start by rewriting this equation as 11 times 3 X minus three. And to do that, 4 I just factored. And I'm going to write that over X minus three is 5 equal to X. Now that the top and the bottom of the left side 6 both have an X minus three dose. 7 Those cancel out and we get X is equal to 8 11, and that is the final answer.