Explanation for Question 17 From the Math (No Calc) Section on the 2018 April Sat
Question 17 asks what is one possible solution to the equation above. 2 So I'm going to start by taking X minus one and multiplying that by 3 X minus one. So I'm just going to rewrite that up here. 4 So X minus one times X minus one, 5 and that gives me X squared plus negative one ax plus negative 6 one X, which is negative two X plus one is 7 equal to three X minus five. 8 Now that I have that written out, I'm going to move everything to the 9 left side of the equation so that, 10 that I can factor. So I'm going to subtract the reaction, 11 both sides, which gives me X squared plus negative five X. 12 And I'm going to add a five to go outsides to get plus six. 13 And now finally, I'm going to factor those factors into X minus three and 14 X minus two. And so I know that in this problem, 15 X can either be equal to two, 16 or it can be equal to three. And since this question asks, 17 what is one possible solution? 18 Either two or three would be the correct answer.