Explanation for Question 28 From the Math (Calc) Section on the 2018 May Sat
Question 28 says the equation has no solutions and a is a constant, 2 what is the value of it? So I'm going to start by distributing to 3 get negative three X plus X plus negative 4 a is equal to seven X minus two, 5 but this problem tells us that the equation above has no solutions, 6 which means we need to figure out what value of X would ensure that 7 we could get rid of the access, but that the remaining constants wouldn't equal 8 each other. So when does three negative three X plus 9 X equals seven X, but negative a not equal negative 10 two. So if we plug in negative two to this top 11 equation, we'd get negative three times negative two, 12 which is positive six X plus X is equal to seven X. 13 And this is true because we have seven X on both sides. 14 But when we plug in negative two for eight on the bottle, 15 we get positive. Two is equal to negative two, 16 which isn't true. Therefore, 17 negative two is the correct answer because that's a case in which the Xs 18 would cancel out. But the constants that were left would not equal each other. 19 And therefore there wouldn't be any solutions. 20 Therefore B's the correct answer.