Explanation for Question 33 From the Math (Calc) Section on the 2020 March Sat
So narrow looking at question number 33, 2 which we can tell is a systems of equations question, 3 because it gives us two equations right off the bat, 4 right? And we're told to solve for the value of X. 5 Now, when we're talking about systems of equations questions, 6 there are two ways to solve them. We can think about doing elimination or 7 substitution. Now you can do either way. 8 Um, but really this question is almost perfectly made for elimination. 9 The reason I say that is because if we look here, 10 both of the Y's have just the one is the coefficient. 11 So it'll be really easy to cancel them out by elimination. 12 All I have to do is multiply one of the equations by negative one. 13 So I'll choose the bottom equation. If I multiply this equation by negative one, 14 I'll get negative eight X minus Y is equal to negative five. 15 I've now switched the signs on all of my terms. 16 So I'll eliminate this equation so that I am not no longer confused, 17 right? Get all that out of the way, 18 move this up here. And now we see that the Y's 19 are going to cancel out perfectly because plus Y minus Y is zero. 20 So now I'll just add straight down to solve for X and a four 21 X minus eight X will give me negative four X again, 22 these whys cancel it out. And then four minus five gives me negative one. 23 So to solve for X, I need to now divide by negative four because 24 that's the coefficient that is on my X. 25 So if I divide by negative four, I get the X is equal to 26 negative. One divided by negative four. Now these negatives will cancel out, 27 right? A negative divided by a negative is a positive. 28 So I get positive one over four. 29 When I solve for X, therefore my best answer is going to be one 30 fourth. However, you don't have to solve it by elimination. 31 You could also solve one of these equations for Y and then plug it 32 in. But either way you should arrive at one fourth.