Explanation for Question 18 From the Math (No Calc) Section on the 2019 March Sat
Question 18 says if X, 2 Y is the solution to the system of equations above what is the value 3 of a hundred X plus 40 Y. So we need to solve for X 4 and Y and then plug them into a hundred X plus 40 Y. 5 So to start, I'm going to take that bottom equation, 6 three X minus Y is equal to 30, 7 and I'm going to try to isolate Y. So I'm going to add Y 8 to both sides and subtract 30 from both sides. 9 And that's going to give me three X minus 30 is equal to Y. 10 Now that I have that I can plug that into this top equation here. 11 And there's only going to be one variable. So I'll be able to sell 12 for X then. So I'm going to rewrite that equation as two X plus 13 three times three X minus 30 is 14 equal to 31. 15 I'm going to distribute that and it becomes two X plus nine X 16 minus 90 is equal to 31. 17 Then I can add like terms that becomes 11 X minus 90 18 is equal to 31 and add 90 to both sides. 19 And I get 1 21 is equal to 11 X. 20 Finally, once I divide both sides by 11, 21 I get X is equal to 11. 22 So now I know what X is equal to, but I still need to 23 solve for Y. So I'm going to take this X value and I'm going 24 to plug it into the bottom equation. So three times 11 is 33, 25 and I'm going to write that minus Y is equal to 30. 26 If I add Y to the right side and subtract 30 from the left 27 side, I get three is equal to Y finally, 28 to find the actual answer, I can take a hundred X plus 40 Y 29 and plug in my known values. So a hundred times 11 plus 30 40 that wasn't writing correctly 40 31 times three. And to find that answer, 32 I get 1100 lost one 20. 33 And finally, the answer is 1,220.