Answer Choices
3 and 5
6 and 35
10 and 21
31 and 41
Explanation for Question 15 From the Math (No Calc) Section on the Official Sat Practice Test 1
So question 15 gives us this expression and asks us for all values 2 of X and a plus B equals eight. 3 What are the two possible values for C? 4 So we can see that C is the coefficient on X. 5 And so the first thing we can think about doing is foiling out this 6 expression. So that way we have an X squared term of next term and 7 then a normal number. So if we foil this out, 8 we'll get a, B X squared plus seven, 9 a X plus two B X plus 14. 10 And now we can see that the 14 is the same here. 11 And we know that a times B should be equal to 15, 12 right? So if a times B has to be equal to 15, 13 we know that because a times B would give us this number right here. 14 And we also know that a plus B is equal to eight. 15 So a and B are going to be either three or five, 16 but there's two possible combinations that we can make here because we could have 17 a, is three and B is five, or we could have a as five 18 and B as three. So with all that in mind, 19 we can now think more closely about this seven, 20 eight X plus two BX part, 21 because those two are the coefficients on our X. 22 So that means that seven, eight X plus two B X will be equal 23 to C, which is what we're trying to solve for. 24 So first we can use combination number one, 25 and we can plug in the a, and we can plug in the B, 26 right? And so if a is three, we have seven times, 27 three times X and be used to, 28 we have two times, five times X that gives us 21 plus 29 X plus 10. 30 Oh, about 21 plus X 21 X plus...