Hello, I came across a question TY9C6-RQKVH while solving inequalities. Here is the question.
If 5x + 5 >55 and xy < 0 then which of the following must be true if y is an integer?
y+1 < 2
5y > y
y-4 < -7
y^2 < 0
y > x
Explanation . If we simplify for x we get x>10. Also, xy < 0 . So y<0
Right choice y + 1 < 2. This means y < 1. Which means y could take values 0 and below on the number line. When y=0 , xy=0 and xy < 0 condition fails.
I chose option y-4< -7 . I wanted to know why this is wrong and the choice y+1 <2 is correct.
Hi @Colonial_olive_marmo, your logic is very close!
Once we have y<0, you can basically just plug into the equations and see what would work for different versions of y.
The choice y-4<-7 works in some cases, but is not always correct, because rearranged, this choice can be expressed as y<-3, which does not completely overlap with y<0. You can see this if you plug in y=-1. You would get (-1)-4 < -7, which simplifies to -5 < -7, which is not true.
The choice y+1<2 can be rearranged to y<1, which does overlap completely with y<0. Plugging in y=-1 does work, as you get (-1)+1 < 2, which simplifies to 0<2, which is true. Even if you use a more negative number like y=-100, the choice is still true: (-100) + 1 < 2 simplifies to -99 < 2, which is still true.
Please let me know if I can clarify further!
I agree with your description. However, I feel you don’t consider the corner case when y=0.
y+1 < 2 implies y < 1 and not y< 0. Since y< 1 , y can take values left of 1 , that includes 0. The issue with choosing y=0 is then it invalidates xy< 0. So, I still don’t feel y+1< 2 is correct. On the other hand if it was y+3< 2 or any other combination where, after simplification we get y<0, I would have no doubt in my mind.
We’ll never be able to have y = 0 because we already determined from the given equations that we must have y < 0 in all cases. It isn’t a possibility for it to be zero or greater because it would violate the constraints of the question.
As y is also defined to be an integer, the greatest it could be is y = -1
I see your point now. With xy< 0 and x> 10, we know that y <0 . We are looking for an answer that gives us all ranges of y. So y+1< 2 is correct answer compared to y-4<-7. Thank you @Justin