the solution makes sense if your equation is x+y > z, but what if your equation is x < y + z. Is there anything wrong x < y + z? For the second statement it would be x < 6 + 3. x < 9 which could be less than 3 but also could be greater than 3.


My question is that how do you know to use x + y > z, when x < y + z, or y < z + x are all valid equations but would give you different results.

-Stan

Stan,

The rule is that, for any three sides of a triangle,

x + y > z

and

z > |x - y| and/or z > | y - x|

- Veritas Help

Sorry I am confused. The rule is that any two sides of a triangle is greater than the third side.
This will actually give us 3 inequalities, and not just one inequality.

If x, y, z are the lengths of each side of a triangle. So we can actually write three equations.

a) x + y > z
b) y + z > x
c) z + x > y

If we applied statement 2, y=3, and z=6

a) x + 3 > 6
b) 3 + 6 > x
c) 6 + x > 3

since we know that for a triangle a side cannot be negative, a) tells us x has to greater than 3. But b), c) tells us that 9 > x > 0. My question is that how did you know to use x + y > z instead of the other two equally valid equations? When I was solving this question I picked equation B) because that was the most convenient for me.

I think to properly solve this problem, you have to actually work through all three inequalities, to find the minimum. And cannot just randomly pick one of the three inequalities, and hope to get lucky.

thanks,
-Stan

You are correct. You do have to consider all three, since 2 of the three (for example x < 9) don't clarify if x < 3.

- Veritas Help