Let me ask you this. What are the solutions if a = 1/2, for example. How many (legal) solutions do we get?
2Let me ask you this. What are the solutions if a = 1/2, for example. How many (legal) solutions do we get?
-Dan
What are they specifically?
okay, but in the answer i need it to go from [1,3/2) so how do i get it then??The statement [imath]a\ge 1[/imath] is false. Test out a few values.
and here is wolfram alphaokay, but in the answer i need it to go from [1,3/2) so how do i get it then??
In the OP.okay, but in the answer i need it to go from [1,3/2) so how do i get it then??
I am a bit confused how you combined those inequalitions.In the OP.
[imath]x \le 2[/imath] and [imath]x\le 2a \implies a\le1[/imath]
Put your domains together.
If [imath]x=2[/imath] and [imath]x=2a[/imath], what is the value of a?I am a bit confused how you combined those inequalitions.
1If [imath]x=2[/imath] and [imath]x=2a[/imath], what is the value of a?
I meant to say [imath]a \ge1[/imath], not [imath]a\le1[/imath]1
but
if x≤2
x could be -1
x≤2a
a could be 0 and it would still work
so is it safe to assume this:
2≤2a
cause it this case it would be wrong
so if x is 0I meant to say [imath]a \ge1[/imath], not [imath]a\le1[/imath]
If x=0, then 2*(something greater or equal to 1) is still greater than 0.so if x is 0
x≤2
x≤2a
a could also be 0 zero
a≥1
so what now?
okayIf x=0, then 2*(something greater or equal to 1) is still greater than 0.
Right. And a < 3/2, so [imath]1 \leq a < 3/2[/imath]. Done!okay
so can i combine them like this
x≤2
x≤2a
2≤2a
1≤a