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Sun 11 Sep, 2005 01:56 am
I am pretty sure my book got an error,, can someone solve this inequality.
1/2X + 3 < 2/3X
The book's answer is X>18 and I got X>18, but the book's steps are wrong.
The book solves it like this:
1/2X + 3 < 2/3X
3X + 18 < 4X -----> 3X -4X + 18 < 4X -4X
-X + 18 < 0 -----> -X + 18 -18 < 0 -18
-X/-1 > -18/-1
X>18
Althoug the answer is right, the problem that I see is that in
3X + 18 < 4X ---> 3X (-4X) + 18 < 4X (-4X)
you cannot subtract 4X from both sides, when you should subtract 3X from both sides.
So please tell me if the book got an error or if I am wrong.
Watch your notation
When you subtracted -4x and put it in parenthesis (-4x) it looks like multiplication
I use the following notation for steps
m(6) means multiplying the equation by 6
d(6)-dividing by 6
a(6)-adding 6 to both sides
s(6)-subtracting 6 from both sides
The book seems to be trying to emphasis that when you multiply by -1 you flip the inequality sign
A simpler solution (IMO) is
1/2x+3<2/3x
m(6) 3x+18<4x
s(3x) 18<x
flip x>18 (put in variable first notation)
The book solution
3x+18<4x
s(4x) -x+18<0
s(18) -x<-18
m(-1) x>18 (the inequality sign flips direction when you multiply by negative numbers)
Another solution
1/2x+3<2/3x
s(1/2x) 3<(2/3-1/2)x
3<1/6x
m(6) 18<x
flip x>18 (put in variable first notation)
They're all result in the same answer
Rap
Thanks.
When I got to this part of an inequality 3X + 18 < 4X ---> 3X s-4X + 18 < 4X s-4X
It's actually subtracting 4 from both sides. But I mean can you really subtract 4 from both sides and leave other two numbers by itself on the other side of the equation?
3X + 18 < 4X
-4X <---> -4X
------------------
-X +18 < 0
so therefore is that actually the right way of doing it? I mean the answer is right but, shouldn't you subtract 3X from both sides of the inequality?
Either way is right. In reality I would subtract 3x from both sides resulting in 18<x and then rewrite this as x>18, but the other way (subtracting 4x) will get you to the same place.
Rap