I need to find all the real numbers for (2x-1)/(4+x) <= 2
what I tried was to move the 2 so that I would get (2x-1)/(4+x)-2 <= 0 and from there I got -9/(4+x) <=0
\[\frac{2x-1}{4+x} \le 2\]Step 1: multiply both sides by 4+x. What do you get?
yea, I'm pretty sure that I'm not allowed to do that
Awww... It seems like...... I'm on the wrong track...
@invidos so because there are x on denominator my opinion that in the first step you need make the restriction for what value not can getting newer x can you do it ?
x != -4
yes x not can being never equal -4 so because do you can tel me now why ?
because you cannot divide by zero and 4+(-4) = 0 or am I missunderstanding you'r question?
yes do you know it right but with ,,mathanswer" so because than this fraction will be undefined ok ?
yea
so now can you continue it ?
ehm, so I should just multiply both sides and than say that it is undefined when x =-4?
yes but this x not can being equal -4 this wann being just the first step when you need to solve inequality like in this case ok ? so than now you can continue it how you have wrote little up @RolyPoly
Nope, the 2x cancels out on both sides!
x=-4 is a ''critical point'' When x> -4, this inequality is true. When x<-4, this inequality is not true.
ehm, I am feeling kind of slow right now, but when multiplying both sides with (4+x) you get -1<=8 that might be true but it doesn't give anything
ok, but isn't there a "smart" way to solve that? I undestand how you got that x>-4 but it seems like you just tested you way there
Because I don't think you can 'solve' it using 'normal' way...
so what will get you if you add to both sides (-2) ?
2x-1-8-2x <= 0 if you then multiply both sides with 4+x
the 2x still cancels out!
I guess that you could rewrite everything as (-9)/(4+x) <= 0 which would make it alot easier to se that it cant be smaller than -4
You guys are smart :D
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