If the number on the bottom is greater than 4 it will result in a negative number on the bottom. Divide a positive by a negative and you get a negative, and we all know a negative number is less than 3. That is how x>4 is an answer too. Watch out for divisions in these problems.
After multiple fast edits, decided to cut everything out but this. Hope I helped.
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Since this was kinda ignored...
Here you go:
x(4 + x)/((4 - x)(4 + x)) < 3
(4x + x^2)/(16 - x^2) < 3
4x + x^2 < 3(16 - x^2)
4x + x^2 < 48 - 3x^2
4x^2 + 4x - 48 < 0
4(x^2 + x - 12) < 0
x^2 + x - 12 < 0
(x + 4)(x - 3) < 0
Therefore, x > 4 or x < 3.
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Personally, I don't think you can SOLVE for x>4, it is more or less proven due to the fact that 4 just makes you divide by 0 and examples that anything greater than 4 works.
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As long as the answer is right, the process must have been right.
I say draw a dishwasher.
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okay, first attempt was a fail, complete version in reply:
multiply both sides by (x^2-8x+16), if you factor out a negative of (4-x) it's -(x-4), so I got (-x)(x-4)<3x^2-24x+48. from here bring everything over to the right, you now have 0<4x^2-28x+48, now factor out a 4 and divide it to the left side (essentially getting rid of it), and factor the trinomial into: 0<(x-3)(x-4). again i haven't done anything after that but you can prove from here that it will only be true if both binomials equate to positve or negative, both are positive when x is greater than 4, and both are negative when x is less than 3
Sorry keep on deleting, but i think it was a false alarm...
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alternately you really just need to find the 0s, since the graph of the x^2-7x+12 shows where the values are true and aren't true, since it's the trinomial is greater than 0, true values are above the x-axis, false values on or below the x-axis, thus the range of x values that make the inequality true are those above the x-axis being <3 and >4
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x/(4-x)<3
(x-4)^2 x/(4-x) < 3 (x-4)^2
(x-4)^2 -x/(x-4) < 3 (x-4)^2
-x (x-4) < 3 (x^2 - 8x + 16)
-x^2 + 4x < 3x^2 - 24x + 48
4x < 4x^2 - 24x + 48
0 < 4x^2 - 28x + 48
0 < 4 (x^2 - 7x + 12)
0 < x^2 - 7x + 12
0 < (x-4)(x-3)
true if:
(x-4)>0 and (x-3)>0
in this case, true only if x>4
or if:
(x-4)<0 and (x-3)<0
in this case, true only if x<3
the less wordy version of what i said
or if you view it the alternative way:
a=1
b=-7
c=12
z=(7+/-sqrt(49-48))/2
z=(7+/-1)/2
z=8/2,6/2
z=4,3
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If something should be bigger than 3 and bigger than 4 same time, we may ignore "bigger than 3" cause if something meet "bigger than 4" requirements it will meet "bigger than 3", isn't it?
So, we use only "bigger than 4".
Sorry, my math is much better than my english.
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x/(4-x)<3
x/(4-x)-3<0
x/(4-x)-3(4-x)/(4-x)<0
(x-3(4-x))/(4-x)<0
(4x-12)/(4-x)<0
(4x-12)(1/(4-x))<0
4x-12<0 AND 1/(4-x)<0
4x-12<0 AND (1/(4-x))(4-x)^2<0*(4-x)^2
4x<12 AND 4-x<0
x<3 AND x>4
Q.E.D.
It's amazing what a good night's sleep can do for your math skills.
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Incorrect. 4-x is the multiplicative inverse (reciprocal) of 1/(4-x), just as 3 is the reciprocal of 1/3. When multiplied together, they give 1. When applying the multiplicative or additive inverse to both sides of an inequality, it reverses the sign.
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I just took a close look at it, and he does indeed have a correct solution. Bravo! That said, if I could figure out exactly where I've gone wrong, I feel like I'm on the verge of having a more elegant solution (in my opinion, of course; in mathematics, as in everything else, when it comes to aesthetics, your mileage may vary).
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Once you're at the (4x-12)(1/(4-x))<0, you can write that as a fraction: (4x-12)/(4-x)<0. In order for a fraction to be negative, one of the following must be true: (top > 0 and bottom < 0) or (top < 0 and bottom > 0). So (4x-12 > 0 and 4-x < 0) or (4x-12 < 0 and 4-x > 0). If you solve those, you get (x < 3 and x < 4) or (x > 3 and x > 4). You can then simplify this to (x < 3) or (x > 4).
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wolfram alpha sucks ass now. Ever since they've introduced the pro features, it's no longer as useful as it was, especially when you wan them to show you the steps to the solution....unless you buy their "pro" subscription.
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When he turns "4x - 12" into "4 . (x - 3)" he's using the simplification method in order to make achieving the result an easier task. Everything else on the paper is just the way he gave the answer.
Click here for more info on basic algebra simplification methods.
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Okay: both sides: *(4-x) you have 2 cases: a) (4-x)<0 and b) (4-x)>0. If it's 0, then division through zero => x!=0. If its negative (x>4) the "<" changes to ">". You get x>12-3x. Both sides +3x => 4x>12. This is true for all x>3, but you have to match x>4, so x>4 is one solution. If (4-x) was positive (x<4) you get x<12-3x. both sides +3x => 4x<12 this is true for all x<3.
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Oh, let me play, too! :D
You obviously can't divide by 0, so 4 is totally out of the picture. The two possible cases are now x > 4 and x < 4. Let's see what happens if we assume x to be larger than 4.
If x > 4, obviously x/(4-x) is negative. Now we multiply the whole thing with (4-x) to remove the denominator. Bur remember that 4-x is negative! We have to invert the inequality sign so we get x(4-x)/(4-x) > 3(4-x).
Divide 4-x by 4-x and you are left with x > 12-3x. Add 3x:
4x > 12
Now divide by 4 and you have your solution: x > 3.
Now remember that we said that this should be the calculation for every x that is bigger than 4. Now we know that every x that is greater than 4 solves as long as it also bigger than 3. That means our solution for this case is "Every x that is simultaneously bigger than 3 and bigger than 4".
I don't think I need to explain that every x that i bigger than 4 is always bigger than 3 and that an x between 3 and 4 is NEVER bigger than 4. So we disregard the x > 3 and the final answer for this one case is x > 4.
You said you solved the other case yourself, so that all fine and dandy. I tried to stay away from mathematical language in the hopes of being more accessible ;)
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I wish the math problems I face now were this simple...
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Derp, pressed enter during in the title bar.
I feel incredibly stupid for asking, but I need help on this problem:
x/(4-x)<3
The solution is x>4 and x<3.
My question is, how do I get to x>4? Can someone show me the steps, because my mind is blanking.
Thanks...
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