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Question 42794: can ANYONE help
x+3>5x-21
b)
|1+2x|>=5
c)
x^2-x-6>=0
d)
x-1/2-x >= 0
e)
3x+1/2plus 4 < x/2
f)
|1-3x|<5
i got all of them wrong today in class
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! a) x+3>5x-21
-4x+3>-21
-4x>-24
x<(-24)/(-4)
x<6
b) |1+2x|>=5
this "absolute" question forms a V-shape if we plot it. The equations of the 2 parts of the V are 1+2x and -(1+2x) so we need to solve both parts ie ask where are both lines greater or equal to 5.
Right then:
and:
so, we have x less than or equal to -3
or we have x is greater than or equal to 2
c) x^2-x-6>=0
first do x^2-x-6=0
(x-3)(x+2)=0
so x-3=0 or x+2=0
so x=3 or x=-2.
Now, you need to visualise this curve. It is a u-shape cutting the x-axis at x=-2 and x=3.
The question asked, where is it greater or equal to zero? The answer is everywhere on the curve above the x-axis. So the answer is values of x less than or equal to -2 and also values of x greater than or equal to 3.
d) x-1/2-x >= 0
This is a tricky one... you need to know that this form of equation has 2 distinct parts, separateed by asymptotes. We need to find the asymptotes and where the curve(s) cross the axis. Then we can sketch it and then we can answer the question.
asymptote is when 2-x=0 --> x=2
Now for the y-asymptote:
asymptote is when 1+y=0 --> y=-1.
Now for where the curves cross the axes:
when x=0,
--> 
--> 
--> 
when y=0,
--> 
--> 
--> x=1
So draw axes.
Draw 2 asymptotes, at x=2 and y=-1
Plot 2 point: (-1/2, 0) and (0,1)
Sketch one part of the curve through these 2 points approaching the asymptotes. The other half of the curve is in the diagonally opposite quadrant.
Now to answer the question:
where is the curve greater or equal to zero. By looking at the sketch, the curve is greater or equal to zero from x>=1 upto x<2.
e) 3x+1/2plus 4 < x/2
not sure what you mean here...you try
f) |1-3x|<5
same approach as other || question. You try it.
jon.
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