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Absolute Vales with fractions HELP PLEASE

  • Thread starter p.ella
  • Start date
  • #1
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Absolute Vales with fractions HELP PLEASE :)

hey everyone,

Homework Statement



find the solution set:

l 1/x -3 l > 6

Homework Equations



none

The Attempt at a Solution



1/x > 9 or 1/x < -3
1 > 9x or 1 < -3x
1/9 > x or -1/3 > x

this answer doesn't seem to make sense, I always get confused when the variable is in the denominator :l any help would be much appreciated please & thankyou! :)
 

Answers and Replies

  • #2
418
28
I think you are sort of on the right track but the key is be careful with your algebra, remembering to flip the inequalities when a negative number multiplies both sides.

l 1/x -3 l > 6 can be changed algebraically into
A) 1/x - 3 > 6 and
B) -[1/x - 3] > 6, which is equivalent to -1/x + 3 > 6

A) gives 1/x>9, and since x must be positive in this expression, we multiply both sides by x>0
1>9x or 1/9 > x.

B) gives -1/x > 3. Now for this expression to be true, x must be negative, so we multiply both sides by x<0,
-1<3x or -1/3<x.

Does this make sense now?
 
  • #3
41
0
I think you are sort of on the right track but the key is be careful with your algebra, remembering to flip the inequalities when a negative number multiplies both sides.

l 1/x -3 l > 6 can be changed algebraically into
A) 1/x - 3 > 6 and
B) -[1/x - 3] > 6, which is equivalent to -1/x + 3 > 6

A) gives 1/x>9, and since x must be positive in this expression, we multiply both sides by x>0
1>9x or 1/9 > x.

B) gives -1/x > 3. Now for this expression to be true, x must be negative, so we multiply both sides by x<0,
-1<3x or -1/3<x.

Does this make sense now?
Hey,
first off thank you so much for responding :) I seem to be okay with everything except part B) of the problem, don't really get the bit at the end to be honest :frown:
 
  • #4
418
28
Okay well, first of all, if you multiply both sides of an inequality by a negative number, you need to flip the inequality sign. For example, we know that 7>2 and -14<-4. The way you get between these two equations is by multiply both sides by -2 and ALSO changing > into <. Do you get this?

Now the second thing you have to keep track of is how it is possible for fractions to satisfy inequalities. Remember that
(positive)/(positive) = (positive)
(positive)/(negative) = (negative)
(negative)/(positive) = (negative), and
(negative)/(negative) = (positive).

So if we had an expression like 1/x > 4, then obviously x cannot be negative because we would then have (negative)>4, and of course all negative numbers are less than all positive numbers.

So when you have -1/x > 3, we can definitely see that x must be negative--otherwise the -1/x would be a negative number, and all negative numbers are less than 3.

Now for doing the algebra with inequalities. A good way to check your logic and to figure out how this works is by plugging in numbers and watching if things make sense. For example, let's try plugging in the value x=-1/4 into the equation -1/x > 3:

-1/x=-1/(-1/4) = 4, and 4>3, so x=-1/4 satisfies the inequality -1/x>3. So the inequality is "true" when x=(-1/4).

So what happens when we try to move the x to the other side of the inequality? Let's take each side independently and see what happens when we multiply it by x = (-1/4). The original inequality is:
-1/(-1/4) > 3
Let's call the left hand side of this equation "LHS" [LHS=-1/(-1/4)] and the right hand side "RHS" [RHS=3], and the inequality would read LHS>RHS.

LHS*x = -1/(-1/4) * (-1/4) = -1
RHS*x= 3*(-1/4) = -3/4
We know of course that -1 < -3/4
So we have concluded LHS*x < RHS*x
Why did the inequality turn around when we multiplied by x? The reason is exactly what I said at the beginning of this post--x is a negative number here, so when we multiply both sides by x, we must flip the inequality signs as well.
 
Last edited:

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