Rational Inequality

J

JSmith

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Algebraically determine when (x+4)/(x-3) > (x-5)/(x+2).

I need help with the steps. First, I know you have to bring the right side over and set the inequality to zero. Then, you subtract the right side from left, and simplify. Then you determine when the numerator and denominators are equal to zero. Then, these are the intervals to put in the interval chart. Can someone show me how they would solve this??
 
You're not following the correct Order of Operations.



For example, you need to multiply the binomial x+4 by the binomial x+2.

Writing (x+4)(x+2) is correct

Writing x + 4(x+2) is not correct



(x+4)(x+2) = x^2 + 6x + 8

But x + 4(x+2) = 5x + 8
 
Multiply (x+4)(x+2), first

Multiply (x-5)(x-3), second

Subtract the latter result from the former result, third
 
This looks mostly good (still missing some grouping symbols, though), but there's an issue with the last line (if you want to completely simplify -- which may not be needed, in this exercise).

How did you change the numerator from 14x-7 to x-1/2 ?



Notice that both 14 and -7 are divisible by 7.

Factor out this 7, and then divide both sides by 7.
 
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mmm4444bot said:
This looks good, until the last line.

How did you change the numerator from 14x-7 to x-1/2 ?



Notice that both 14 and -7 are divisible by 7.

Factor out this 7, and then divide both sides by 7.
Click to expand...

I assume the OP factored out 14 instead of 7 - strange but "legal"!
 
Subhotosh Khan said:
I assume the OP factored out 14 instead of 7
Click to expand...

I was thinking that perhaps x=1/2 was known by Smith in advance to be one of the boundary points, and so Smith simply changed 14x-7 to x-1/2 to "make it work". :cool:
 
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JSmith said:
I wont show all of my work, but the end result was: (x+5.75)/(x-3)(x+2) >0
Click to expand...

JSmith,

regarding this previous post of yours, the binomial factors for the intended
denominator are correct, but you must have grouping symbols around the
denominator when the inequality is typed out horizontally, such as:

(binomial numerator)/((x - 3)(x + 2)) or

(binomial numerator)/[(x - 3)(x + 2)].


Otherwise, because of the Order of Operations,
what you typed is equivalent to:


\(\displaystyle \bigg( \dfrac{binomial \ \ numerator}{x - 3} \bigg)\bigg(x + 2 \bigg) \ > \ 0.\)
 
Initially, yes.

Do you remember my instruction to now divide both sides of the inequality by 7?

That makes the 7 go away, and you're left with 2x-1 on top. :cool:


Edit: Actually, for the purposes of this exercise, I supposed that you could leave the numerator as 7(2x-1) -- in other words, not fully simplified -- because you're simply trying to find the boundary points of the solution set for the inequality AND having that 7 up there does not change the fact that x=1/2 causes the numerator to become zero (thus 1/2 is a boundary point in the solution).

Even so, it's good to know how to fully simplify (until ubiquitous software eliminates the need for anybody to do algebra other than as a novelty demonstration).
 
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Ok, does this satisfy everything people have mentioned. I ended up factoring by 7 instead of 14.
math.jpg
 
Yes.



Here's a side note about notation: Beginning with your third line above, the square brackets around the denominators are unnecessary.

You would only need those particular grouping symbols, if you were texting the expressions, instead of taking the time to format everything pretty. Like this texting:

(2x-1)/[(x-3)(x+2)] > 0

As you have chosen to post an image of "pretty-print" mathematical formatting, the fraction bar serves as the grouping symbol. In other words, when you use a fraction bar versus a forward slash (to denote a ratio), it's obvious what's on top and what's on bottom. There is no need for grouping symbols around denominators which appear underneath a fraction bar.

Placing a blank line between the inequalities will also help improve readability.



So, what's your next step, in this exercise? :cool:
 
Ok, thanks for the clarification on the brackets, I was a little confused about their purpose. Now, I think I should be ok, I was just a little confused about the earlier steps. Now, I can create an interval chart that looks like this:
X< -2-2 < x < ½½ < x < 3x>3
2x-1--++
x-3---+
X+2-+++
-+-+

And from this I can conclude that (x+4)/(x-3) > (x-5)/(x+2) on the intervals -2 < x < 1/2 and x>3.

Look good?
 
Last edited by a moderator:
I just noticed your missing grouping symbols above. (I inserted them into your post in red.)

If you would like to understand the grouping-symbol issue better, then please tell us whether you understand the reasoning below (quoted from our posting guidelines). Otherwise, you may ignore this post.

mmm4444bot said:
ֺ

NOTE: If you've learned about Order of Operations, then be sure to type grouping symbols in your posts where needed to prevent confusion.

For example, typing this: x+80/x-10 means \(\displaystyle x + \dfrac{80}{x} - 10\)

But typing (x+80)/(x-10) means \(\displaystyle \dfrac{x + 80}{x - 10}\)
Click to expand...
 
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