algebra help?

Question

im not too sure how to do this problem:
http://img441.imageshack.us/img441/7658/algebrabd3.jpg

are you supposed to factor out the denominator, switch the fraction over to the other side and make the inequality less than 0? and then after that are you supposed to sign graph it? im a little confused, help pls

Answer 1

x/(x + 2) - x/(x + 3) <= 7/(x^2 + 5x + 6)

Your first step would be to bring the fraction on the right hand side to the left hand side.

x/(x + 2) - x/(x + 3) - 7/(x^2 + 5x + 6) <= 0

Now, factor the denominator x^2 + 5x + 6.

x/(x + 2) - x/(x + 3) - 7/[(x + 2)(x + 3]) <= 0

*** At this point it may be tempting to multiply both sides by (x + 2)(x + 3) to eliminate the denominator. DON'T DO IT ... it eliminates values and is an incorrect step ***

It's easy now to put these under a common denominator. The lowest common denominator is (x + 2)(x + 3), so we have

[x(x + 3) - x(x + 2) - 7] / [(x + 2)(x + 3)] <= 0

Simplify the top,

[x^2 + 3x -x^2 - 2x - 7] / [(x + 2)(x + 3)] <= 0

[x - 7] / [(x + 2)(x + 3)] <= 0

At this point, we obtain our critical values by determining what makes the left hand side equal to 0, or what makes the left hand side undefined.
What makes the left hand side 0 would be x = 7.
What makes the left hand side undefined would be what makes the denominator undefined; that would be x = {-2, -3}

Your critical numbers are x = {7, -2, -3}

You're going to have four intervals to deal with:

1) (-infinity, -3)
2) (-3, -2)
3) (-2, 7]
4) [7, infinity)

Note that there are *square* brackets at the 7; this is because the inequality is less than _or equal to_ 0.

What we have to do at this point is make a number line, and TEST each region by plugging in one value in that region; if that one value is positive, the whole region is positive. If that one value is negative, that whole region is negative. What we want are negative regions, since the inequality is less than or equal to 0.

Our number line looks like as follows:

. . . . . . (-3) . . . . . (-2) . . . . . . (7) . . . . . .

First, we test a value to the left of (-3). Test -10; then,

[x - 7] / [(x + 2)(x + 3)] = (-17) / [(-8) (-7)], which is negative.
We want negative intervals, so we retain the interval (-infinity, -3).

. . . { - } . . (-3) . . . . . (-2) . . . . . . (7) . . . . . .

Test x = -5/2. Then [x - 7] / [(x + 2)(x + 3)] =
(negative) / [(negative)(positive)] = positive.

. . . { - } . . (-3) . . { + } . . (-2) . . . . . . (7) . . . . . .

Test x = 0. Then [x - 7] / [(x + 2)(x + 3)] = (-7) / (2)(3) = negative.

. . . { - } . . (-3) . . { + } . . (-2) . . . { - } . . (7) . . . . . .

Test x = 10000000. {Note, we can do this because all we need is ONE value!}. Then [x - 7] / [(x + 2)(x + 3)] =
[positive] / (positive * positive) = positive.

. . . { - } . . (-3) . . { + } . . (-2) . . . { - } . . (7) . . { + } . . .

From our number line, we now discern where we have negative intervals. It is clear that, from the number line, it is negative at

(-infinity, -3) U (-2, 7]

and that's precisely our answer. In set notation, the answer is

{x | x < -3 OR -2 < x <= 7 }

[Note the x <= 7 portion, to correspond to the fact that x may in fact equal 7].

Answer 2

x / (x + 2) - x / (x + 3) < = 7 / (x + 3)(x + 2)
x (x + 3) - x (x + 2) < = 7
x² + 3x - x² - 2x < = 7
x < = 7

I`m not sure what you mean by "sign graph".
It may simply mean draw the vertical line x= 7 and the required answer is everything on or to the left of this line.

Answer 3

I think you are close. I'd multiply top and bottom of first term by (x+3), then second term top and bottom by (x+2), which will allow you to get rid of (x^2+5x+6) in the denominator.

This leaves you with x^2+3x-x^2-2x <= 7, or x<=7.

Good luck.....