# Solving the inequalitites in term of x, applying rules

• rohan03
I will work on it and re post it later.In summary, the conversation discussed the process of solving an inequality and finding the appropriate range for the solution. The conversation included a discussion on how to handle multiplying inequalities by unknowns and the importance of considering both positive and negative values. Steps for finding the range were shared, including plotting the roots of the quadratics on a number line and checking the sign of the expression in each range. The final solution was determined to be (-∞,-4)U(-2,-1), and it was recommended to provide a start-to-finish argument for the method used.
rohan03
I am working through this chapter and trying out exercises - I am stuck on this one- this is what I have done so far but since this is my first experience with this topic I am just not sure what to do next
i have typed it and attached it here with

Thanks

## Homework Statement

I have typed it all as I find it easier to use equations in words

## Homework Equations

Its attached in PDF file

## The Attempt at a Solution

is also attached

Last edited:
It is certainly always the case that, given ##x>0##, it is true that ##-x<0##. More generally, ##x>y \implies -x<-y##.

Note that this means that you cannot always multiply inequalities by unknowns, because you don't know whether the unknowns are positive or negative.

I'm assuming that the factorization is not an issue for you.

No factorising is fine but that makes all positive fraction smaller than zero and that's what I was worried about.

rohan03 said:
No factorising is fine but that makes all positive fraction smaller than zero and that's what I was worried about.
Starting from
$$\frac{-x^2-3x-2}{x^2+7x+12} > 0$$
If you multiply both sides by -1, the inequality sign must change direction.
$$\frac{x^2+3x+2}{x^2+7x+12} < 0$$

I think you are assuming that both fractions are positive, since all the coefficients of the terms are positive. That isn't necessarily true, since x could be negative in value.

rohan03 said:
No factorising is fine but that makes all positive fraction smaller than zero and that's what I was worried about.

If $$\frac{-x^2 - 3x - 2}{x^2 + 7x + 12} > 0,$$ then either both the numerator and denominator are > 0 or both are < 0.

(1) Assuming the numerator and denominator are both > 0 we have $-x^2 - 3x - 2 > 0 \Longrightarrow x^2 + 3x < - 2, \text{ and } x^2 + 7x + 12 > 0 \Longrightarrow x^2 + 7x > -12.$ You can figure out what the values of x must be to have that.

(2) Assuming the numerator and denominator are both < 0 we have $-x^2 - 3x - 2 < 0 \Longrightarrow x^2 + 3x > -2, \text{ and } x^2 + 7x + 12 < 0 \Longrightarrow x^2 + 7x < -12.$ Again, you can figure out what (if any) x are allowed.

RGV

Ray Vickson said:
If $$\frac{-x^2 - 3x - 2}{x^2 + 7x + 12} > 0,$$ then either both the numerator and denominator are > 0 or both are < 0.

(1) Assuming the numerator and denominator are both > 0 we have $-x^2 - 3x - 2 > 0 \Longrightarrow x^2 + 3x < - 2, \text{ and } x^2 + 7x + 12 > 0 \Longrightarrow x^2 + 7x > -12.$ You can figure out what the values of x must be to have that.
I think it's better to leave the three terms on the left side of the inequality, and then factor it, rather than move the constant to the other side.
Ray Vickson said:
(2) Assuming the numerator and denominator are both < 0 we have $-x^2 - 3x - 2 < 0 \Longrightarrow x^2 + 3x > -2, \text{ and } x^2 + 7x + 12 < 0 \Longrightarrow x^2 + 7x < -12.$ Again, you can figure out what (if any) x are allowed.

RGV

Maybe you would like to write the equation like this:-
$$\frac{(x+1)(x+2)}{(x+4)(x+3)}<0$$

Plot the roots of the quadratics on the number line. For example, what sign you get when you substitute a number greater than -3 and less than -2 in x. Make sure you first plot the numbers on the number line. When you substitute -2.5, you get a positive sign. Similarly check what happens when you substitute a number less than -3 and greater than -4. Check for all the possible ranges. If you get a positive sign in a range, put a plus sign in that range on the number line. The range having the negative sign will be your answer.

Hope that helped!

Thank you all. Finding the range is my next step. I will post my finding today and see what I get

after workign out the quadratic roots I get solution set as
{x:(x^(2 )+11x+22)/(x^(2 )+7x+12)>2} = (- ∞,-2) u (-1,∞) is this correct

No.

One free check you can do is look at the value at x=zero. Another is to think about trends to both infinities.

But basically once you had the equation recast into an inequality against zero, you can simply find ranges where the numerator and denominator have the appropriate sign.

Now I am lost!

I think that this should be {x:-2<x<-1}

No. Can you show the inequality you are considering to derive these?

Useful tip: Press "Quote" next to one of the existing messages on the thread if you want to see how to format expressions on this board.

I am typing out my sign table and will reattached my attachment. But any help in between be great!

Joffan said:
No. Can you show the inequality you are considering to derive these?

Useful tip: Press "Quote" next to one of the existing messages on the thread if you want to see how to format expressions on this board.

I pressed quote but can't see any input format!

May be I am no mobile- time to switch pc on

rohan03 said:
May be I am on mobile- time to switch pc on
Yes, that would probably help; easier to review the material already discussed too.

You should have your expression as a fraction consisting of a polynomial in the numerator (top) and another polynomial in the denominator limited by inequality to zero.

Finding the roots of those two polynomials shows you where they change sign.

x (-∞,-4) -4 (-4,-3) -3 (-3,-2) -2 (-2,-1) -1 (-1,∞
(x+2) -ve -ve -ve -ve -ve 0 + + +
(x+1) -ve -ve -ve -ve -ve - 0 +
(x+3) -ve -ve -ve 0 +ve + +
(x+4) -ve 0 +ve -ve +ve + +

gives + 0 - * + 0 - 0 +

Now what? this I think gives me (-∞,-4)U(-2,-1) hope I am right now

sorry I think its (-4,-3)U(-2,-1)

Yes, that's the right range. I recommend that you try to make a start-to-finish argument for how you produced the answer; otherwise in any significant exam you would be losing a lot of marks for method.

do you mean I write out reasoing for all my steps? Yes I would and thank you so much.

## 1. How do I solve inequalities in terms of x?

To solve inequalities in terms of x, you can use the basic rules of algebra. You can add or subtract the same number to both sides of the inequality, multiply or divide both sides by a positive number, or switch the direction of the inequality if you multiply or divide both sides by a negative number.

## 2. Can I solve inequalities in terms of x using graphing?

Yes, you can solve inequalities in terms of x using graphing. You can graph the inequality on a number line and determine the values of x that satisfy the inequality. The solution will be the shaded region on the number line.

## 3. How do I know when to switch the direction of the inequality?

You should switch the direction of the inequality when you multiply or divide both sides by a negative number. This is because multiplying or dividing by a negative number flips the sign of the inequality. For example, if you divide both sides of the inequality by -3, the inequality x > 5 becomes x < -5.

## 4. Are there any special cases when solving inequalities in terms of x?

Yes, there are a few special cases to keep in mind when solving inequalities in terms of x. If you multiply or divide both sides by a variable, you need to consider the sign of the variable. Additionally, when dealing with absolute value inequalities, you may need to set up two separate inequalities and find the intersection of their solutions.

## 5. Can I solve inequalities in terms of x using substitution?

Yes, you can solve inequalities in terms of x using substitution. This involves substituting a value for x and determining if the resulting inequality is true or false. If it is true, then the value is a solution to the inequality. If it is false, then the value is not a solution.

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