# how to solve 8/x + 4/(x+1) > 10 ?

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Aleksander Krol

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#1

I 've tried, I was looking for it's answer and I couldn't find anywhere to see if my working is right. I've checked on websites, but they all have different answers.

Question: Find the set of values of x:

8/x + 4/(x+1) > 10

I got, x<-1 and -1<x<-0.8 and -0.8<x<-1

Is it right?

My working:

Question: Find the set of values of x:

8/x + 4/(x+1) > 10

I got, x<-1 and -1<x<-0.8 and -0.8<x<-1

Is it right?

My working:

Last edited by Aleksander Krol; 3 months ago

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mqb2766

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#2

(Original post by

I 've tried, I was looking for it's answer and I couldn't find anywhere to see if my working is right. I've checked on websites, but they all have different answers.

Question: Find the set of values of x:

8/x + 4/(x+1) > 10

I got, x<-1 and -1<x<-0.8 and -0.8<x<-1

Is it right?

**Aleksander Krol**)I 've tried, I was looking for it's answer and I couldn't find anywhere to see if my working is right. I've checked on websites, but they all have different answers.

Question: Find the set of values of x:

8/x + 4/(x+1) > 10

I got, x<-1 and -1<x<-0.8 and -0.8<x<-1

Is it right?

You can always sub a few values in to check.

Uploading your working would help.

Last edited by mqb2766; 3 months ago

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Aleksander Krol

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#3

(Original post by

https://www.desmos.com/calculator/6ihhumvber

You can always sub a few values in to check.

Uploading your working would help.

**mqb2766**)https://www.desmos.com/calculator/6ihhumvber

You can always sub a few values in to check.

Uploading your working would help.

and this is the final answer i got, x<-1 and -1<x<-0.8 and -0.8<x<-1

not sure if it's right.

Last edited by Aleksander Krol; 3 months ago

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mqb2766

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#4

(Original post by

I attached images of my workings a few seconds ago

**Aleksander Krol**)I attached images of my workings a few seconds ago

* Id sketch each of the recriprocals and their sum. That should give you an idea of the answer. In fact, it pretty much solves it.

* When multiplying through an equation, squaring it, ... you can get extraneous solutions which do not satisfy the original equation. You have to be careful and check if the final solutions actually satisfy the original equation. For your x<-1, hopefully you can see the left hand side is alway negative and so certainly not > 10?

Why not upload your sketch first?

A common way to not fall into the inequalities "trap" is to replace the inequality with an equality and solve for the critical values, then use the sketch to determine which region.

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Aleksander Krol

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#5

(Original post by

Ok, not read through it carefully, but the question has a few gotchas in you have to be careful about and it appears you've fallen into them? A few general points.

* Id sketch each of the recriprocals and their sum. That should give you an idea of the answer. In fact, it pretty much solves it.

* When multiplying through an equation, squaring it, ... you can get extraneous solutions which do not satisfy the original equation. You have to be careful and check if the final solutions actually satisfy the original equation. For your x<-1, hopefully you can see the left hand side is alway negative and so certainly not > 10?

Why not upload your sketch first?

A common way to not fall into the inequalities "trap" is to replace the inequality with an equality and solve for the critical values, then use the sketch to determine which region.

**mqb2766**)Ok, not read through it carefully, but the question has a few gotchas in you have to be careful about and it appears you've fallen into them? A few general points.

* Id sketch each of the recriprocals and their sum. That should give you an idea of the answer. In fact, it pretty much solves it.

* When multiplying through an equation, squaring it, ... you can get extraneous solutions which do not satisfy the original equation. You have to be careful and check if the final solutions actually satisfy the original equation. For your x<-1, hopefully you can see the left hand side is alway negative and so certainly not > 10?

Why not upload your sketch first?

A common way to not fall into the inequalities "trap" is to replace the inequality with an equality and solve for the critical values, then use the sketch to determine which region.

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mqb2766

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#6

(Original post by

you mean sketching graphs of 8/x and 4/(x+1) ?

**Aleksander Krol**)you mean sketching graphs of 8/x and 4/(x+1) ?

You want to identify the regions which are then > 10.

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#7

(Original post by

Yes and their sum. Note a rough sketch is the key thing, it doesn't have to be perfect. So the left hand side of the inequality.

You want to identify the regions which are then > 10.

**mqb2766**)Yes and their sum. Note a rough sketch is the key thing, it doesn't have to be perfect. So the left hand side of the inequality.

You want to identify the regions which are then > 10.

i shaded the regions which are y>10

but i'm still lost, how do i find the set of vaues of x from the sketch ?

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#8

(Original post by

i did a rough sketch of the two graphs.

i shaded the regions which are y>10

but i'm still lost, how do i find the set of vaues of x from the sketch ?

**Aleksander Krol**)i did a rough sketch of the two graphs.

i shaded the regions which are y>10

but i'm still lost, how do i find the set of vaues of x from the sketch ?

The sketch should show the vertical asymptotes (-1 and 0) and should be clear about the sign of each term either side of the vertical asymptote. It also indicates the 3 regions where the quadratic x(x+1) flips sign. These are important in solving the problem algebraically.

If you're happy with the two regions, then there are two critical values which determine the other end of each interval and these can be solved using the quadratic

8(x+1) + 4x = 10x(x+1)

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#9

(Original post by

What did you get? Im presuming you got two regions, one to the right of (greater than) -1 and the other to the right of (greater than) 0. If nothing else it should show the x<-1 is incorrect and the third inequality you solved -0.8<x<-1 is empty/impossible.

The sketch should show the vertical asymptotes (-1 and 0) and should be clear about the sign of each term either side of the vertical asymptote. It also indicates the 3 regions where the quadratic x(x+1) flips sign. These are important in solving the problem algebraically.

If you're happy with the two regions, then there are two critical values which determine the other end of each interval and these can be solved using the quadratic

8(x+1) + 4x = 10x(x+1)

**mqb2766**)What did you get? Im presuming you got two regions, one to the right of (greater than) -1 and the other to the right of (greater than) 0. If nothing else it should show the x<-1 is incorrect and the third inequality you solved -0.8<x<-1 is empty/impossible.

The sketch should show the vertical asymptotes (-1 and 0) and should be clear about the sign of each term either side of the vertical asymptote. It also indicates the 3 regions where the quadratic x(x+1) flips sign. These are important in solving the problem algebraically.

If you're happy with the two regions, then there are two critical values which determine the other end of each interval and these can be solved using the quadratic

8(x+1) + 4x = 10x(x+1)

Attachment 1041466

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#12

Note for the x intervals > 10, you'd get -1 < x < ? and 0 < x < ? where the ? are the limits to be found using the quadratic.

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old_engineer

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**Aleksander Krol**)

I 've tried, I was looking for it's answer and I couldn't find anywhere to see if my working is right. I've checked on websites, but they all have different answers.

Question: Find the set of values of x:

8/x + 4/(x+1) > 10

I got, x<-1 and -1<x<-0.8 and -0.8<x<-1

Is it right?

1) Re-state the LHS as (12x + 8) / (x(x + 1)).

2) Multiply both sides by x^2(x + 1)^2 (as you did in your working).

3) DO NOT multiply out both sides of the equation. Instead, note that x(x + 1) is a factor common to both sides, and keep it as a factor when you rearrange all terms to one side of the inequality sign.

4) You should quite easily get to something like x(x + 1)(quadratic) < 0. If you now factorise the quadratic, you can identify the four places where the curve crosses the x axis. You know that the expression is a quartic with a positive x^4 coefficient, so it will have a curved "W" shape. You now just need to interpret the axis crossings in the light of the general shape of the curve to identify the regions where (quartic) < 0.

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#14

(Original post by

I'm hesitating slightly before adding anything here, as I fully support the curve sketching and visualisation approach advocated by mqb2766. However, what I will add is that the algebraic approach you've attempted in your first page of working is basically sound, but you've done it in a way that makes it very difficult to proceed any further. My suggestions for making it much easier to continue with this approach are:

1) Re-state the LHS as (12x + 8) / (x(x + 1)).

2) Multiply both sides by x^2(x + 1)^2 (as you did in your working).

3) DO NOT multiply out both sides of the equation. Instead, note that x(x + 1) is a factor common to both sides, and keep it as a factor when you rearrange all terms to one side of the inequality sign.

4) You should quite easily get to something like x(x + 1)(quadratic) < 0. If you now factorise the quadratic, you can identify the four places where the curve crosses the x axis. You know that the expression is a quartic with a positive x^4 coefficient, so it will have a curved "W" shape. You now just need to interpret the axis crossings in the light of the general shape of the curve to identify the regions where (quartic) < 0.

**old_engineer**)I'm hesitating slightly before adding anything here, as I fully support the curve sketching and visualisation approach advocated by mqb2766. However, what I will add is that the algebraic approach you've attempted in your first page of working is basically sound, but you've done it in a way that makes it very difficult to proceed any further. My suggestions for making it much easier to continue with this approach are:

1) Re-state the LHS as (12x + 8) / (x(x + 1)).

2) Multiply both sides by x^2(x + 1)^2 (as you did in your working).

3) DO NOT multiply out both sides of the equation. Instead, note that x(x + 1) is a factor common to both sides, and keep it as a factor when you rearrange all terms to one side of the inequality sign.

4) You should quite easily get to something like x(x + 1)(quadratic) < 0. If you now factorise the quadratic, you can identify the four places where the curve crosses the x axis. You know that the expression is a quartic with a positive x^4 coefficient, so it will have a curved "W" shape. You now just need to interpret the axis crossings in the light of the general shape of the curve to identify the regions where (quartic) < 0.

My reservation about trying to patch the original solution was that the understanding about why the problems were occurring seemed to be missing. Once you understand its essentially a quadratic equality with a couple of vertical asymptotes, its not hard to make any of several approaches work including what the OP originally started with.

Last edited by mqb2766; 3 months ago

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#15

**old_engineer**)

I'm hesitating slightly before adding anything here, as I fully support the curve sketching and visualisation approach advocated by mqb2766. However, what I will add is that the algebraic approach you've attempted in your first page of working is basically sound, but you've done it in a way that makes it very difficult to proceed any further. My suggestions for making it much easier to continue with this approach are:

1) Re-state the LHS as (12x + 8) / (x(x + 1)).

2) Multiply both sides by x^2(x + 1)^2 (as you did in your working).

3) DO NOT multiply out both sides of the equation. Instead, note that x(x + 1) is a factor common to both sides, and keep it as a factor when you rearrange all terms to one side of the inequality sign.

4) You should quite easily get to something like x(x + 1)(quadratic) < 0. If you now factorise the quadratic, you can identify the four places where the curve crosses the x axis. You know that the expression is a quartic with a positive x^4 coefficient, so it will have a curved "W" shape. You now just need to interpret the axis crossings in the light of the general shape of the curve to identify the regions where (quartic) < 0.

(Original post by

Or note the sign of x(x+1) and multply through flipping the inequality as appropriate and rejecting two of the four solutions (solving a pair of quadratic inequalities) which will be invalid in the relevant intervals.

My reservation about trying to patch the original solution was that the understanding about why the problems were occurring seemed to be missing. Once you understand its essentially a quadratic equality with a couple of vertical asymptotes, its not hard to make any of several approaches work including what the OP originally started with.

**mqb2766**)Or note the sign of x(x+1) and multply through flipping the inequality as appropriate and rejecting two of the four solutions (solving a pair of quadratic inequalities) which will be invalid in the relevant intervals.

My reservation about trying to patch the original solution was that the understanding about why the problems were occurring seemed to be missing. Once you understand its essentially a quadratic equality with a couple of vertical asymptotes, its not hard to make any of several approaches work including what the OP originally started with.

Is this right?

I got,

-1<x<-0.8 and

0<x<1

as the final answer.

Last edited by Aleksander Krol; 3 months ago

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#16

If you multply tihrough by the positive expression x^2(x+1)^2 you get

8x(x+1)^2 + 4x^2(x+1) > 10x^2(x+1)^2

x(x+1)[8(x+1) + 4x - 10x(x+1)] > 0

x(x+1)(8+10x)(1-x) > 0

Then its just about reasoning about the sign of each term so we want 0, 2 or 4 +ive terms. A quick sketch of the number line with the 4 roots and their +/- signs on (and hence the number of +ive terms in each interval) gives

-1 < x < -0.8

0 < x < 1

Two of your intervals are too broad (one sided) as is easy to verify by subbing values in - this seems to be mirrored in your sketch. Its easy to make the problem too complex, expanding out the quartic, unless you've got a good idea of what you;re trying to do. Note you;ve edited your post (after this post) and removed these, so this comment is now useless.

If you sketched it (find the vertical asymptotes at 0 and -1), then you also want to solve

(8+10x)(1-x) = 0

giving -0.8 and 1, so ... The sketch should simply note the vertical asymptotes and the sign on each side, as per the previous approach.

If you noted the sign of x(x+1), so negative between -1 and 0, you'd have by multiplying through by x(x+1)

(8+10x)(1-x) > 0 for x<-1 and x>0

(8+10x)(1-x) < 0 for -1<x<0

again giving the two intervals

https://www.desmos.com/calculator/uc1ma8h3il

Personally, I'd say the latter two approaches are simpler as they keep the problem as a quadratic and represent the vertical asymptotes values as interval end points, rather than transforming them into factors and then later analyzing them as interval end points as in the the first approach, though none are that complex which is why I was trying to avoid patching your initial (and latest) solution (too much algebra) and get you to understand the problem.

If you thought about what you've been trying to do, you're multiplying through by each denominator squared in order to make it positive and not flip the inequality. This immediately imples you will always have two common factors (the denominators) which can be directly factorized as these represent the vertical asymptote points and "must" be factors (at least for simple reciprocal terms). The remaining expression is a quadratic which obviously pops up in all three approaches. The main difference is how you represent the vertical asymptotes in the analysis.

8x(x+1)^2 + 4x^2(x+1) > 10x^2(x+1)^2

x(x+1)[8(x+1) + 4x - 10x(x+1)] > 0

x(x+1)(8+10x)(1-x) > 0

Then its just about reasoning about the sign of each term so we want 0, 2 or 4 +ive terms. A quick sketch of the number line with the 4 roots and their +/- signs on (and hence the number of +ive terms in each interval) gives

-1 < x < -0.8

0 < x < 1

Two of your intervals are too broad (one sided) as is easy to verify by subbing values in - this seems to be mirrored in your sketch. Its easy to make the problem too complex, expanding out the quartic, unless you've got a good idea of what you;re trying to do. Note you;ve edited your post (after this post) and removed these, so this comment is now useless.

If you sketched it (find the vertical asymptotes at 0 and -1), then you also want to solve

(8+10x)(1-x) = 0

giving -0.8 and 1, so ... The sketch should simply note the vertical asymptotes and the sign on each side, as per the previous approach.

If you noted the sign of x(x+1), so negative between -1 and 0, you'd have by multiplying through by x(x+1)

(8+10x)(1-x) > 0 for x<-1 and x>0

(8+10x)(1-x) < 0 for -1<x<0

again giving the two intervals

https://www.desmos.com/calculator/uc1ma8h3il

Personally, I'd say the latter two approaches are simpler as they keep the problem as a quadratic and represent the vertical asymptotes values as interval end points, rather than transforming them into factors and then later analyzing them as interval end points as in the the first approach, though none are that complex which is why I was trying to avoid patching your initial (and latest) solution (too much algebra) and get you to understand the problem.

If you thought about what you've been trying to do, you're multiplying through by each denominator squared in order to make it positive and not flip the inequality. This immediately imples you will always have two common factors (the denominators) which can be directly factorized as these represent the vertical asymptote points and "must" be factors (at least for simple reciprocal terms). The remaining expression is a quadratic which obviously pops up in all three approaches. The main difference is how you represent the vertical asymptotes in the analysis.

Last edited by mqb2766; 3 months ago

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