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# Inequalities trick

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Intern
Joined: 30 Nov 2012
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24 Oct 2016, 19:02
VeritasPrepKarishma wrote:
Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:
doc.jpg

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.

Karishma,
Pls tell me it is always be that signs will change like +-+- or it is possible the case when it will be the same repeats. for example, ++-+ or +--+

thank you
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25 Oct 2016, 02:20
1
1
alesia257 wrote:
Karishma,
Pls tell me it is always be that signs will change like +-+- or it is possible the case when it will be the same repeats. for example, ++-+ or +--+

thank you

Signs will always change in the pattern discussed above. Understand the reason why this is so - check out these posts:

https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/
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Karishma
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25 Oct 2016, 04:18
VeritasPrepKarishma wrote:
alesia257 wrote:
Karishma,
Pls tell me it is always be that signs will change like +-+- or it is possible the case when it will be the same repeats. for example, ++-+ or +--+

thank you

Signs will always change in the pattern discussed above. Understand the reason why this is so - check out these posts:

https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/

So I can put plus in the right segment and then just put -+-+ without trying numbers,arent I ?
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25 Oct 2016, 07:20
1
alesia257 wrote:
VeritasPrepKarishma wrote:
alesia257 wrote:
Karishma,
Pls tell me it is always be that signs will change like +-+- or it is possible the case when it will be the same repeats. for example, ++-+ or +--+

thank you

Signs will always change in the pattern discussed above. Understand the reason why this is so - check out these posts:

https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/

So I can put plus in the right segment and then just put -+-+ without trying numbers,arent I ?

Yes, absolutely but after you bring the factors in the form (ax +/- b)
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Karishma
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07 Nov 2016, 01:01
1
1
VeritasPrepKarishma wrote:
Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression.
e.g. $$(x-4)^2(x - 9)(x+11) < 0$$
We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this:

positive... -11 ... negative ... 9 ... positive

Since we need the region where x is negative, we get -11 < x < 9.
Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression.

I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them.

Responding to a pm:

Quote:
I have a question on this particular thing, wouldn't the range for <0 exclude 4? I understand that it does not change the sign of the graph but it does = 0:
$$(x-4)^2(x - 9)(x+11) < 0$$ if x=4, then the whole thing goes to 0 and it would not be inside the 'valid' range.

Am I correct?

We are given that $$(x-4)^2(x - 9)(x+11)$$ is less than 0.
We need to find the range of values that x can take in that case.
Note that the expression IS LESS THAN 0. This means that it cannot be 0. So x cannot be 4 because that will make the expression 0. So we don't plot 4 on the number line. (x-4)^2 is positive only and hence doesn't affect our signs.
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12 Nov 2016, 00:12
1
sushantarora wrote:
hey ,
can u please tel me the solution for this ques

a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?

ans -11

Check solution as attached
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02 Apr 2017, 01:43
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Thanks a lot for this post , it's very helpful. Would be great if you clarify the below query:

Given:
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

Question: What if we have 2 factors? do we need to start out interpretation as + - + and for 5 factors, do we need to have it in this way? - + - + - +

Uma
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03 Apr 2017, 04:34
1
umabharatigudipalli wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Thanks a lot for this post , it's very helpful. Would be great if you clarify the below query:

Given:
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

Question: What if we have 2 factors? do we need to start out interpretation as + - + and for 5 factors, do we need to have it in this way? - + - + - +

Uma

Yes, that is correct. Usually, n factors will divide the number line into (n+1) regions. You start out by giving + to the rightmost region (provided all your factors are of the form (ax + b) or (ax - b)) and then alternating the signs on the left.

It will be good if you take a look at this post which gives links to explanations on why this happens:
https://gmatclub.com/forum/inequalities ... l#p1753431
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02 Jun 2017, 05:24
Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts

I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?

Thanks
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02 Jun 2017, 05:33
Shiv2016 wrote:
Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts

I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?

Thanks

Hi Shiv2016 ,

The concept is very easy.

1. Find out the zero points.
2. Arrange them in ascending order.
3. Draw them on a number line.
4. Take the right most as +ve and proceed towards left taking alternate signs.
5. If the inequality is of > form, then take all +ve ranges.
6. if the inequality is of lesser form, then take all -ve ranges.

E.g.

(x-2) (x-3 ) > 0

Equality form is greater than(>).

Zero points = 2 and 3

Draw them on number line. You will get 3 ranges. x<2; 2<x<3; and x >2.

Here, right most will be +ve, or x>2 will be +ve.
then 2<x<3 will be -ve
then x<2 will be +ve.

Since inequality is of (>) form, we will take all the ranges which have +ve sign.

Hence, the answer will be x>2 and x<2.

I hope it makes sense.
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02 Jun 2017, 07:46
Shiv2016 wrote:
Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts

I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?

Thanks

The links of the three posts given here should clear all your doubts: https://gmatclub.com/forum/inequalities ... l#p1753431
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02 Jun 2017, 08:32
Shiv2016 wrote:
Hi

I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?

Thanks

Dear Shiv2016,

I'm happy to respond.

I see that abhimahna already gave I wonderful explanation. I will just add a few points.

The post at the beginning of the thread is not 100% accurate and it is way to general for the GMAT. Inequalities with powers of x is a exceeding rare topic on the GMAT, and when there is a power in an inequality, it almost always is just x. I think you could take 50 GMATs in a row and not see a power of x higher than 2 in an inequality.

You may find this blog helpful:

I definitely agree that finding the roots, the x-values that make the function zero, is the first step. Find these, and put these on a number line.

If the highest power is simply 2, then you should know that graph is a parabola. If the coefficient of $$x^2$$ is positive, then the parabola opens upward, like the letter U. If the coefficient of $$x^2$$ is negative, then the parabola opens downward, like an upside-down U. Right there, that should tell you where the function is positive or negative.

Higher powers of x ($$x^3$$, $$x^4$$, etc.) appear infrequently. The alternating pattern doesn't always work: in particular, the signs don't alternate if there's a repeated root, e.g. $$x^3 - 4x^2 + 4x > 0$$). That's a caveat for higher mathematics, but I don't think anyone would ever need to know this for the GMAT.

My friend, part of the reason you were confused is because the person who opened the thread was talking about more advanced math than you need for the GMAT. People who study engineering learn way more math than is needed for the GMAT, and some of them like to show off, but this doesn't really help anyone.

Does all this make sense?
Mike
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01 Dec 2017, 01:08
1
VeritasPrepKarishma wrote:
Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:
doc.jpg

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.

Responding to a pm:
Quote:
hii can u plz elaborate what happens when we have <= or >=, in division as you said the solution will differ.

Note that in case of division, the denominator cannot be 0.

So given $$\frac{(x+2)(x-1)}{(x-7)(x-4)} <= 0$$

We will use the same method to find that the solution will be -2 < x < 1 or 4< x < 7 but here, we have an equal to sign too. So the expression can be equal to 0 too. So x can be equal to -2 and it can be equal to 1 too. But x cannot be 4 and neither can it be 7 since that will make the denominator 0.

So the actual solution here will be -2 <= x <= 1 or 4< x < 7
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04 Jul 2018, 16:46
[quote="gurpreetsingh"]I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

$$f(x) = (x-a)(x-b)(x-c)(x-d) < 0$$

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. $$a<b<c<d$$

So for f(x) < 0 consider "-" curves and the ans is: $$(a < x < b)$$, $$(c < x < d)$$
and for f(x) > 0 consider "+" curves and the ans is: $$(x < a)$$, $$(b < x < c)$$, $$(d < x)$$

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

Hi, if the function is as such (-x-a)(-x-b), would we just flip the direction of the curve, and hence the order of signs? Thanks!
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31 Jul 2018, 14:11
1

Some further clarification on the role of Odd (i.e. x) vs Even (i.e. x^2) powers on the Wavy Curve Method

Thought I'll add some more explanation below for better understanding on this method.

Explanation:

Suppose, if we have "x^3(x-1)<0", then that can be transformed to "x.x^2.(x-1)<0". Notice that "x" and "x^2" work differently here. At "x^2", the line bounces back so the graph doesn't actually change the sign, and hence we can always omit the even powers such as x^2 or (x-1)^2 from the inequality, leading to this: "x(x-1)<0". Using the wavy curve plot, we get 0<x<1.

More examples for transformation needed for Wavy Curve Method:

Example 1:
(x-1)^3 <0 => (x-1).(x-1)^2 => (x-1) < 0 [We ignore even power (x-1)^2 for the reason explained above]
Example 2:
x^2-x^3>0 => x^2(1-x)>0 => (1-x)>0 => x<1 [We ignore even power x^2 for the reason explained above]
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12 Jan 2020, 13:04
Hi,

I have a question.
(1-2x)(1+x)<0 solving by curve method gives x in the range -1 and 1/2 for <0

but (2x-1)(x-1)>0 gives x<-1 and x>1/2.

I am not sure what I am doing wrong. This changes the answer completely.
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24 Jun 2020, 08:57
alesia257 wrote:
VeritasPrepKarishma wrote:
alesia257 wrote:
Karishma,
Pls tell me it is always be that signs will change like +-+- or it is possible the case when it will be the same repeats. for example, ++-+ or +--+

thank you

Signs will always change in the pattern discussed above. Understand the reason why this is so - check out these posts:

https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/

So I can put plus in the right segment and then just put -+-+ without trying numbers,arent I ?

hi

Yes, absolutely but after you bring the factors in the form (ax +/- b)

with reference to your post here https://www.veritasprep.com/blog/2012/0 ... ns-part-i/

there is an inequality with your solution ( see below)

$$-2x^3 + 17x^2 – 30x > 0$$

$$x(-2x^2 + 17x – 30) > 0$$ (taking x common)

$$x(2x – 5)(6 – x) > 0$$ (factoring the quadratic)

$$2x(x – 5/2)(-1)(x – 6) > 0$$ (take 2 common)

$$2(x – 0)(x – 5/2)(x – 6) < 0$$ (multiply both sides by -1)

i dont get how you got $$x(2x – 5)(6 – x) > 0$$

here is what i did

given : $$-2x^3 + 17x^2 – 30x > 0$$

$$x(-2x^2 + 17x – 30) > 0$$

$$x(-2x^2 -12x - 5x – 30) > 0$$

$$x(-2x^2 -12x) - (5x – 30) > 0$$

$$(x) (-2x) (x+6) - 5(x-6)$$

now what do do ? kinda confused

and how you got " - 1 " here ---> $$2x(x – 5/2)(-1)(x – 6) > 0$$

and how you got "x-0 " here --- > $$2(x – 0)(x – 5/2)(x – 6) < 0$$

so i thought i would follow similar pattern

so again given $$-2x^3 + 17x^2 – 30x > 0$$

$$-x(2x^2 - 17x + 30x > 0)$$

$$-x(2x^2-12x-5x+30)>0$$

$$-x((2x^2-12x)-(5x+30))>0$$

$$-x((2x)(x-6)-5(x-6))>0$$

$$(-x)(2x-5)(x-6)>0$$

okay now what to do ? again stuck

thanks
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24 Jun 2020, 22:05
2
dave13 wrote:
alesia257 wrote:
VeritasPrepKarishma wrote:
alesia257 wrote:
Karishma,
Pls tell me it is always be that signs will change like +-+- or it is possible the case when it will be the same repeats. for example, ++-+ or +--+

thank you

Signs will always change in the pattern discussed above. Understand the reason why this is so - check out these posts:

https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/

So I can put plus in the right segment and then just put -+-+ without trying numbers,arent I ?

hi

Yes, absolutely but after you bring the factors in the form (ax +/- b)

with reference to your post here https://www.veritasprep.com/blog/2012/0 ... ns-part-i/

there is an inequality with your solution ( see below)

$$-2x^3 + 17x^2 – 30x > 0$$

$$x(-2x^2 + 17x – 30) > 0$$ (taking x common)

$$x(2x – 5)(6 – x) > 0$$ (factoring the quadratic)

$$2x(x – 5/2)(-1)(x – 6) > 0$$ (take 2 common)

$$2(x – 0)(x – 5/2)(x – 6) < 0$$ (multiply both sides by -1)

i dont get how you got $$x(2x – 5)(6 – x) > 0$$

here is what i did

given : $$-2x^3 + 17x^2 – 30x > 0$$

$$x(-2x^2 + 17x – 30) > 0$$

$$x(-2x^2 -12x - 5x – 30) > 0$$

$$x(-2x^2 -12x) - (5x – 30) > 0$$

$$(x) (-2x) (x+6) - 5(x-6)$$

now what do do ? kinda confused

and how you got " - 1 " here ---> $$2x(x – 5/2)(-1)(x – 6) > 0$$

and how you got "x-0 " here --- > $$2(x – 0)(x – 5/2)(x – 6) < 0$$

so i thought i would follow similar pattern

so again given $$-2x^3 + 17x^2 – 30x > 0$$

$$-x(2x^2 - 17x + 30x > 0)$$

$$-x(2x^2-12x-5x+30)>0$$

$$-x((2x^2-12x)-(5x+30))>0$$

$$-x((2x)(x-6)-5(x-6))>0$$

$$(-x)(2x-5)(x-6)>0$$

okay now what to do ? again stuck

thanks

The step in red is incorrect. It should be

$$-x((2x^2-12x)-(5x-30))>0$$

If you are taking -5 common, the sign inside the bracket becomes - too so that when you open the bracket, you get a + back.

$$-x((2x)(x-6)-5(x-6))>0$$

$$(-x)(2x-5)(x-6)>0$$

Now multiply both sides by -1 to get

$$x(2x-5)(x-6) < 0$$

Now simply plot 0, 5/2 and 6 on the number line and the negative sections are x < 0 and 5/2 < x< 6
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24 Jun 2020, 22:43

one question, if i multiply whole inequity $$(-x)(2x-5)(x-6)>0$$ by -1

then i get $$(x) (-2x+5) (-x+6)<0$$

how is it possible that signs inside of brackets didnt change in your case after multiplying by -1
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24 Jun 2020, 22:56
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dave13 wrote:

one question, if i multiply whole inequity $$(-x)(2x-5)(x-6)>0$$ by -1

then i get $$(x) (-2x+5) (-x+6)<0$$

how is it possible that signs inside of brackets didnt change in your case after multiplying by -1

3 * 5 > 0

Multiply both sides by 2 to get:

2 * 3 * 5 > 0

Does it become 2*3 * 2*5 > 0? No.
They are all factors. When you multiply by 2, one more factor comes in.

You can write it as 2*3 * 5 > 0 or as 3 * 2*5 > 0 but not as 2*3 * 2*5 > 0

Similarly,
(-x) (2x-5)(x-6)>0 has three factors. When you multiply by -1, only one term needs to get multiplied by -1, not all.

You are confusing it with

(3 + 5) > 0
Now if you multiply by 2, you will multiply both terms with 2. This happens only when you have a + or a - sign.
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Re: Inequalities trick   [#permalink] 24 Jun 2020, 22:56

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