Rational Inequality (common mistake)

[@cmilkau]

You can totally multiply by the denominator, you just need to make two cases depending on whether it is positive or negative.

[@lordcezar4657]

I usually just multiply by the square of the denominator as it's always positive. It works for rational Inequalities like this. You'll get a quadratic or in some cases a cubic inequality which you can then solve by graphing

[@adw1z]

True but it will become a mess of cases if the denominators of both sides are different and u have more than 1 term on each side - so in general it won't be great, but for this example it works perfectly with only 2 cases to consider :)

[@pureatheistic]

While your correct, in the most general case, it's more trouble than it's worth. And this video seems to be targeted at first year algebra students, so it's not helpful to confuse the subject. Note he didn't say it's impossible. Only that you should never do so. And in general that's correct. There are very few situations where doing so is better than using simpler ahebraic methods.

[@fangliren]

Nice! Another approach which has helped some of my students would be to separately consider the case where x+2 is positive or negative. Then the multiplication works, as long as you remember to flip the inequality for the negative case!

[@fangliren]

Definitely a valid operation, but it leaves you with a quadratic to solve, which in this particular case happens to be (x+2)^2 >= 0, but in other cases will leave you with quite a lot of work still to do when you have two different roots (and therefore two changes of sign) to consider

[@adw1z]

This is a really good method for a question like this, and sometimes this is the only way to do inequalities! - but it can be cumbersome if u have many terms on both sides all with different unknown denominators and u have a massive mess of cases to consider

[@somenathjana2510]

The way which is told in this video is actually the shortest and easiest to understand for students.I am an 11th grade student and i found it rather intresting when i first learnt about the inequalities.Nice video!

[@AbouTaim-Lille]

We have rarely dealt with inqualities in the classic analysis. Except for the ones we got in the functional analysis like Hölder, Minkowski and Bessel inqualities.

[@malifalitiko495]

Bro i just found your channel and its amazing keep up the good work.

[@chow4444]

3:00 the most underrated laugh

[@pureatheistic]

I subbed just for the energy and smile while doing math. Love it.

[@jimbobago]

We have different definitions of the term "cross multiply". Multiplying both sides of an equality (or inequality) by an expression is *not* what I've called cross multiplying. Starting from a/b = c/d, cross multiplying would be to convert the equation to ad = bc.

[@valentineshades7235]

i think thats what he meant too

[@cmilkau]

better strategy in this case is to add 5/(x+2), leaving you with 1≥0, so equation is satisfied for all x for which it is defined, i e. x≠-2.

[@rimantasri4578]

2:55 that evil laugh 🤣🤣🤣 it's so good

[@Justiin_rm]

Need more series of inequalities.

[@JayTemple]

What I taught is: Look for values where the related equation is true. Look for values where the denominator is 0. Set those values in order and determine whether it's true within each interval. In this case the only intervals are (-inf, -2) and (-2, inf).

[@jonmoore8995]

Really good.

[@Kipngetich-h4q]

wow, that's prime... big up!

[@jackkalver4644]

If you multiply by (x+2)^2, the inequality still applies. Just add the condition that x+2≠0.

[@Abby-hi4sf]

I love his short cut, why long run?

[@AngryEgg6942]

You can cross multiply. However you will need to check 2 options. 1 option is when x+2 is positive and the other is when it’s negative, so when x>-2 or x

[@algirdasltu1389]

Love this channel

[@musluktandokulenkarabatak]

I love your hats too much lol

[@jamesharmon4994]

Thank you for this reminder. My coffee deprived mind made the mistake.

[@SG49478]

I would disagree with your statement to never cross multiply on rational inequalities. On this example here you can avoid it, however there are other problems, where you will have no other choice. You just have to do it the right way and remember that the inequality symbol reverses if you multiply both sides with a negative number. Therefor you have to explore all the possible cases for the term you multiply with becoming greater or less than 0. For this example we see x+2 can not equal 0, means x-2. For x+2 >0, means x>-2 after multiplying both sides with x+2 the inequality symbol does not change means we get x-3>=-5 that means x>=-2 after adding 3 on both sides. Since both of our conditions need to be fulfilled for this case we get x>-2. Second case x+2

[@cmilkau]

very first step, multiply by x+2 turns the inequality around when x+2

[@gamalieljeanbaptiste6462]

I made the same mistake. Breaking the first rule of inequalities gives you wrong answers.

[@sayemsaad8692]

Thank You

[@tamilselvanrascal5956]

❤❤❤🎉🎉🎉

[@tamilselvanrascal5956]

👍

[@KingGisInDaHouse]

The way I have been taught was to simply treat inequality as an equals sign and solve for x and 1/0 points and test the intervals it gives. For example here, (x-3)/(x+2>)=-5/(x+2) Where denominator=0 x+2=0 x=-2 Solving for x ignoring the inquality (x-3)/(x+2)=-5/(x+2) We've already ruled out the denominator equalling zero x-3=-5 x=-5+3=-2 So we need to check before after and the given points which are only x=-2 -2 Intervals to test (-inf,-2),x=-2,(-2,inf) -3 is in (-inf,-2) -6/-1 >= -5/-1 6>5 TRUE x=-2 -5/0>=-5/0 There are no limit signs here so were going to exclude x=-2. (-2,inf) x=3 satisfies this 0>=-5/5 0>=-1 TRUE so x is in (-inf,-2)U(-2,inf)

[@stephenlesliebrown5959]

That is by far the easiest way to do inequalities! 😃 Best wishes to all 🎉

[@williamspostoronnim9845]

Экселенц!

[@stealthgamer4620]

I’m confused, at what point is 1>=0 always true? because how does it ever equal zero when x≠-2. I feel like i’m missing something here.

[@PrimeNewtons]

In logic, when a statement has 'or', you only need one part to be true. For example, "you are dead or alive" is always true.

[@MrBesmir7]

it mean for all of x result is the same 1>=0

[@surjosamanta2190]

Sir,What about 3?

[@surjosamanta2190]

how come 0>= -1

[@Madosatoshist]

Plug in 3 and you have 0 > -1, which is True.

[@bilalghammat8614]

it will come 0>=-1 , which is true.

[@szymonkauza6092]

You have 68 thousand subs and 260 000 likes in 50 min. Wtf yt. THE ALGORYTHM must like you. Wait, 110 views?

[@cynicviper]

I'm assuming it was 26 likes because now it's 27.

[@Madosatoshist]

By now it's 68 likes, 468 views. It's fine.

[@pajthefaj9396]

Hey what is the name of the notation that you used at 6:47

[@PrimeNewtons]

Complement of -2 in the Reals. However, I should have put {} around the -2

[@PatrickAndrewsMacphee]

1>=0 Always true, can be confusing. Maybe better to emphasise that [(1 is >0) OR (1=0)] is true, because one of these inner brackets is always true.

[@cynicviper]

But that is exactly the definition of 1>=0.

[@adw1z]

I don't know if this comment is a joke or not, but I'm pretty sure 1 being greater than 0 is obvious? xD Anyways, a>b ==> a>=b, so u never need to justify equality if it is always strict