You can totally multiply by the denominator, you just need to make two cases depending on whether it is positive or negative.
@lordcezar46579 ай бұрын
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
@adw1z9 ай бұрын
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 :)
@fangliren9 ай бұрын
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!
@fangliren9 ай бұрын
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
@adw1z9 ай бұрын
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
@somenathjana25109 ай бұрын
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!
@jimbobago9 ай бұрын
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.
@valentineshades72359 ай бұрын
i think thats what he meant too
@AbouTaim-Lille9 ай бұрын
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.
@cmilkau9 ай бұрын
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.
@rimantasri45789 ай бұрын
2:55 that evil laugh 🤣🤣🤣 it's so good
@malifalitiko4959 ай бұрын
Bro i just found your channel and its amazing keep up the good work.
@jackkalver46449 ай бұрын
If you multiply by (x+2)^2, the inequality still applies. Just add the condition that x+2≠0.
@chow44449 ай бұрын
3:00 the most underrated laugh
@Justiin_rm8 ай бұрын
Need more series of inequalities.
@SG494789 ай бұрын
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
@JayTemple9 ай бұрын
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).
@AngryEgg69429 ай бұрын
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
@user-il1iz5si9v4 ай бұрын
wow, that's prime... big up!
@cmilkau9 ай бұрын
very first step, multiply by x+2 turns the inequality around when x+2
@jonmoore89959 ай бұрын
Really good.
@KingGisInDaHouse9 ай бұрын
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)
@stephenlesliebrown59598 ай бұрын
That is by far the easiest way to do inequalities! 😃 Best wishes to all 🎉
@algirdasltu13899 ай бұрын
Love this channel
@sayemsaad86929 ай бұрын
Thank You
@jamesharmon49949 ай бұрын
Thank you for this reminder. My coffee deprived mind made the mistake.
@musluktandokulenkarabatak9 ай бұрын
I love your hats too much lol
@gamalieljeanbaptiste64628 ай бұрын
I made the same mistake. Breaking the first rule of inequalities gives you wrong answers.
@tamilselvanrascal59569 ай бұрын
👍
@tamilselvanrascal59569 ай бұрын
❤❤❤🎉🎉🎉
@williamspostoronnim98459 ай бұрын
Экселенц!
@pajthefaj93969 ай бұрын
Hey what is the name of the notation that you used at 6:47
@PrimeNewtons9 ай бұрын
Complement of -2 in the Reals. However, I should have put {} around the -2
@surjosamanta21909 ай бұрын
Sir,What about 3?
@surjosamanta21909 ай бұрын
how come 0>= -1
@Madosatoshist9 ай бұрын
Plug in 3 and you have 0 > -1, which is True.
@bilalghammat86149 ай бұрын
it will come 0>=-1 , which is true.
@stealthgamer46209 ай бұрын
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.
@PrimeNewtons9 ай бұрын
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.
@MrBesmir77 ай бұрын
it mean for all of x result is the same 1>=0
@szymonkauza60929 ай бұрын
You have 68 thousand subs and 260 000 likes in 50 min. Wtf yt. THE ALGORYTHM must like you. Wait, 110 views?
@cynicviper9 ай бұрын
I'm assuming it was 26 likes because now it's 27.
@Madosatoshist9 ай бұрын
By now it's 68 likes, 468 views. It's fine.
@PatrickAndrewsMacphee9 ай бұрын
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.
@cynicviper9 ай бұрын
But that is exactly the definition of 1>=0.
@adw1z9 ай бұрын
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