How to really solve a rational inequality?

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[UnbreakablePickaxe]

you can multiply the equation by x+3 and consider two cases where in case 1, x+3 is positive and in case 2, x+3 is negative. In case of negative you can flip the sign of inequality.

[nikhilnagaria]

Simply because the sign of the quantity you multiply by in an inequality matters. We could instead multiply by (x+3)^2 which is always positive so the inequality becomes quadratic, and has the same solutions as the given rational inequality.

[Usiek]

There is a much simpler method for such inequalities. Just treat them as fractions a/b (where b≠0). When is this fraction greater than zero? When product of a and b is greater than zero.

Therefore we can just write (x-2)(x+3)≥0, keeping in mind that x≠-3. Then proceed like in quadratic or polynomial inequalities to get the same result.

[cocolcool]

I’m in my first year of a math major and i have literally never seen it done in this way, very cool

[tomatrix]

It basically that you can’t multiply or divide by a negative without changing the inequality (making the current one untrue), so let’s say we can only multiply by x+3 when it is positive, which means x > -3 and as u did we get x>=2 which means for all x > -3 ONLY x>= 2 work, i.e. everything inbetween does not, we must now check when x-3 is negative which is when x < -3 which it is then clear that the numerator is negative for akk x < -3 and thus -/- = + which satisfies inequality so all x < -3 work too.

[fizixx]

Didn't know you had this channel, so I subscribed. Love these kinds of problems, but I did not know about this kind of difference.

[michaelempeigne]

I rather assume x + 3 > 0 and work out the inequality. Then repeat for x + 3 < 0

if x + 3 > 0, ( x + 3 ) * [ ( 2x + 1 ) / ( x + 3 ) ] > or = 1 ( x + 3 )
2x + 1 > or = x + 3
x > or = 2
and x > - 3

if x + 3 < 0, ( x + 3 ) * [ ( 2x + 1 ) / ( x + 3 ) ] < or = 1 * ( x + 3 )
2x + 1 < or = x + 3
x < or = 2
and x < - 3

so x belongs to ( - inf, - 3 ) U [ 2, inf )

[theepicfailguy]

My number one tip while solving linear inequality
If it's positive like a whole square then you can cross multiply it

[ZipplyZane]

What I've always wondered is if there is a way to solve these without test points. It always seemed so inelegant.

[vatrqx]

Thank you so much for all the content you provided 🖤❤.
but can I ask you to do a video about absolute function 🙏, i mean finding limit/solve and some examples
I feel selfish to ask that so if you feel like doing it I will appreciate it so much 🙂

[mathmancalc]

This could have been done on your fast channel, Quick Attack Style.
In the first one, x cannot equal 3. In the second one, x can equal 3. Therefore, they are not the same. I'm done in like 5 seconds.

[ArtemisPlaysGamez]

you can multiply both sides by (x+3)^2, then you will be able to solve it like a quadratic inequality

[shenal]

Can you integrate square root of cos2x/cosx

[graeme]

Thanks I'm writing admaths in a week and this is probably going to save me an extra 5 marks

[Nylspider]

My thoughts before watching the video: so, I think the answer is no since x+3 hypothetically could be negative depending on the value of x, meaning they describe different inequalities, I think

Edit: ok

[Firefly]

I thought you do these inequalities by considering x<0 and x>0. Then flip the sign when doing x<0

[alberteinstein]

My gut is telling me no because why would math be that easy?

[alberteinstein]

Who else comments while they’re watching the video as opposed to after watching the video?

[aswinr]

Just try putting x = -4



You will see that they are not the same