how to solve 1,2 questions?

[eternal ridge]

q1=A
q2=B
q3=B

[sinful arrow]

ignore question3. just 1,2 how to know what should i use x < 4 and x >= 4 or both

can you explain please

[eternal ridge]

f(4-p) and p >0 ==> 4-p<4

and you should use x+1/x-4 because 4-p<4

[sinful arrow]

how to know p value?

is it 1 or 2 or 3?

[eternal ridge]

question

its p>0

and it can be 1 , 2 , 3 , .... any way Whichever it is 4-p is always is smaller than 4

do you understand ?

[sinful arrow]

not really, i understood that 4-p is greater than 0, but why it couldn't be 3,4,etc... how can i know

[eternal ridge]

there is no way to find p

but from Question you solve the problem

for etc q1 doesnt want to find p

or value of

The question gives you a hypothesis (here p>0)

[sinful arrow]

it just mentioned that p > 0. I just dont know why we used the x+1/x-4

[eternal ridge]

because if p>0 always ==> 4-p is smaller than 4

[sinful arrow]

how to know if 4-p is smaller or greater than 4

[eternal ridge]

Inequality

[sinful arrow]

i can say that 4-p is 4-0 then its gonna be 4 so i use the second method?

what makes p letting me use the first method

[eternal ridge]

p>0 ==>.-1 ==> -p<0 ==>+4 ==> -p+4<4

just an Inequality

[eternal ridge]

yes

but not 0-4

[sinful arrow]

hold on i feel i understood it. I'll solve it and show you

[eternal ridge]

-infinity to 4

[sinful arrow]

can you solve it all for q1

i tried didn't get it correctly

[eternal ridge]

at the end The numerator and denominator of the fraction multiplied by -1

[sinful arrow]

why didn't use the other method. there's no value for p at all?

[eternal ridge]

no

[sinful arrow]

whats the difference between q1,2 cuz q2 is 4+p but diff answer

[eternal ridge]

we just know p>0

[eternal ridge]

yes q2 you should use The second rule

x+1

[sinful arrow]

why

[eternal ridge]

cause 4+p > 4

The solution to this question is the same as the previous question

[sinful arrow]

if i says p =1 cuz its p > 0. p+4> 4 = 5 > 4 so i use the second method?

[eternal ridge]

yes

[sinful arrow]

for the second question why i cant say the same and use the second or cuz its minus ? so it will never be 4 or greater

4-p > 4 anything i put for p will not be 4 or greater still?

[eternal ridge]

4-p isnt bigger than 4

[sinful arrow]

cuz its minus?

so it will never be bigger than 4?

or even 4?

[eternal ridge]

its just an Inequality

[eternal ridge]

you need to solve like this

[sinful arrow]

why you put 4 out of nowhere?

[eternal ridge]

do you know Inequality
?

[sinful arrow]

yes

[eternal ridge]

when you have an Inequality you can + - . to both sides

[sinful arrow]

like this?

[eternal ridge]

you know your problem is

when you mulitmultiplied

wehn you mulitmultiplied by Negative numbers like -1 you should change direction

for etc

x>4

mulitmultiplied by -1 ==> -x<-4

[sinful arrow]

i did change the direction in the last step

[eternal ridge]

thats wrong

[sinful arrow]

whats the right then 😭

[eternal ridge]

[sinful arrow]

its easy asf why cant my brain understand this shit

💀

[sinful arrow]

how can i solve it ugh

[sinful arrow]

bro please i dont want to use this out of nowhere which is x-1 i hate this. i prefer to solve from what given as i wrote in that pic. i just dont know how to fix my mistakes

[eternal ridge]

that mulitmultiplied

notx

[sinful arrow]

why did u use p > 0 . why didn't use 4-p first

[eternal ridge]

always we should start with Assumption of the question

here is p>0 than...

[sinful arrow]

why didn't u multiply by -1 why not 1

[eternal ridge]

becuase Question need -p not p

on q 1

4-p

[sinful arrow]

yes q1

for q2 is there a difference if i wrote p+5 or 5+p as long as its + not minus?

cuz there's no p+5 in the answers

so if there's p+5

p+5 same as 5+p?

[sinful arrow]

wym +?

all the same?

[eternal ridge]

yes

[sinful arrow]

tysm

[eternal ridge]

❤️

good luck and try more at math ❤️

[sinful arrow]

tysm!

that's actually one of the easiest questions i have. just got confused a lilttle lmao

[zealous shuttle]

q1. f(4-p) -> 4-p is x. -p is always smaller so find which one is fulfilling the condition (smaller than or equal to), in this case it’s x<4 so you’ll use f(x)=(x+1)/(x-4) replace every x with 4-p, you’ll get it solved (but it results in -p, so you’ll have to change it to positive since p>0).

q2. is slightly different than q1, so I’ll let you try it :) (remember +p is always greater so find the greater fulfilling condition)

[sinful arrow]

ur genius

[ivory jewel]

.solved