is their a simpler way to do these | Mathematics | Page 1

true hamlet May 12, 2026, 11:30 PM

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Im going crazy is their a way to solve more easily than expanding out everything?

sinful raptorBOT May 12, 2026, 11:30 PM

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sinful raptorBOT May 12, 2026, 11:30 PM

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sinful raptor

true hamlet May 12, 2026, 11:30 PM

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i know how to do it i just think the way im doing it is too long

dense temple May 12, 2026, 11:31 PM

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How did you start solving it?

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eplain in broad terms the details are not that important here

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@true hamlet

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$\frac{x^2 - 3}{2x^2 + 12x + 18} - \frac{x^2 - 2x + 1}{x^2 - 9} = \frac{-x - 5}{2x + 6}$

hoary pelicanBOT May 12, 2026, 11:33 PM

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TheVinkler

true hamlet May 13, 2026, 1:27 AM

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dense temple How did you start solving it?

multiply both denomiator a/ b - c/d = a.d - cb / bd

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you still there

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but it gives me 3rd degreee polinomials

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@dense temple

dense temple May 13, 2026, 1:32 AM

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I mean facorize first

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@true hamlet like

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$2x^2+12x+18 = 2(x^2+6x+9)=2(x+3)^2$

hoary pelicanBOT May 13, 2026, 1:33 AM

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TheVinkler

dense temple May 13, 2026, 1:34 AM

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but it will result in a simple (relatively speaking) polynomial

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first you can factorize it in the following way:
$\frac{x^2 - 3}{2(x+3)^2} - \frac{x^2 - 2x + 1}{(x-3)(x+3)} = \frac{-(x + 5)}{2(x+3)}$
and you just go from there

hoary pelicanBOT May 13, 2026, 1:34 AM

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TheVinkler

dense temple May 13, 2026, 1:36 AM

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You need to make a common divisor, which will be $2(x+3)^2(x-3)$

hoary pelicanBOT May 13, 2026, 1:36 AM

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TheVinkler

true hamlet May 13, 2026, 1:40 AM

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dense temple You need to make a common divisor, which will be $2(x+3)^2(x-3)$

there most be a more simpler way

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where the commun divisors align

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its a 1 hour examn with 5 questions like these

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there is no way

dense temple May 13, 2026, 1:41 AM

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It is quite simple

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you do it like this

true hamlet May 13, 2026, 1:41 AM

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you need to expand the brackets

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then multiply the resut by 2

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you are not doing it correctly I beleve

dense temple May 13, 2026, 1:42 AM

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No ofc I am not actually computing that

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but the procces looks like this

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$(x^2 - 3)(x-3) - 2(x^2 - 2x + 1)(x+3) = -(x+5)(x+3)(x-3)$

hoary pelicanBOT May 13, 2026, 1:43 AM

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TheVinkler

true hamlet May 13, 2026, 1:43 AM

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dense temple

but these is for the denominator only

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then you need to multiply the (x^2 -3). (2(x+3)^2(x-3)

dense temple May 13, 2026, 1:44 AM

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No NO NO

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Hell no

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there is a 2(x+3)^2 in the denominator

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so it cancels out

true hamlet May 13, 2026, 1:45 AM

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once you find the comun denominator you need to multiply it by the upper ones

dense temple May 13, 2026, 1:46 AM

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yes but if the denomentor already has it it cancles out

true hamlet May 13, 2026, 1:46 AM

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what?

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you need to do a/ b - c/d = a.d - cb / bd

dense temple May 13, 2026, 1:46 AM

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Yes but why is that

true hamlet May 13, 2026, 1:46 AM

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bd= (2(x+3)^2(x-3)

dense temple May 13, 2026, 1:47 AM

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The reason is the following

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$\frac{a}{b}-\frac{c}{d}\\
\frac{ad}{bd}-\frac{cb}{bd}\\
\frac{ad-bc}{bd}$

hoary pelicanBOT May 13, 2026, 1:48 AM

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TheVinkler

true hamlet May 13, 2026, 1:48 AM

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yeah i get that

dense temple May 13, 2026, 1:48 AM

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ok

true hamlet May 13, 2026, 1:48 AM

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thats litteraly what i told you

dense temple May 13, 2026, 1:48 AM

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so now what we are doing is instead of multiplying the entire bd

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multiplying by 2(x+3)^2(x-3)

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we can take each fraction and multply it by exactly what it needs to become this 2(x+3)^2(x-3)

true hamlet May 13, 2026, 1:50 AM

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dense temple multiplying by 2(x+3)^2(x-3)

but im telling you you need to do (x^2 -3)(x-3)(x+3)

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(x^2 -3)(x-3)(x+3) = ad

dense temple May 13, 2026, 1:51 AM

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$\frac{(x^2 - 3) \cdot {\color{red}(x-3)}}{2(x+3)^2 \cdot {\color{red}(x-3)}}

\frac{(x^2 - 2x + 1) \cdot {\color{red}2(x+3)}}{(x-3)(x+3) \cdot {\color{red}2(x+3)}}
= \frac{(-x - 5) \cdot {\color{red}(x+3)(x-3)}}{2(x+3) \cdot {\color{red}(x+3)(x-3)}}$

hoary pelicanBOT May 13, 2026, 1:51 AM

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TheVinkler

dense temple May 13, 2026, 1:51 AM

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This is what I mean

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@true hamlet

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ok?

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after this all the denotators cancel out and it much much more managble

true hamlet May 13, 2026, 1:53 AM

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hoary pelican **TheVinkler**

why are you multiplying by (x-3) if the full term is (x-3) (x+3)

dense temple May 13, 2026, 1:53 AM

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In the first 2nd or 3rd fractions?

true hamlet May 13, 2026, 1:54 AM

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in the first

dense temple May 13, 2026, 1:54 AM

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Because it already has (x+3)^2

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and 2

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it is only missing the (x-3)

true hamlet May 13, 2026, 1:54 AM

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so you multpily by the missing denominators?

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all terms

dense temple May 13, 2026, 1:55 AM

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exactly,

true hamlet May 13, 2026, 1:55 AM

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then you multpily the upper part by the same part missing

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and then its more managable right

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?

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i got it?

dense temple May 13, 2026, 1:55 AM

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Yes

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Now since they all have the same denominators we can cancel all of them out

true hamlet May 13, 2026, 1:56 AM

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if you cancel all of them out you get to the original point

dense temple May 13, 2026, 1:56 AM

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$\frac{(x^2 - 3)(x-3)}{{\color{red} \cancel{2(x+3)^2(x-3)}}}

\frac{2(x^2 - 2x + 1)(x+3)}{{\color{red} \cancel{2(x+3)^2(x-3)}}}
= \frac{(-x - 5)(x+3)(x-3)}{{\color{red} \cancel{2(x+3)^2(x-3)}}} \
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&\text{Leaving us with:} \
&(x^2 - 3)(x-3) - 2(x^2 - 2x + 1)(x+3) = (-x - 5)(x+3)(x-3)$

true hamlet May 13, 2026, 1:56 AM

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ah okay

hoary pelicanBOT May 13, 2026, 1:56 AM

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TheVinkler
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true hamlet May 13, 2026, 1:56 AM

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i thought you meant between the fracion cancel out

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okay got it

dense temple May 13, 2026, 1:57 AM

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No, in that case you are right and we will get to the orignal equation

true hamlet May 13, 2026, 1:57 AM

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its simpler than i thought

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i just forget that you multply by the missing term instead of the whole denominator

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thank you

dense temple May 13, 2026, 1:57 AM

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My pleasure

true hamlet May 13, 2026, 1:57 AM

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algo

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also

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can i ask you a question

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not related

dense temple May 13, 2026, 1:58 AM

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Sure

true hamlet May 13, 2026, 1:58 AM

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you do engineering right?

dense temple May 13, 2026, 1:58 AM

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Hell no

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why did you guess that?

true hamlet May 13, 2026, 1:58 AM

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idk

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because you are in a math discord

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you do STEM right?

dense temple May 13, 2026, 1:58 AM

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I mean it's fair, but I am a math student currently

dense temple May 13, 2026, 1:58 AM

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true hamlet you do STEM right?

Yep

true hamlet May 13, 2026, 1:59 AM

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is it worth it an ipad for STEM

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because i run out of paper

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constantly

dense temple May 13, 2026, 2:00 AM

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I personally love pen and paper, but a lot of people do prefer the ipads...
I suggest you buy like a 2nd hand from someone else to try it for it bit, see if it works for you

true hamlet May 13, 2026, 2:00 AM

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okay

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thank you

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im majoring in computer science

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idk how hard it will be

dense temple May 13, 2026, 2:01 AM

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My pleasure, feel free to dm me if you have any other math related question

true hamlet May 13, 2026, 2:01 AM

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because im practicing a lot

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but they are always new techniques being use

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in each excercise

dense temple May 13, 2026, 2:02 AM

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Computer science is great

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It has a lot of variety

true hamlet May 13, 2026, 2:02 AM

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dense temple It has a lot of variety

i like math

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but its not similar to high school

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in anyway

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its much harder

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well thats it

dense temple May 13, 2026, 2:03 AM

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Glad I could help, your answer for your question should be -21

true hamlet May 13, 2026, 2:04 AM

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well thank you

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bye bye

dense temple May 13, 2026, 2:04 AM

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Cya around

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just close this post

sinful raptorBOT May 14, 2026, 3:00 AM

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@true hamlet

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#is their a simpler way to do these