megannicole51
  • megannicole51
solve the integral ((x)/(x^2-9x+20))dx I solved it all the way to the bottom and now i am at x(A+B)+A(-4)+B(-5)=x I was told that A+B can only equal 0 when the number after the equal sign (in this case "x" cannot be a variable). So i tried solving it out anyways and got A=x and B=-x. Then I plugged it into the original ((A)/(x-5))+((B)/(x-4)) and it doesn't work. @pgpilot326
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
is this partial fractions?
anonymous
  • anonymous
\[x^2-9x+20=(x-4)(x-5)\] \[\frac{x}{(x-4)(x-5)}=\frac{A}{x-4}+\frac{b}{x-5}\] there is a real quick snappy way to solve
anonymous
  • anonymous
that is what you are looking for, right?

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anonymous
  • anonymous
A and B are numbers, not variables. one way to find A is this take \[\frac{x}{(x-4)(x-5)}\] put your hand over the part of the denominator that is the denominator for A \[\frac{x}{\cancel{(x-4)}(x-5)}\] replace \(x\) by \(4\) and get \[\frac{4}{4-5}=-4\]
megannicole51
  • megannicole51
yes it is....sorry my laptop died
anonymous
  • anonymous
ok so now we have \[\frac{-4}{x-4}\] for the A part
anonymous
  • anonymous
\[ \frac{x}{x^{2}-9x+20}=\frac{A}{x-5}+\frac{B}{x-4} \Rightarrow A(x-4)+B(x-5)=x\]
megannicole51
  • megannicole51
yup ive got that....then i have x(A+B)+A(-4)+B(-5)=x
anonymous
  • anonymous
for B repeat and get \[\frac{x}{(x-4)\cancel{(x-5)}}\]
anonymous
  • anonymous
you get 2 equations in 2 unknowns... this is a longer way as it should yield the same thing @satellite73 was saying.
anonymous
  • anonymous
put in \(x=5\) to get \[\frac{5}{5-4}=5\]
megannicole51
  • megannicole51
@satellite73 i dont understand why you are crossing them out of the denominator
anonymous
  • anonymous
ok lets go back to the beginning i was just showing off a snappy trick
megannicole51
  • megannicole51
so @pgpilot326 what do i do after..x(A+B)+A(-4)+B(-5)=x
anonymous
  • anonymous
\[A(x-5)+B(x-4)=x\] replace \(x\) by \(5\)
anonymous
  • anonymous
@satellite73 you show off...
anonymous
  • anonymous
you get \[B(5-4)=5\] sp \[B=5\]
megannicole51
  • megannicole51
i dont need to go to the beginning i just need to know what to do after x(A+B)+A(-4)+B(-5)=x please:)
anonymous
  • anonymous
that is not correct though
anonymous
  • anonymous
like i said... you get 2 equations in 2 unknowns... A + B = 1 -4A-5B=0 solve the system
anonymous
  • anonymous
\[\frac{x}{x^{2}-9x+20}=\frac{A}{x-5}+\frac{B}{x-4}\] so \[A(x-4)+B(x-5)=x\] for the numerator now this has to be true for all \(x\) in particular it must be true of \(x=5\)
megannicole51
  • megannicole51
why is A+B=1?
anonymous
  • anonymous
Ax+Bx=x so this is the same as A+B=1
anonymous
  • anonymous
i will shut up except to say this is a long and hard way to do it
anonymous
  • anonymous
i know
anonymous
  • anonymous
would you like to do it the easy way? without solving a system of equations?
anonymous
  • anonymous
the main thing is that @megannicole51 understand and can do it herself... it doesn't matter to me
megannicole51
  • megannicole51
i just need to know how to do it after x(A+B)+A(-4)+B(-5)=x.....ive been doing them like this all day so i dont want to get confused
anonymous
  • anonymous
ok then as @pgpilot326 said, this means \(A+B=1\) since \((A+B)x=1x\) and also that \(-4A-5B=0\) since there is no constant on the right hand side
megannicole51
  • megannicole51
so if there is no constant A+B will always equal 1 by default?
anonymous
  • anonymous
x terms with x terms and constant terms with constant terms...
anonymous
  • anonymous
no it is because \(x=(A+B)x\)
anonymous
  • anonymous
if you had for example \(3x\) on the right, then \(A+B=3\)
megannicole51
  • megannicole51
oooooh that makes sense
megannicole51
  • megannicole51
okay let me see if i can finish it by myself
anonymous
  • anonymous
ok, then if you like, i can show you how to do it in your head without writing anything down
megannicole51
  • megannicole51
let me do it real fast then okay!
anonymous
  • anonymous
k
anonymous
  • anonymous
you're in good hands @megannicole51 , I'll leave you be
megannicole51
  • megannicole51
thank you for your help ill message you if i need help! :D
megannicole51
  • megannicole51
so why does -4A-5B=0?
anonymous
  • anonymous
because A and B are numbers so \(-4 A-5B\) is a number (constant) but there is no constant in the numerator, only \(x\)
megannicole51
  • megannicole51
oh okay
anonymous
  • anonymous
if the numerator was \(x+7\) then \(-4A-5B=7\)
megannicole51
  • megannicole51
gotcha!
megannicole51
  • megannicole51
okay so in order to solve for A and B i just do simple algebra and substitute correct?
anonymous
  • anonymous
you solve the system however you like elimination or in this case substitution which will probably be easier
megannicole51
  • megannicole51
so i have -4A+5(4A/5)=0....am i on the right track?
megannicole51
  • megannicole51
im doing substitution
anonymous
  • anonymous
no i don't think so \[A+B=1\iff B=1-A\] substitute in to \[-4A-5B=0\] and solve for \(A\)
anonymous
  • anonymous
clear or no?
megannicole51
  • megannicole51
yeah it is hold on
megannicole51
  • megannicole51
im having pencil issues lol
anonymous
  • anonymous
k
megannicole51
  • megannicole51
A=5 B=-4
anonymous
  • anonymous
whew!!
megannicole51
  • megannicole51
yay!
anonymous
  • anonymous
you see how long that took? all that work? multiply out, solve a system yechh
megannicole51
  • megannicole51
well i have to do it for tests and quizzes and homework :/
anonymous
  • anonymous
then lets to it the snappy way because this is ridiculous
megannicole51
  • megannicole51
hahah okay show me!
anonymous
  • anonymous
do you have this \[\frac{x}{x^{2}-9x+20}=\frac{A}{x-4}+\frac{B}{x-5}\] somewhere on your paper?
megannicole51
  • megannicole51
yes
megannicole51
  • megannicole51
no
megannicole51
  • megannicole51
my 4 and 5 are switched but it doesnt really matter does it?
anonymous
  • anonymous
oh ok is it switched? just want to make sure so we don't screw up
megannicole51
  • megannicole51
lol yeah bt keep goin!
anonymous
  • anonymous
\[\frac{x}{x^{2}-9x+20}=\frac{A}{x-5}+\frac{B}{x-4}\]
megannicole51
  • megannicole51
:) good
anonymous
  • anonymous
so also on your paper you have \[A(x-4)+B(x-5)=x\] right?
megannicole51
  • megannicole51
yup
anonymous
  • anonymous
ok now here is the deal: \[A(x-4)+B(x-5)=x\] must be true NO MATTER WHAT \(x\) IS !
anonymous
  • anonymous
in particalular it must be true if \(x=5\) and if \(x=5\) you get \[A(5-4)+B(5-5)=5\] i.e. \[A=5\]
anonymous
  • anonymous
similarly it must be true if \(x=4\) which, skipping the stupid step of writing \(4-4\) we go right two \[B(4-5)=4\] or \(-B=4\) and so \(B=-4\) done finished no system etc
megannicole51
  • megannicole51
yeah thats how you do it when there isnt a coefficient after the equal sign
megannicole51
  • megannicole51
dang! i like it:)
megannicole51
  • megannicole51
thank you for showing me!
anonymous
  • anonymous
and there is an even easier way, which amounts to the same thing
megannicole51
  • megannicole51
i love math lol
anonymous
  • anonymous
\[\frac{x}{(x-5)(x-4)}=\frac{A}{x-5}+\frac{B}{x-4}\]
megannicole51
  • megannicole51
just find the zeros?
anonymous
  • anonymous
find \(A\) the denominator of \(A\) is \(x-5\) so replace \(x\) by \(5\) \[\frac{x}{x-4}\]
anonymous
  • anonymous
that is what i meant by this \[\frac{x}{\cancel{(x-5)}(x-4)}\] put your hand over the denominator of \(A\) and then replace \(x\) by \(5\)
anonymous
  • anonymous
you get \(A=5\) in your head
anonymous
  • anonymous
much much easier than multiplying out, collecting terms, equating coefficients, solving linear systems, it makes me exhausted just to think about it
megannicole51
  • megannicole51
lol agreed but thts wat i love about math
anonymous
  • anonymous
glad you love it too btw the method i just showed obviously doesn't work all the time for example if you are going to end up with something like \[\frac{2x+1}{x^2+1}+\frac{x-2}{x^2+7}\] however you should always check to see if you can plug in anything for \(x\) to drop out all terms but one, making the solution to one of the constants simple
megannicole51
  • megannicole51
oh okay! good to know! thank you for your help!!!
anonymous
  • anonymous
yw

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