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I am required to find numbers a and b so that: $$\frac{2x+5}{x^2+x-6}=\frac{a}{x+3}+\frac{b}{x-2}$$

$$\frac{ax-2a\:+\:bx+3b}{x^2+x-6}$$

$$\therefore ax-2a\:+\:bx+3b\:=\:2x+5 $$

To this step i understand my process and the rest i believe would be like a simultaneous equation, or I could even use a trial and error method to find the numbers, but I know there is a much simpler way to solve this, what step is next? or have I made a mistake?

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  • $\begingroup$ You have a typo here: $ax\color{red}{+}2a$. It is called partial fraction decomposition. See Example 1 here: en.wikipedia.org/wiki/Partial_fraction_decomposition $\endgroup$
    – farruhota
    May 13, 2018 at 7:48
  • $\begingroup$ @farruhota thanks i fixed them $\endgroup$
    – Utsav
    May 13, 2018 at 7:57
  • $\begingroup$ Sorry, it was the other part: $x\color{red}{-}2$. I edited. $\endgroup$
    – farruhota
    May 13, 2018 at 8:01

3 Answers 3

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$$x^2+x-6=(x+3)(x-2)$$

We have $$a(x-2)+b(x+3)=2x+5$$

Set $x-2=0$ and $x+3=0$ one by one

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  • $\begingroup$ Can you please explain why you set x + 2 and x + 3 equal to 0? I do not understand that $\endgroup$
    – Utsav
    May 13, 2018 at 7:41
  • $\begingroup$ @Utsav, Sorry for the typo. If we set $x-2=0,$ the coefficient of $a$ becomes zero, right? $\endgroup$
    – lab bhattacharjee
    May 13, 2018 at 7:46
  • $\begingroup$ yes but how can we just decide to make x - 2 = 0? if we do this, dont we have to do the same to the other side? $\endgroup$
    – Utsav
    May 13, 2018 at 7:58
  • $\begingroup$ @Utsav - that's exactly what you do - you make $x-2=0$ - that is, make $x=2$. If you let $x=2$, then the equation becomes $5b=9$, easy to solve for $b$. $\endgroup$
    – Glen O
    May 13, 2018 at 13:19
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Here's the most basic method. Group the $x$ coefficients $$ (a+b)x + (3b-2a) = 2x + 5 $$

this needs to be true for $\forall x \in \mathbb R$. Therefore

$$ a + b = 2 $$ $$ 3b - 2a = 5 $$

Solve this system for $a,b$

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  • $\begingroup$ Can you please explain your process as I am having trouble understanding how you came to a + b = 2 and 3b - 2a = 5 $\endgroup$
    – Utsav
    May 13, 2018 at 8:03
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    $\begingroup$ @Utsav if $Ax + B = Cx + D$ must hold for all $x$, then $A = C$ and $B=D$, hence the equations. $\endgroup$
    – Henno Brandsma
    May 13, 2018 at 8:05
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    $\begingroup$ The coefficients in $x$ must be equal on both sides $\endgroup$
    – Dylan
    May 13, 2018 at 8:06
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Substitute $x=0$ to get $\frac{a}{3}+\frac{b}{2}=\frac{-5}{6}$ and $x=-1$ to get $\frac{a}{2}+b=\frac{-1}{2}$ and solve them to get $a=-7, b=3$

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  • $\begingroup$ There may be many values of a and b. How can you say there is only one value satisfying the equation. $\endgroup$
    – Love Invariants
    May 13, 2018 at 7:44
  • $\begingroup$ It is a question of partial fractions and only unique answer exists $\endgroup$
    – Abhishek Choudhary
    May 13, 2018 at 7:44
  • $\begingroup$ Ooops. Didn't notice this. thanks. $\endgroup$
    – Love Invariants
    May 13, 2018 at 7:46
  • $\begingroup$ @LoveInvariants you are welcome $\endgroup$
    – Abhishek Choudhary
    May 13, 2018 at 7:47
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    $\begingroup$ @user477343 MSE is for clarification only and btw you are welcome $\endgroup$
    – Abhishek Choudhary
    May 13, 2018 at 8:18

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