8

The following exercise was a resolution to this problem

Let $\displaystyle\frac{2x+5}{(x-3)(x-7)}=\frac{A}{(x-7)}+\frac{B}{(x-3)}\space \forall \space x \in \mathbb{R}$. Find the values for $A$ and $B$

The propose resolution was:

In order to isolate $A$ on the right side, multiply all the equation by $x-7$

$\displaystyle\frac{(2x+5)(x-7)}{(x-3)(x-7)}=\frac{A(x-7)}{(x-7)}+\frac{B(x-7)}{(x-3)}$

Now is my doubt. The resolution suggests that $x-7$ cancel out.

$\displaystyle\frac{(2x+5)}{(x-3)}=A+\frac{B(x-7)}{(x-3)}$

But, $x-7$ can be equal zero. In this situation, is allowed to perform this operation? One can say "for every $x\neq7$", but on the next step the resolution says "for $x=7$ we have".

$\displaystyle\frac{(14+5)}{(7-3)}=A+\frac{B(0)}{(7-3)} \Leftrightarrow A=\frac{19}{4}$

I think there is a contradiction is this resolution.

  • 1
    Because the LHS has the same exact issue at $x = 3$ and $x = 7$. Is that clear? Also, I would multiply and equate the LHS numerator with $A(x-3) + B(x-7)$ and get $A = 19/4$ and $B = -11/4$. – Amzoti Jun 05 '13 at 04:31
  • 1
    Related: http://math.stackexchange.com/questions/165118/how-does-partial-fraction-decomposition-avoid-division-by-zero – lab bhattacharjee Jun 05 '13 at 04:34
  • ya,the number is right. but the word seems not good. why not use $A+B=2,-3A-7B=5$, it is more straightforward. – chenbai Jun 05 '13 at 04:41

3 Answers3

6

The incorrect bit is "$\forall x\in\Bbb R$." We must indeed rule out $x=7$ and $x=3$ in order to avoid problems. What we can still do in that case is take the limit as $x$ approaches $7$ (we can't just plug $x=7$ in, if we've declared that $x\ne 7$), and get the same result.

Cameron Buie
  • 102,994
  • @CameronBlue, if $\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=b$ but $f(a)$ is $\frac00,$ can we say $f(a)=b?$ – lab bhattacharjee Jun 05 '13 at 04:38
  • @lab: No, we can't, but that is irrelevant. $A$ and $B$ are constants, so finding the limit is sufficient for our purposes. – Cameron Buie Jun 05 '13 at 11:16
  • @lab: As a better analogy, suppose we had functions $f$ and $g$ with $$f(x)=t+g(x)$$ for all $x\ne a,$ and we had constants $b,c$ such that $$\lim_{x\to a}f(x)=b$$ and $$\lim_{x\to a}g(x)=c.$$ How could we use this to find the constant $t$ in terms of $b$ and $c$? – Cameron Buie Jun 05 '13 at 17:03
1

If $\frac{2x+5}{x-3}=A+B\frac{x-7}{x-3}$ for all $x\ne 7$ (or $3$), then we can take the limit as $x\to 7$ on both sides to get the given result, and since both sides are continuous, it is equivalent to just plugging in $x=7$.

JLA
  • 6,442
  • 2
  • 24
  • 40
1

I would do it in another way. the equalation is $$\frac {2x+5}{(x-3)(x-7)}=\frac A {x-3}+\frac B {x-7}$$. what we can do is to take the common domniator on the right side getting $$\frac {2x+5}{(x-3)(x-7)}=\frac {A(x-7)+B(x-3)} {(x-7)(x-3)}$$. since the numerator in both sides is condcuted from numbers and x, we can conclude that: $$\begin{cases} A+B=2 & \\ -7A-3B=5 & \ \end{cases} $$. solving this system is relatively simple and by substituting $A=2-B$ in the second equaltion you get $$B=4.75,A=-2.75$$

  • A way this is "streamlined" nowadays (not the way I was taught when I was first learning this) is that your second equation requires that the two sides be equal where defined. So we can just work with $ \ A(x - 7) + B(x - 3) = 2x + 5 \ . $ Setting $ \ x = 3 \ $ produces $ \ A \cdot (-4) = 11 \ $ and using $ \ x = 7 \ $ yields $ \ B \cdot 4 = 19 \ $ . This looks a bit hokey mathematically, but it a "shorthand" for the process of dealing with the limits (that sounds a bit nicer than the frequent justification that "it just works"...). [I now see that this is similar to JLA's approach.] – colormegone Jun 05 '13 at 18:53