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I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.

We want to solve:

$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$

Moving the things in RHS to LHS:

$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$

Writing everything above a common denominator:

$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$

Multiplying both sides with the denominator to cancel the denominator:

$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$

Multiplying the first two factors in every term:

$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$

Simplifying the first factors in every term:

$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$

Multiplying factors again:

$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$

Removing the parenthesis yields:

$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$

Which results in:

$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$ which is not correct. The correct answer is $x = \frac{5}{2}$.

AccidentalFourierTransform
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saner
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    I haven't looked, but probably a simple arithmetic error (sign error most likely). But you made it harder by getting everything on one side. Instead, start by simplifying each side separately. Then cross-multiply. – quasi Aug 10 '17 at 04:47
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    +1, nice question, not because of the mistake you did but because if you find the equation symmetric about 5/2 and substitute x-(5/2)=y, see where it leads – Vikram Aug 10 '17 at 05:53
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    Hint. For you sanity simplify before removing the parenthises in the second to last step. You just so many like terms you are bound to make an error somewhere. – fleablood Aug 10 '17 at 06:00
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    A big tip : simplify LHS and RHS separately, you will greatly reduce the number of steps and number of parenthesis – Vikram Aug 10 '17 at 06:03
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    In the second to last line you have a -12x when it should be a +12x. – fleablood Aug 10 '17 at 06:03
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    So you end up with 24x when it should be 4x. – fleablood Aug 10 '17 at 06:05
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    You can easily check that you just made a mistake in some calculation: http://m.wolframalpha.com/input/?i=%28x-2%29%28x-3%29%28x-4%29+-+%28x-1%29%28x-3%29%28x-4%29+-+%28x-2%29%28x-1%29%28x-4%29+%2B+%28x-1%29%28x-2%29%28x-3%29&x=0&y=0 Now do the same for the following lines to see at which point you get a different result. – Carsten S Aug 10 '17 at 10:32
  • With the help of a CAS, $(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)=4x-10$. –  Aug 11 '17 at 07:14

7 Answers7

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The error is in the third group of parentheses in the "multiplying factors again" step. What you have as $-12x$ should be positive.

symplectomorphic
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    Also, it is easy to see that $2.5$ must be a solution by exploiting the symmetry of the problem: we obviously get a solution if $x-1=4-x$ and $x-2=3-x$, which both say that $2x=5$. – symplectomorphic Aug 10 '17 at 05:31
  • I see how $x−1=4−x$ and $x−2=3−x$ are $2x=5$, but how did you come up with $x−1=4−x$ and $x−2=3−x$? I don't seem to understand the symmetry of the problem. – saner Aug 11 '17 at 02:46
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    @eriksan: your equation has the form $a-b=c-d$. Certainly a sufficient condition for this to hold is to have $a=-d$ and $b=-c$. (It would also be sufficient to have $a=c$ and $b=d$, but you can see that in your problem that is not possible.) Of course, these are not the only two possibilities, so all my comment gives is a simple argument to find that $2.5$ is a solution; you would need another argument to conclude it is the only solution. – symplectomorphic Aug 11 '17 at 02:59
  • @eriksan: please accept an answer to close the question. – symplectomorphic Aug 11 '17 at 15:34
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A hint to make your life much simpler.

Always simplify your variables to start with. Here you can easily make the substitution $x-4 = y$ to get an easier equation in $y$.

Then noting that $\frac 1y - \frac 1{y+1} = \frac 1{y(y+1)}$, your can very easily see that the numerators on both sides when combining the rational expressions is just $1$ (to avoid confusion, note that I transposed the terms on each side to get a positive numerator). Taking the reciprocal on both sides and expanding, you get quadratics on both sides where the square terms immediately cancel, giving you a simple linear equation. Solve for $y$, then add $4$ to get $x$.

Deepak
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    the numerators on both sides when combining the rational expressions is just 1 That should rather be $,-1,$ since the two sides are of the form $,\frac{1}{y+1}-\frac{1}{y},$. – dxiv Aug 10 '17 at 05:09
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    @dxiv I transposed the terms on both sides first to keep both sides' numerators positive. Note the order of the terms in my answer. – Deepak Aug 10 '17 at 05:27
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    I think that's worth spelling out in the answer. I didn't say it's outright wrong, but it may be confusing to the OP who appears to have already missed some clues in the problem itself. – dxiv Aug 10 '17 at 05:32
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    @dxiv Thanks, you're quite right. I've done so. I tend to do these mentally so I neglect to mention every step. It's my failing. – Deepak Aug 10 '17 at 09:47
  • If I substitute $x-4 = y$ in $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$ I get $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{y}$. How do you get to $\frac 1y - \frac 1{y+1} = \frac 1{y(y+1)}$? – saner Aug 11 '17 at 05:14
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    @eriksan $x-4 =y \implies x=y+4$. Now replace every $x$ with $y+4$. – Deepak Aug 11 '17 at 05:38
  • Ingenious. It was a while since I did variable substitution, I got it now! – saner Oct 09 '17 at 16:46
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But as I suggested in my comment, the better approach is to first simplify each side separately . . . \begin{align*} &\frac{1}{x-1} - \frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}\\[4pt] \implies\; &\frac{(x-2)-(x-1)}{(x-1)(x-2)}=\frac{(x-4)-(x-3)}{(x-3)(x-4)}\\[4pt] \implies\; &\frac{-1}{(x-1)(x-2)}=\frac{-1}{(x-3)(x-4)}\\[4pt] \implies\;&(-1)(x-3)(x-4)=(-1)(x-1)(x-2)\\[4pt] \implies\;&(x-3)(x-4)=(x-1)(x-2)\\[4pt] \implies\;&x^2-7x+12=x^2-3x+2\\[4pt] \implies\;&4x=10\\[4pt] \implies\;&x=\frac{5}{2}\\[4pt] \end{align*} Also note, as warned about in Mark Bennet's reply, we have to worry about canceling algebraic factors that might actually be equal to zero. In the steps above, the factors $$(x-1),\;(x-2),\;(x-3),\;(x-4)$$ were canceled in the cross-multiplication step, but that was safe since, based on the original equation, none of those factors had the potential to be zero.

quasi
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    Yes, it's continuing from OP's question of what went wrong. Besides the OP's sign error, the key message here is that the choice of approach was not the best. – quasi Aug 10 '17 at 05:07
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    What went wrong is the choice of approach -- it was a recipe for disaster! – quasi Aug 10 '17 at 05:13
  • No, I'll leave it as it is. – quasi Aug 10 '17 at 05:21
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    Anyone can make a sign error. Identifying where a sign error occurred does not teach the OP much. Showing a simpler, more standard way provides a better learning experience. – quasi Aug 10 '17 at 05:26
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    @quasi: Your answer comes across very oddly. It would read much better if, instead of starting "But as I suggested in my comment..." it started something like "The approach you're taking is bound to cause problems..." – psmears Aug 10 '17 at 11:07
  • It's a style choice. I like the form of my answer, so I'll stay with it. – quasi Aug 10 '17 at 11:10
  • @quasi Is the implication of "cancelling algebraic terms that might actually be equal to zero" that we get a non-answer as in Mark Bennet's example?

    Also, how did you conclude that the factors $(x-1),;(x-2),;(x-3),;(x-4)$ did not have the potential to be zero in the original equation?

     – saner Aug 11 '17 at 06:37
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    @eriksan: They are denominators of the original equation. – quasi Aug 11 '17 at 07:16
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Here is another way, which also illustrates how easy it is to make a mistake with these kinds of problems. If we take the negative fractions to the other side of the equation we get $$\frac 1{x-1}+\frac 1{x-4}=\frac 1 {x-3}+\frac 1{x-2}$$ which becomes $$\frac {2x-5}{(x-1)(x-4)}=\frac {2x-5}{(x-2)(x-3)}$$Now cancel and clear fractions to get $$(x-2)(x-3)=(x-1)(x-4)$$ or $$6=4$$

What went wrong? - Well in cancelling $2x-5$ I didn't check to make sure I was not dividing by zero - so the answer I want is $2x-5=0$ or $x=\frac 52$

Mark Bennet
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The arithmetic error is already pinpointed. The result is a special case of the Theorem below

since we have $\ \overbrace{x\!-\!1\, +\, x\!-\!4}^{\Large a\ \ +\ \ b} \, =\, \overbrace{x\!-\!2\, +\, x\!-\!3}^{\Large c\ \ +\ \ d}\ =\ \overbrace{\color{#c00}{2x\!-\!5}}^{\LARGE\color{#c00}s}\ \ $ and $\ \ \color{#0a0}{\overbrace{{x\!-\!2\neq x\!-\!1,\,x\!-\!4}}^{\Large c\ \neq\ a,\,b\quad\ \ }}\,$

therefore $\quad\ \ \ \dfrac{1}{x\!-\!1}+\dfrac{1}{x\!-\!4} = \dfrac{1}{x\!-\!2}+\dfrac{1}{x\!-\!3}\!\iff\! \color{#c00}{2x\!-\!5} = 0$

Theorem $\ \ $ If $\ a\!+\!b = c\!+\!d =: \color{#c00}s\ $ then $\ \dfrac{1}a+\dfrac{1}b =\dfrac{1}c+\dfrac{1}d \iff \color{#c00}{s = 0}\,$ or $\,\color{#0a0}{c = a}\,$ or $\,\color{#0a0}{c=b}$

$\begin{align}{\bf Proof}\qquad \dfrac{1}a+\dfrac{1}b\, &=\,\dfrac{1}c+\dfrac{1}d\\[.3em] \iff \dfrac{s}{ab}&=\dfrac{s}{cd}\quad{\rm by} \ \ \ \color{#c00}s = a+b = c+d\\[.3em] \iff\, \ \ 0\, &=\, s(ab-cd)\\[.3em] &=\, s(ab-c(a+b-c))\\[.3em] &=\, \color{#c00}s\color{#0a0}{(a-c)(b-c)}\\ \end{align}$

Bill Dubuque
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To avoid errors or pitfalls it can sometimes be helpful to try an initial 'lazy approach', where you still don't drop mathematical circumspection.

When looking at the equation a third degree polynomial can be sought, but that is no fun and we are really hoping that a quadratic equation 'pops-up'. The equation is ugly but perhaps by inverting the LHS and RHS things will simplify. We check that there is no $x$ with both the LHS and RHS equal to $0$ (see next section), so we can safely invert.

For the LHS,

$\frac {1} { \frac {1}{x-1} - \frac {1}{x-2}} = \frac{(x-1)(x-2)}{x-2-(x-1)} = -(x^2-3x+2)$

For the RHS,

$\frac {1} { \frac {1}{x-3} - \frac {1}{x-4}} = \frac{(x-3)(x-4)}{x-4-(x-3)} = -(x^2-7x+12)$

Combining our work,

$\tag 1 x^2 -3x +2 = x^2 -7x +12$

OMG! It is turning into a linear equation!

$\tag 2 4x = 10$

So $x = 2.5$

Examining

$\tag 3 \frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$

the LHS, $\frac{1}{(x-1)} - \frac{1}{(x-2)}$ is zero if and only if $x - 1 = x - 2$, which is silly. So we can forget about looking for any $x$ solutions that would make the LHS and RHS of (3) both equal to 0.

CopyPasteIt
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  • Cool way to reach the solution. Would have never thought about starting with inversion. How do you "check that there is no x with both the LHS and RHS equal to 0" systematically? – saner Oct 09 '17 at 16:59
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There is a shortcut way to evaluate $$p(x):=(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4)\\ - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3).$$

It is clear that this is a polynomial of at most second degree, as all cubic term will cancel out*.

Then we perform the easy evaluations

$$p(1)=-6,\ p(2)=-2,\ p(3)=2,\ p(4)=6$$ and clearly $$p(x)=4x-10.$$

*Actually we can also find the quadratic coefficient to be $$-2-3-4+1+3+4+2+1+4-1-2-3=0,$$ but we needn't use this fact, direct evaluation is anyway required to get the linear terms.

  • Cool, but that's a ninja move I don't understand, could you elaborate on a basic level how it relates to my problem? – saner Oct 09 '17 at 17:03