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If this is true : $(x - 1) - x = x - (x+1)$ Why isn't $$\frac{2}{x-1}-\frac{2}{x}=\frac{2}{x}-\frac{2}{x+1}$$ also true?

  • Flew in a plane in still air to a place $900$ km away. On the return flight, a tailwind increased the speed by $45$ km/h. The return trip was $1.5$ h less than the flight to the camp. Determine the speed of the plane on the return.

I did $\frac{900}{x}-\frac{900}{x+45}=\frac{3}{2}$ but the solution says the correct to be $\frac{900}{x-45}-\frac{900}{x}=\frac{3}{2}$. I don't understand what difference it makes which speed we make the variable $x$.

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    Because $\frac{1}{a+b}\neq\frac{1}{a}+\frac{1}{b}$ in general. – 2017-01-24
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    Why should it be true? Division is not the same as subtraction. That's like saying "a cat has four legs" is true, why isn't "a person has four legs" true? – 2017-01-24
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    Plus, if $x\in\{-1,0,1\}$, then $(x-1)-x=x-(x+1)$ is true but one of the fractions $\frac{1}{x-1}$, $\frac{1}{x}$ or $\frac{1}{x+1}$ won't even make sense... – 2017-01-24
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    The numerators are equal. The denominators are not. – 2017-01-24
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    Because the reciprocal function is not a homomorphism from the *additive* group $\mathbf R$ into itself. – 2017-01-24
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    @Bernard I'm sure someone who hasn't noticed that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$ isn't true will understand your comment. – 2017-01-24
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    @NeedForHelp: I know. It was just for fun. I'm from a generation which learnt some basic notions of group theory in high school. – 2017-01-25

2 Answers 2

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For the first question:

We have: $$(x-1)-x=x-(x+1)$$ This is therefore true (power of $-1$ on both sides): $$\frac{1}{(x-1)-x}=\frac{1}{x-(x+1)}$$ Hence, this is also true (multiplying both sides by $2$): $$\frac{2}{(x-1)-x}=\frac{2}{x-(x+1)}$$ However, you cannot seperate the denominator as follows: $$\frac{2}{(x-1)-x}\neq \frac{2}{x-1}-\frac{2}{x}$$ You can verify that this does not work with a counter-example. For example, try $x=6$: $$\frac{2}{(6-1)-6}\neq \frac{2}{6-1}-\frac{2}{6}$$ $$-2 \neq \frac{1}{15}$$ Or, you can try it with $x=0$, and realise that $\frac{2}{0}$ is undefined.

For the second question: It does not matter whichever speed you define as $x$. For example, I deduce that from the answer you were given that they've denoted $x$ as the speed with the tailwind (return trip). You've denoted $x$ as the speed without the tailwind (on the forward trip). Therefore, when you solve for $x$ with your equation, you will get the speed without the tailwind. From that value, you can solve for the other speed (with tailwind).

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    In a word problem such as - flew in a plane in still air to a place 900 km away. On the return flight, a tailwind increased the speed by 45 km/h. The return trip was 1.5 h less than the flight to the camp. Determine the speed of the plane on the return. How would we know which variable to make x since 900/x - 900/x+45 = 3/2 would be different of 900/x-45 - 900/x = 3/2 – 2017-01-24
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    @padapa I'll add this to my answer too. – 2017-01-24
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It doesn't matter whether $x$ stands for the starting speed, or the returning speed of the plane. Pick one, be clear about what $x$ means, and use it consistently in your work.

You are defining $x$ to be the starting speed of the plane. When you solve your equation, the return speed will be $x+45$, which you will report as your answer to the problem.

If you use the approach outlined in the solution, then $x$ is defined as the returning speed of the plane, and you will report $x$ as the answer to the problem.

The answer should be the same under both approaches. Both approaches are correct. But this doesn't mean you can equate the two expressions $\frac{900}{x}-\frac{900}{x+45}$ and $\frac{900}{x-45}-\frac{900}{x}$, since $x$ has a different meaning in the two approaches.