Rational Equations why aren't both conditions true

Question

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 $x$.

Answer 1

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).

Answer 2

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.