If you have the equation $$ \left(\frac{1}{x}\right) + 1 = 0, $$ and you solve it like this $$ \left(\frac{1}{x}\right) + 1 = 0 $$ $$ \left(\frac{1}{x}\right) = -1 $$ $$ -x = 1 $$ $$ x = -1, $$ everything's good. But if you do it like this $$ \left(\frac{1}{x}\right) + 1 = 0 $$ $$ \left(\frac{1}{x}\right) = -1 $$ multiply both sides by x $$ x = -x $$ divide by x $$ 1 = -1, $$ you get nonsense. What is going on? What am I missing here.
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3When you multiply both sides by $x$ , $ \frac{1}{x} * x =1$, not $x$. – GBQT Oct 07 '15 at 16:41
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The incorrect line is "multiply both sides by x: $x=-x$." It should have been $1=-x$. Remember that $(\frac1x)x=\frac xx=1$. – Akiva Weinberger Oct 07 '15 at 16:55
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Is this a joke? – Mark Watson Oct 07 '15 at 17:11
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You've made a computational error.
$$\frac{1}{x}=-1$$
multiply both sides by $x$ gives
$$1=-x$$
Math is still safe to use.
wythagoras
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QuantumEyedea
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There are two things wrong. The first is that you have multiplied $\frac{1}{x}$ by $x$ and got $x$ instead of $1$.
The second is that you have divided by a variable without first determining that the variable is not equal to zero.
Wildcard
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If you have x = -x, the only time when that's true is when x=0, but if x=0, you can't solve the equation by dividing by x. Am I correct? – user265554 Oct 07 '15 at 16:49
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Right, you would solve it by adding x to both sides and then dividing by 2. – Wildcard Oct 07 '15 at 16:57
