Question

Describe the error in solving the equation. 1x+y+1x1=2(x+1)(x1),\frac{1}{x+y}+\frac{1}{x-1}=\frac{2}{(x+1)(x-1)}, (x1)+(x+1)=2,(x − 1) + (x + 1) = 2, 2x=2,2x = 2, x=1x = 1 The solution is x = 1.

Solution

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Answered 1 year ago
Answered 1 year ago

The main error in this solution is that the original equation has not been evaluated. We needed to see what we get if we take x=1x=1 as input value.

We can see that the input value x=1x=1 yields a denominator of zero and any kind of fraction cannot be defined in that situation. This kind of solution is called extraneous solution, and based on that, the original equation has no real solutions.

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