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Bunuel
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May I know is this ok if x=-1 then we cannot multiple both side?
kungfury42
\(\frac{8-x}{x+1} = x\)

\(8-x = x*(x+1)\)
\(8-x = x^2+x\)
\(x^2+2x = 8\)

Therefore, \(x^2+2x-3 = 8-3 = 5\) option D.
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May I know is this ok if x=-1 then we cannot multiple both side?
kungfury42
\(\frac{8-x}{x+1} = x\)

\(8-x = x*(x+1)\)
\(8-x = x^2+x\)
\(x^2+2x = 8\)

Therefore, \(x^2+2x-3 = 8-3 = 5\) option D.

If x = -1, the expression is undefined because the denominator (x + 1) becomes zero. Therefore, (8 − x)/(x + 1) = x does not hold for x = -1.
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