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Hello, can somebody help me out with this problem?
(x)/(x-2)+(x-1)/(x+1)=1

User EricR
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1 Answer

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\frac x{x-2}+(x-1)/(x+1)=1


\frac x{x-2}\cdot(x+1)/(x+1)+(x-1)/(x+1)\cdot(x-2)/(x-2)=1

So long as
x\\eq-1 and
x\\eq2, we can carry out the manipulation above. Then


(x(x+1))/((x-2)(x+1))+((x-1)(x-2))/((x+1)(x-2))=1


(x(x+1)+(x-1)(x-2))/((x-2)(x+1))=1


(x^2+x+x^2-3x+2)/(x^2-x-2)=1


(2x^2-2x+2)/(x^2-x-2)=1


(2x^2-2x+2)/(x^2-x-2)\cdot(x^2-x-2)=1\cdot(x^2-x-2)


2x^2-2x+2=x^2-x-2


x^2-x+4=0

We can complete the square to solve:



x^2-x+\frac14+\frac{15}4=0


\left(x-\frac12\right)^2=-\frac{15}4

However,
y^2\ge0 for all (real) values of
y, which means there is no real solution to this equation.

If you're solving over the complex numbers, we can take the square root of both sides to get


x-\frac12=\pm i\frac{√(15)}2


\implies x=\frac{1\pm i√(15)}2
User Ryan Hill
by
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