Solve for $x$: $\dfrac{1}{x} + \dfrac{1}{x + 2} = 1$

Jhun Vert's picture

Determine the value of x from the following equation:
$\dfrac{1}{x} + \dfrac{1}{x + 2} = 1$

esmilitar's picture

X= the square root of 2
Simplify the given equation
1/x + 1/(x+2)=1
=>(( x+2) + x )/(x (x+2)) =1
=>(2x+2)/x(x+2)=x(x+2)/x(x+2)
=>2x +2=x^2+2x
=>x^2+2x-2x=2
=>x^2=2
Thus X= the square root of 2 ans.
To verify the answer substitute the value of x=the square root 2 from the given equation if it satisfies equals to 1
Using the calculator by typing the equation using Alpha X and Shift solve it will gives you an answer x= 1.414213562 which is the value of square root of 2.

(1/x) + (1/(x+1)) = 1 multiply by (x * (x+1))
(X+1) + x = x(x+1)
2x + 1 = ×^2 + x
x^2 - x - 1 = 0
x^2 - x = 1
x^2 - x + (1/2)^2 = 1 + (1/2)^2
(x - (1/2))^2 = ( 5/4 )
x - (1/2) = sqrt ( 5/4 )
x = (1/2) +- ((sqrt( 5 ))/2)

X1 = (1 + sqrt ( 5 ))/2
X2 = (1 - sqrt ( 5))/2

Jonas Cayanan1's picture

+sqrt2 , -sqrt2

Infinitesimal's picture

$$ \begin{eqnarray}
\frac{1}{x} + \frac{1}{x+2} &=& 1\\
(x + 2) + x &=& x(x+2) \\
2x + 2 &=& x^2 + 2x\\
x^2 &=& 2\\
x &=& \boxed{\pm \sqrt{2}}
\end{eqnarray} $$

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