Question

$\frac{x - 1}{x} + \frac{x}{x - 1} = 4$
solve for x ; x does't =0/1

Answer 1

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$\frac{x - 1}{x} + \frac{x}{x - 1} = 4$
The L.C.M. of x and x-1 is x^2 -x . So,
$\frac{x - 1(x - 1) + x(x - 1)}{ {x}^{2} - x } = 4$
Multiply both sides by x^2 - x
$(x - 1)(x - 1) + {x}^{2} - x = 4( {x}^{2} - x)$
Use the identity (a-b)^2
${x}^{2} - 2x + 1 + {x}^{2} - x = 4 {x}^{2} - 4x$
$2 {x}^{2} - 4 {x}^{2} - 3x + 4x + 1 = 0$
$- 2 {x}^{2} + x + 1 = 0$
The formula to solve a quadratic equation
$a {x}^{2} + bx + c = 0$
is
$x = \frac{ - b + or - \sqrt{ ( {b}^{2} - 4ac)} }{2a}$
So here ,
$x = \frac{ - 1 + or - \sqrt{1 - \: (4 \times - 2 \times 1)} }{ - 4}$
$x = \frac{ - 1 + or - \sqrt{9} }{ - 4}$
Now taking + sign answer is
$x = \frac{ - 1 + 3}{ - 4} \\ x = - 0.5$
Now taking the - sign answer is
$x = \frac{ - 1 -3}{4} \\ x = 1$