Question

Please explain with steps​

Answer 1

$x = \dfrac{1}{2 - \dfrac{1}{2 -\dfrac{1}{2 - x} } }$

In these types of sums, we have to go step by step from the bottom most denominator.

LCM = 2-x.

$x = \dfrac{1}{2 - \dfrac{1}{ \dfrac{4 - 2x - 1}{2 - x} } }$

$x = \dfrac{1}{2 - (1 \div \dfrac{3 - 2x}{2 - x})}$

$x = \dfrac{1}{2 - (1 \times \dfrac{2 - x}{3 - 2x})}$

$x = \dfrac{1}{2 - \dfrac{2 - x}{3 - 2x} }$

LCM = 3 - 2x

$x = \dfrac{1}{ \dfrac{6 - 4x - 2 + x}{3 - 2x} }$

$x = \dfrac{1}{\dfrac{4-3x}{3-2x} }$

$x = 1 \div \dfrac{4 - 3x}{3 - 2x}$

$x = \dfrac{3 - 2x}{4 - 3x}$

By cross multiplication,

$x(4-3x) = 3-2x$

$4x - 3x^{2} = 3 - 2x$

By transposing 4x - 3x² to the RHS (Right Hand Side)

$0 = 3 - 2x - 4x + 3x^{2}$

$0 = 3 - 6x + 3x^{2}$

$0 = 3x^{2} - 6x + 3$

By factorizing using splitting the middle term method,

$0 = 3x^{2} - 3x - 3x + 3$

$0 = 3x(x-1) - 3(x-1)$

$0 = (3x-3)(x-1)$

Therefore,

$0 = 3x - 3$

$0 + 3 = 3x$

$3 = 3x$

$\dfrac{3}{3} = x = 1$

(or)

$0 = x - 1$

$0 + 1 = x$

$1 = x$

Verification :-

substituting x for 1,

$1 = \dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-1} } }$

$1 = \dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{1} } }$

$1 = \dfrac{1}{2-\dfrac{1}{1} }$

$1 = \dfrac{1}{1}$

LHS = RHS

Hence verified.