Question

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Answer 1

$\Large{\underline{\underline{\text{Question:}}}}$

$\rm If \ \dfrac{37}{13} = 2 + \dfrac{1}{x+\dfrac{1}{y+\frac{1}{z}}}, \ find \ the \ value \ of \ z.$

$\Large{\underline{\underline{\text{Explanation:}}}}$

Consider the given expression,

$\rm\implies \dfrac{37}{13} = 2 + \dfrac{1}{x+\dfrac{1}{y+\frac{1}{z}}}$

Can be rewritten as,

$\rm\implies 2 + \dfrac{11}{13} = 2 + \dfrac{1}{x+\dfrac{1}{y+\frac{1}{z}}}$

$\rm\implies 2 - 2+ \dfrac{11}{13} = \dfrac{1}{x+\dfrac{1}{y+\frac{1}{z}}}$

$\rm\implies \dfrac{11}{13} = \dfrac{1}{x+\dfrac{1}{y+\frac{1}{z}}}$

$\rm\implies \dfrac{1}{ \dfrac{13}{11} } = \dfrac{1}{x+\dfrac{1}{y+\frac{1}{z}}}$

On cross multiplication, we get:

$\rm\implies \dfrac{13}{11} =x+\dfrac{1}{y+\frac{1}{z}}$

$\rm\implies \dfrac{13}{11} - x = \dfrac{1}{y+\frac{1}{z}}$

$\rm\implies \dfrac{13 - 11x}{11} = \dfrac{1}{y+\frac{1}{z}}$

$\rm\implies \dfrac{1}{ \frac{11}{13 - 11x}} = \dfrac{1}{y+\frac{1}{z}}$

Again on cross multiplication, we get:

$\rm\implies\dfrac{11}{13 - 11x} = y+\dfrac{1}{z}$

$\rm\implies\dfrac{11}{13 - 11x} - y = \dfrac{1}{z}$

$\rm\implies\dfrac{11 - 13y + 11xy}{13 - 11x} = \dfrac{1}{z}$

$\purple { \boxed{ \rm \implies z = \dfrac{13 - 11x}{11 - 13y + 11xy} }}$

Hence this is the required answer.