"Spoooooooky" Russian rational expressions 4

Algebra Level 2

E = 5 y 2 1 y 2 ÷ ( 1 1 1 y ) \mathscr{E } = \dfrac{5y^2}{1-y^2} \div \left( 1- \dfrac{1}{1-y} \right)

If y = 2016 y = 2016 , and the value of E \mathscr{E} can be represented in the form a b -\dfrac{a}{b} , where a a and b b are coprime positive integers , find a + b a+b .


Credit: My former Trig teacher's worksheet of Russian rational expressions


The answer is 12097.

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1 solution

Chew-Seong Cheong
Jun 30, 2016

E = 5 y 2 1 y 2 ÷ ( 1 1 1 y ) = 5 y 2 1 y 2 ÷ ( y 1 y ) = 5 y 2 ( 1 y ) ( 1 + y ) ( y 1 y ) = 5 y 1 + y = 5 × 2016 1 + 2016 = 10080 2017 \begin{aligned} \mathscr E & = \frac {5y^2}{1-y^2} \div \left(1-\frac 1{1-y}\right) \\ & = \frac {5y^2}{1-y^2} \div \left(\frac {-y}{1-y}\right) \\ & = \frac {5y^2}{(1-y)(1+y)} \left(\frac {y-1}{y}\right) \\ & = - \frac {5y}{1+y} = - \frac {5\times 2016}{1+2016} = - \frac {10080}{2017} \end{aligned}

a + b = 10080 + 2017 = 12097 \implies a + b = 10080 + 2017 = \boxed{12097}

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