Question

If 1/9! +1/10! =x/11! , find x.

Answer 1

Hope you understand this.

Answer 2

$\displaystyle \sf{ If \: \frac{1}{9!} + \frac{1}{10!} = \frac{x}{11!} \: \: then \: \: x = 121 }$

Given :

$\displaystyle \sf{ \frac{1}{9!} + \frac{1}{10!} = \frac{x}{11!} }$

To find :

The value of x

Solution :

Step 1 of 2 :

Write down the given expression

The expression is

$\displaystyle \sf{ \frac{1}{9!} + \frac{1}{10!} = \frac{x}{11!} }$

Step 2 of 2 :

Find the value of x

$\displaystyle \sf{ \frac{1}{9!} + \frac{1}{10!} = \frac{x}{11!} }$

$\displaystyle \sf{ \implies \frac{11 \times 10}{11 \times 10 \times 9!} + \frac{11}{11 \times 10!} = \frac{x}{11!} }$

$\displaystyle \sf{ \implies \frac{110 }{11 !} + \frac{11}{11 !} = \frac{x}{11!} }$

$\displaystyle \sf{ \implies \frac{110 + 11 }{11 !} = \frac{x}{11!} }$

$\displaystyle \sf{ \implies \frac{121 }{11 !} = \frac{x}{11!} }$

$\displaystyle \sf{ \implies \frac{x }{11 !} = \frac{121}{11!} }$

$\displaystyle \sf{ \implies x = 121}$

Hence the required value of x = 121

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