Question

If we have an equation as $\dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{x}{{\left| \!{\underline {\,8 \,}} \right. }}$ , find x .

Answer 1

Hint: The given question involves the expansion and simplification of a factorial . The formula of factorial of number n is
$\left| \!{\underline {\,n \,}} \right. = n \times (n - 1) \times (n - 2) \times (n - 3).........3 \times 2 \times 1$
First we expand the factorial of each number into the form containing factorial of smallest number present in problem , that is if we want factorial of 3 in the form involving factorial of 2 that implies , $\left| \!{\underline {\,3 \,}} \right. = 3 \times \left| \!{\underline {\,2 \,}} \right. $ . Later remove the common factorial of small number and then simplify the left part and right part .Then we can acquire the value of x by dividing left part by coefficient of x


Complete step-by-step solution:
To solve this we opt the formula of factorial of 6 , 7 , 8 .
$\dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{x}{{\left| \!{\underline {\,8 \,}} \right. }} \\
\Rightarrow \dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{7 \times \left| \!{\underline {\,6 \,}} \right. }} = \dfrac{x}{{8 \times \left| \!{\underline {\,7 \,}} \right. }} \\
\Rightarrow \dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{7 \times \left| \!{\underline {\,6 \,}} \right. }} = \dfrac{x}{{8 \times 7 \times \left| \!{\underline {\,6 \,}} \right. }} \\
\Rightarrow \dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }}\left( {1 + \dfrac{1}{7}} \right) = \dfrac{x}{{8 \times 7 \times \left| \!{\underline {\,6 \,}} \right. }} \\
\Rightarrow \left( {1 + \dfrac{1}{7}} \right) = \dfrac{x}{{8 \times 7}} \\
\Rightarrow \dfrac{{7 + 1}}{7} = \dfrac{x}{{8 \times 7}} \\
\Rightarrow \dfrac{8}{7} = \dfrac{x}{{8 \times 7}} \\
\Rightarrow 8 = \dfrac{x}{8} \\
\Rightarrow x = 8 \times 8 \\
\Rightarrow x = 64 \\$
Verification :If the value acquired by solving the equation is x = 64 to the equation $\dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{x}{{\left| \!{\underline {\,8 \,}} \right. }}$ .
We substitute the value of x in the equation for verifying L. H. S = R. H . S .
$\dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{{64}}{{\left| \!{\underline {\,8 \,}} \right. }} \\
\Rightarrow \dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }} + \dfrac{1}{{7 \times \left| \!{\underline {\,6 \,}} \right. }} = \dfrac{{64}}{{\left| \!{\underline {\,8 \,}} \right. }} \\
\Rightarrow \dfrac{1}{{\left| \!{\underline {\,6 \,}} \right. }}\left( {1 + \dfrac{1}{7}} \right) = \dfrac{{64}}{{\left| \!{\underline {\,8 \,}} \right. }} \\
\Rightarrow \dfrac{{7 + 1}}{{7 \times \left| \!{\underline {\,6 \,}} \right. }} = \dfrac{{64}}{{\left| \!{\underline {\,8 \,}} \right. }} \\
\Rightarrow \dfrac{8}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{{64}}{{8 \times \left| \!{\underline {\,7 \,}} \right. }} \\
\Rightarrow \dfrac{8}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{8}{{\left| \!{\underline {\,7 \,}} \right. }} \\$
Hence proved ( L .H .S = R. H .S ). The value of x = 64 for $\dfrac{1}{{\left| \!{\underline {\,
6 \,}} \right. }} + \dfrac{1}{{\left| \!{\underline {\,7 \,}} \right. }} = \dfrac{x}{{\left| \!{\underline {\,8 \,}} \right. }}$ .

Note: The problems involving factorial simplification just need to be solved by expanding factorial of higher number into the form containing factorial of common simpler number. They are generally factorial induced algebraic equations. The important thing to note when it comes to topic of factorial are
$\left| \!{\underline {\,0 \,}} \right. = 1 \\
\left| \!{\underline {\,1 \,}} \right. = 1 \\$
$\left| \!{\underline {\,0 \,}} \right. = 1 \\
\left| \!{\underline {\,1 \,}} \right. = 1 \\
$ and factorial of negative numbers are complex numbers which are difficult to solve so we generally avoid taking factorials of negative numbers.