Question:
If $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !^{\prime}}$, find the value of $x$
Solution:
Given Equation :
$\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$
To Find : Value of x
Formula: $n !=n \times(n-1) !$
By given equation
$\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$
$\therefore \frac{8 \times 7}{8 \times 7 \times 6 !}+\frac{8}{8 \times 7 !}=\frac{x}{8 !}$
By using the above formula we can write,
$\therefore \frac{56}{8 !}+\frac{8}{8 !}=\frac{x}{8 !}$
$\therefore \frac{64}{8 !}=\frac{x}{8 !}$
Cancelling (8!) from both the sides,
$\therefore \mathrm{x}=64$
Conclusion : Value of x is 64.
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