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So I was scrolling through the homepage of Youtube to see if there were any math equations that I would think that I could solve, when I came across this supposedly easy factorial problem by SyberMath:$$\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$$which I wanted to try and solve on my own. Here are my steps that I took to get my solution:$$\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$$$$\implies\frac{7!}{6!7!}+\frac{6!}{6!7!}=\frac{x}{8!}$$$$\iff\frac{6!+7!}{6!7!}=\frac{x}{8!}$$$$\iff\frac{6!+7!}{10!}=\frac{x}{8!}$$$$\implies\frac{6!8!+7!8!}{8!10!}=\frac{10!x}{8!10!}$$$$\implies\frac{6!8!+7!8!-10!x}{8!10!}=0$$$$\implies\frac{6!\cancel{8!}+7!\cancel{8!}-\cancel{10!}x}{\cancel{8!}\cancel{10!}}=0$$$$\iff\frac{6!}{10!}+\frac{7!}{10!}=\frac{x}{8!}$$$$\iff\frac{6!8!+7!8!}{10!}=x$$$$\iff64=x$$My question
Is my solution correct, or if I am not, what steps can I take to more easily attain the correct solution?
Best Answer
Just factor it $$\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}\Rightarrow\\ \frac{1}{6!}\Big(1+\frac{1}{7}\Big)=\frac{1}{6!}\frac{x}{7\cdot 8}\Rightarrow\\\frac87=\frac{x}{7\cdot 8}\Rightarrow\\64=x$$