Question 427792
The sum of a number and four times its reciprocal is 29/5.Find the number 

Let's call the number n.

The reciprocal would be {{{1/n}}}

So the formula should look like.
{{{n+4*(1/n)=29/5}}}
now multiply every term on both sides by n, the {{{1/n}}} in the {{{4*(1/n)}}} will cancel to become 1, leaving you with 4*1=4
{{{n^2+4=(29/5)*n}}} multiply both sides by 5.
{{{5n^2+20=29n}}} subtract {{{29n}}} from both sides
{{{5n^2-29n+20=0}}} now you have a solvable quadratic equation, that factors into.
{{{(5n-4)(n-5)=0}}}
So {{{n=5}}} or {{{5n-4=0}}} -> {{{5n=4}}} divide both sides by 5 {{{n=5/4}}}