Question 136258
{{{ 1/ (2n^2) }}} + {{{ 5 / 2n }}} = {{{ (n-2) / n^2 }}}
:
Multiply equation 2n^2 to get rid of the denominators
2n^2*{{{1/(2n^2) }}} + 2n^2*{{{ 5 /(2n)}}} = 2n^2{{{((n-2))/n^2}}}
cancel out the denominators and you have:
1 + 5n = 2(n-2)
1 + 5n = 2n - 4
:
5n - 2n = -4 - 1
3n = -5
n = {{{-5/3}}}
:
Check solution in original equation using decimals; -5/3 = -1.67:
{{{ 1/ (2(-1.67)^2) }}} + {{{ 5 / 2(-1.67) }}} = {{{ ((-1.67)-2) / (-1.67)^2 }}}
{{{ 1/5.56}}} + {{{ 5/(-3.34)}}} = {{{ (-3.67) / 2.8 }}}
.18 - 1.50 = -1.31; close enough to confirm our solution