SOLUTION: {{{ 1/ (2n^2) }}} + {{{ 5 / 2n }}} = {{{ (n-2) / n^2 }}}

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Question 136258: +1%2F+%282n%5E2%29+ + +5+%2F+2n+ = +%28n-2%29+%2F+n%5E2+
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
+1%2F+%282n%5E2%29+ + +5+%2F+2n+ = +%28n-2%29+%2F+n%5E2+
:
Multiply equation 2n^2 to get rid of the denominators
2n^2*1%2F%282n%5E2%29+ + 2n^2*+5+%2F%282n%29 = 2n^2%28%28n-2%29%29%2Fn%5E2
cancel out the denominators and you have:
1 + 5n = 2(n-2)
1 + 5n = 2n - 4
:
5n - 2n = -4 - 1
3n = -5
n = -5%2F3
:
Check solution in original equation using decimals; -5/3 = -1.67:
+1%2F+%282%28-1.67%29%5E2%29+ + +5+%2F+2%28-1.67%29+ = +%28%28-1.67%29-2%29+%2F+%28-1.67%29%5E2+
+1%2F5.56 + +5%2F%28-3.34%29 = +%28-3.67%29+%2F+2.8+
.18 - 1.50 = -1.31; close enough to confirm our solution