SOLUTION: THE SUM OF A NUMBER AND ITS RECIPORCAL IS 10 OVER 3 WHAT IS THAT NUMBER? USE FIVE STEP METHOD I NEED MY DIPLOMA!! HELP!~!

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Question 21082: THE SUM OF A NUMBER AND ITS RECIPORCAL IS 10 OVER 3 WHAT IS THAT NUMBER? USE FIVE STEP METHOD I NEED MY DIPLOMA!! HELP!~!
Found 2 solutions by algebrapro18, Earlsdon:
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
THE SUM OF A NUMBER AND ITS RECIPORCAL IS 10 OVER 3

writing as an algebraic expression we get

x+1/x=10/3

cross multiplying

3(x+1)=10x

distributing

3x+3=10x

collecting like terms

3=7x

dividing by 7

x=3/7

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure what the "five-step" method is but perhaps you can adapt the following to fit. Let n be the unknown number.
Write the equation:
n+%2B+1%2Fn+=+10%2F3 Add the terms on the left side.
Solve the equation:
%28n%5E2+%2B+1%29%2Fn+=+10%2F3 Multiply both sides by n.
n%5E2+%2B+1+=+10n%2F3 Multiply both sides by 3.
3n%5E2+%2B+3+=+10n} Subtract 10n from both sides.
3n%5E2+-+10n+%2B+3+=+0 Solve this quadratic equation for n by factoring.
%283n+-+1%29%28n+-+3%29+=+0 Apply the zero products principle.
3n+-+1+=+0 and/or n+-+1+=+0
If 3n+-+1+=+0, then 3n+=+1 and n+=+1%2F3
If n+-+3+=+0, then n+=+3
So there are two answers to this problem:
n = 3 and n = 1/3
Check:
n = 3: 3+%2B+1%2F3+=+9%2F3+%2B+1%2F3 = 10%2F3 OK
n = 1/3: 1%2F3+%2B+1%2F%281%2F3%29+=+1%2F3+%2B+3 = 1%2F3+%2B+9%2F3+=+10%2F3 OK