SOLUTION: Solve. If equation has no solution, write no solution. x over x+1 (+) x+1 over x = 5 over 2 {{{2x^2+2(x+1)(x+1) = 5[x(x+1)}}} 2x^2+2(x^2+x+x+2) = 5(x^2+x) 2x^2 + 2x^

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Question 69184This question is from textbook Algebra Structure and Method
: Solve. If equation has no solution, write no solution.
x over x+1 (+) x+1 over x = 5 over 2
AMP Parsing Error of [2x^2+2(x+1)(x+1) = 5[x(x+1)]: closing square bracket expected instead of at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 227. .

2x^2+2(x^2+x+x+2) = 5(x^2+x)
2x^2 + 2x^2 + 2x + 2x+4 = 5x^2+5x
0 = x^2 + x-4
that is as far as I got, please help! This question is from textbook Algebra Structure and Method

Answer by Smirnov(15)   (Show Source): You can put this solution on YOUR website!
Well, 0 = x^2 + x-4 would be
Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: x=-2.56155281280883,x=1.56155281280883.
  • Graphical form: Equation was fully solved.
  • Text form: 0=x^2+x-4 simplifies to 0=0
  • Cartoon (animation) form:
    For tutors: simplify_cartoon( 0=x^2+x-4 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at .
Moved these terms to the left highlight_green%28+-x%5E2+%29,highlight_green%28+-x+%29,highlight_green%28+4+%29
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Moved 4 to the right of expression
It becomes .
Look at .
Equation highlight_red%28+-x%5E2-x%2B4=0+%29 is a quadratic equation: -x^2-x+4 =0, and has solutions -2.56155281280883,1.56155281280883
It becomes .
Result:
This is an equation! Solutions: x=-2.56155281280883,x=1.56155281280883.

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Done!
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