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On line Solution of Monovariate Equation：
    Input any unary equation directly, and then click the "Next" button to obtain the solution of the equation.
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    Current location：Equations > Monovariate Equation > The history of univariate equation calculation > Answer

    Overview: 1 questions will be solved this time.Among them
           ☆1 equations

[ 1/1 Equation]
    Work: Find the solution of equation 2/(d-0.4)+1/(d+9) = 1/(d+0.5) .
    Question type: Equation
    Solution:Original question:
     2 ÷ ( d −
2
5
) + 1 ÷ ( d + 9) = 1 ÷ ( d +
1
2
)
     Multiply both sides of the equation by:( d −
2
5
) ,  ( d +
1
2
)
     2( d +
1
2
) + 1 ÷ ( d + 9) × ( d −
2
5
)( d +
1
2
) = 1( d −
2
5
)
    Remove a bracket on the left of the equation:：
     2 d + 2 ×
1
2
 + 1 ÷ ( d + 9) × ( d −
2
5
)( d +
1
2
) = 1( d −
2
5
)
    Remove a bracket on the right of the equation:：
     2 d + 2 ×
1
2
 + 1 ÷ ( d + 9) × ( d −
2
5
)( d +
1
2
) = 1 d − 1 ×
2
5
    The equation is reduced to :
     2 d + 1 + 1 ÷ ( d + 9) × ( d −
2
5
)( d +
1
2
) = 1 d −
2
5
     Multiply both sides of the equation by:( d + 9)
     2 d ( d + 9) + 1( d + 9) + 1( d −
2
5
)( d +
1
2
) = 1 d ( d + 9) −
2
5
( d + 9)
    Remove a bracket on the left of the equation:
     2 d d + 2 d × 9 + 1( d + 9) + 1( d −
2
5
)( d +
1
2
) = 1 d ( d + 9) −
2
5
( d + 9)
    Remove a bracket on the right of the equation:：
     2 d d + 2 d × 9 + 1( d + 9) + 1( d −
2
5
)( d +
1
2
) = 1 d d + 1 d × 9 −
2
5
( d + 9)
    The equation is reduced to :
     2 d d + 18 d + 1( d + 9) + 1( d −
2
5
)( d +
1
2
) = 1 d d + 9 d −
2
5
( d + 9)
    Remove a bracket on the left of the equation:
     2 d d + 18 d + 1 d + 1 × 9 + 1( d −
2
5
)( d +
1
2
) = 1 d d + 9 d −
2
5
( d + 9)
    Remove a bracket on the right of the equation:：
     2 d d + 18 d + 1 d + 1 × 9 + 1( d −
2
5
)( d +
1
2
) = 1 d d + 9 d −
2
5
 d −
2
5
 × 9
    The equation is reduced to :
     2 d d + 18 d + 1 d + 9 + 1( d −
2
5
)( d +
1
2
) = 1 d d + 9 d −
2
5
 d −
18
5
    The equation is reduced to :
     2 d d + 19 d + 9 + 1( d −
2
5
)( d +
1
2
) = 1 d d +
43
5
 d −
18
5
    Remove a bracket on the left of the equation:
     2 d d + 19 d + 9 + 1 d ( d +
1
2
) − 1 ×
2
5
( d +
1
2
) = 1 d d +
43
5
 d −
18
5
    The equation is reduced to :
     2 d d + 19 d + 9 + 1 d ( d +
1
2
) −
2
5
( d +
1
2
) = 1 d d +
43
5
 d −
18
5
    Remove a bracket on the left of the equation:
     2 d d + 19 d + 9 + 1 d d + 1 d ×
1
2
 = 1 d d +
43
5
 d −
18
5
    The equation is reduced to :
     2 d d + 19 d + 9 + 1 d d +
1
2
 d −
2
5
 = 1 d d +
43
5
 d −
18
5
    The equation is reduced to :
     2 d d +
39
2
 d + 9 + 1 d d −
2
5
( d +
1
2
) = 1 d d +
43
5
 d −
18
5
    Remove a bracket on the left of the equation:
     2 d d +
39
2
 d + 9 + 1 d d −
2
5
 d −
2
5
 = 1 d d +
43
5
 d −
18
5
    The equation is reduced to :
     2 d d +
39
2
 d + 9 + 1 d d −
2
5
 d −
1
5
 = 1 d d +
43
5
 d −
18
5
    The equation is reduced to :
     2 d d +
191
10
 d +
44
5
 + 1 d d = 1 d d +
43
5
 d −
18
5

    The solution of the equation:
        d1≈-3.456039 , keep 6 decimal places
        d2≈-1.793961 , keep 6 decimal places    
    There are 2 solution(s).

解程的详细方法请参阅：《方程的解法》


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