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    Current location：Equations > Monovariate Equation > The history of univariate equation calculation > Answer

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           ☆1 equations

[ 1/1 Equation]
    Work: Find the solution of equation 70×3.55%+（1720-70+69+2%V）×[（1+3.55%）0.5-1] = 0 .
    Question type: Equation
    Solution:Original question:
     70 ×
71
2000
 + (1720 − 70 + 69 +
2
100
 V )((1 +
71
2000
) ×
1
2
 − 1) = 0
     Left side of the equation =
497
200
 + (1720 − 70 + 69 +
2
100
 V )((1 +
71
2000
) ×
1
2
 − 1)
    The equation is transformed into :
     
497
200
 + (1720 − 70 + 69 +
2
100
 V )((1 +
71
2000
) ×
1
2
 − 1) = 0
    Remove the bracket on the left of the equation:
     Left side of the equation =
497
200
 + 1720((1 +
71
2000
) ×
1
2
 − 1) − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             =
497
200
 + 1720(1 +
71
2000
) ×
1
2
 − 1720 × 1 − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V
                                             =
497
200
 + 860(1 +
71
2000
) − 1720 − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
343503
200
 + 860(1 +
71
2000
) − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
343503
200
 + 860 × 1 + 860 ×
71
2000
 − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
343503
200
 + 860 +
3053
100
 − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
165397
200
 − 70((1 +
71
2000
) ×
1
2
 − 1) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
165397
200
 − 70(1 +
71
2000
) ×
1
2
 + 70 × 1 + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
165397
200
 − 35(1 +
71
2000
) + 70 + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
151397
200
 − 35(1 +
71
2000
) + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
151397
200
 − 35 × 1 − 35 ×
71
2000
 + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
151397
200
 − 35 −
497
400
 + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
317291
400
 + 69((1 +
71
2000
) ×
1
2
 − 1) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
317291
400
 + 69(1 +
71
2000
) ×
1
2
 − 69 × 1 +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
317291
400
 +
69
2
(1 +
71
2000
) − 69 +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
344891
400
 +
69
2
(1 +
71
2000
) +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
344891
400
 +
69
2
 × 1 +
69
2
 ×
71
2000
 +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
344891
400
 +
69
2
 +
4899
4000
 +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
3306011
4000
 +
2
100
 V ((1 +
71
2000
) ×
1
2
 − 1)
                                             = -
3306011
4000
 +
2
100
 V (1 +
71
2000
) ×
1
2
 −
2
100
 V × 1
                                             = -
3306011
4000
 +
1
100
 V (1 +
71
2000
) −
1
50
 V
                                             = -
3306011
4000
 +
1
100
 V × 1 +
1
100
 V ×
71
2000
 −
1
50
 V
                                             = -
3306011
4000
 +
1
100
 V +
71
200000
 V −
1
50
 V

    
        V≈-85692.353551 , keep 6 decimal places    
    There are 1 solution(s).

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