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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 341.96×（（1+0.031）1-1）+（0+10.26+0.01V）×（（1+0.031）0.5-1） = 0 .
    Question type: Equation
    Solution:Original question:
     
8549
25
((1 +
31
1000
) × 1 − 1) + (0 +
513
50
 +
1
100
 V )((1 +
31
1000
) ×
1
2
 − 1) = 0
    Remove the bracket on the left of the equation:
     Left side of the equation =
8549
25
(1 +
31
1000
) × 1 −
8549
25
 × 1 + (0 +
513
50
 +
1
100
 V )((1 +
31
1000
) ×
1
2
 − 1)
                                             =
8549
25
(1 +
31
1000
) −
8549
25
 + (0 +
513
50
 +
1
100
 V )((1 +
31
1000
) ×
1
2
 − 1)
                                             =
8549
25
 × 1 +
8549
25
 ×
31
1000
 −
8549
25
 + (0 +
513
50
 +
1
100
 V )((1 +
31
1000
) ×
1
2
 − 1)
                                             =
8549
25
 +
265019
25000
 −
8549
25
 + (0 +
513
50
 +
1
100
 V )((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + (0 +
513
50
 +
1
100
 V )((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + 0((1 +
31
1000
) ×
1
2
 − 1) +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + 0(1 +
31
1000
) ×
1
2
 − 0 × 1 +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + 0 (1 +
31
1000
) − 0 +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + 0 (1 +
31
1000
) +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + 0 × 1 + 0 ×
31
1000
 +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 + 0 + 0 +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 +
513
50
((1 +
31
1000
) ×
1
2
 − 1) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 +
513
50
(1 +
31
1000
) ×
1
2
 −
513
50
 × 1 +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
265019
25000
 +
513
100
(1 +
31
1000
) −
513
50
 +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
8519
25000
 +
513
100
(1 +
31
1000
) +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
8519
25000
 +
513
100
 × 1 +
513
100
 ×
31
1000
 +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)
                                             =
8519
25000
 +
513
100
 +
15903
100000
 +
1
100
 V ((1 +
31
1000
) ×
1
2
 − 1)

    
        V≈1161.979360 , keep 6 decimal places    
    There are 1 solution(s).

解一元一次方程的详细方法请参阅：《一元一次方程的解法》


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