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           ☆1 equations

[ 1/1 Equation]
    Work: Find the solution of equation 70×3.65%+(1720-70+69+2%V)×[(1+3.65%)0.5-1] = 33.64+0.00036V .
    Question type: Equation
    Solution:Original question:
     70 ×
73
2000
+ (172070 + 69 +
2
100
V )((1 +
73
2000
) ×
1
2
1) =
841
25
+
9
25000
V
     Left side of the equation =
511
200
+ (172070 + 69 +
2
100
V )((1 +
73
2000
) ×
1
2
1)
    The equation is transformed into :
     
511
200
+ (172070 + 69 +
2
100
V )((1 +
73
2000
) ×
1
2
1) =
841
25
+
9
25000
V
    Remove the bracket on the left of the equation:
     Left side of the equation =
511
200
+ 1720((1 +
73
2000
) ×
1
2
1)70((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             =
511
200
+ 1720(1 +
73
2000
) ×
1
2
1720 × 170((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V
                                             =
511
200
+ 860(1 +
73
2000
)172070((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
343489
200
+ 860(1 +
73
2000
)70((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
343489
200
+ 860 × 1 + 860 ×
73
2000
70((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
343489
200
+ 860 +
3139
100
70((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
165211
200
70((1 +
73
2000
) ×
1
2
1) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
165211
200
70(1 +
73
2000
) ×
1
2
+ 70 × 1 + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
165211
200
35(1 +
73
2000
) + 70 + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
151211
200
35(1 +
73
2000
) + 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
151211
200
35 × 135 ×
73
2000
+ 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
151211
200
35
511
400
+ 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
316933
400
+ 69((1 +
73
2000
) ×
1
2
1) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
316933
400
+ 69(1 +
73
2000
) ×
1
2
69 × 1 +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
316933
400
+
69
2
(1 +
73
2000
)69 +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
344533
400
+
69
2
(1 +
73
2000
) +
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
344533
400
+
69
2
× 1 +
69
2
×
73
2000
+
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
344533
400
+
69
2
+
5037
4000
+
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
3302293
4000
+
2
100
V ((1 +
73
2000
) ×
1
2
1)
                                             = -
3302293
4000
+
2
100
V (1 +
73
2000
) ×
1
2
2
100
V × 1
                                             = -
3302293
4000
+
1
100
V (1 +
73
2000
)
1
50
V
                                             = -
3302293
4000
+
1
100
V × 1 +
1
100
V ×
73
2000
1
50
V
                                             = -
3302293
4000
+
1
100
V +
73
200000
V
1
50
V

    
        V≈-85964.307154 , keep 6 decimal places
    
    There are 1 solution(s).


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