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           ☆1 equations

[ 1/1 Equation]
    Work: Find the solution of equation (2.5-0.5+0.14m)×(120-80-(1/40)m)+(2.8-0.8)×(80+2.4m) = 324-14 .
    Question type: Equation
    Solution:Original question:
     (
5
2
1
2
+
7
50
m )(12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m ) = 32414
    Remove the bracket on the left of the equation:
     Left side of the equation =
5
2
(12080(1 ÷ 40) m )
1
2
(12080(1 ÷ 40) m ) +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             =
5
2
× 120
5
2
× 80
5
2
(1 ÷ 40) m
1
2
(12080(1 ÷ 40) m ) +
7
50
m (12080(1 ÷ 40) m )
                                             = 300200
5
2
(1 ÷ 40) m
1
2
(12080(1 ÷ 40) m ) +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 100
5
2
(1 ÷ 40) m
1
2
(12080(1 ÷ 40) m ) +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 100
5
2
× 1 ÷ 40 × m
1
2
(12080(1 ÷ 40) m ) +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 100
1
16
m
1
2
(12080(1 ÷ 40) m ) +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 100
1
16
m
1
2
× 120 +
1
2
× 80 +
1
2
(1 ÷ 40) m +
7
50
m
                                             = 100
1
16
m 60 + 40 +
1
2
(1 ÷ 40) m +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)
                                             = 80
1
16
m +
1
2
(1 ÷ 40) m +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80
1
16
m +
1
2
× 1 ÷ 40 × m +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80
1
16
m +
1
80
m +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80
1
20
m +
7
50
m (12080(1 ÷ 40) m ) + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80
1
20
m +
7
50
m × 120
7
50
m × 80
7
50
m (1 ÷ 40)
                                             = 80
1
20
m +
84
5
m
56
5
m
7
50
m (1 ÷ 40) m + (
14
5
4
5
)
                                             = 80 +
111
20
m
7
50
m (1 ÷ 40) m + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80 +
111
20
m
7
50
m × 1 ÷ 40 × m + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80 +
111
20
m
7
2000
m m + (
14
5
4
5
)(80 +
12
5
m )
                                             = 80 +
111
20
m
7
2000
m m +
14
5
(80 +
12
5
m )
4
5
(80 +
12
5
m )
                                             = 80 +
111
20
m
7
2000
m m +
14
5
× 80 +
14
5
×
12
5
m
4
5
                                             = 80 +
111
20
m
7
2000
m m + 224 +
168
25
m
4
5
(80 +
12
5
m )
                                             = 304 +
1227
100
m
7
2000
m m
4
5
(80 +
12
5
m )
                                             = 304 +
1227
100
m
7
2000
m m
4
5
× 80
4
5
×
12
5
m
                                             = 304 +
1227
100
m
7
2000
m m 64
48
25
m
                                             = 240 +
207
20
m
7
2000
m m
    The equation is transformed into :
     240 +
207
20
m
7
2000
m m = 32414

    
        m≈6.778825 , keep 6 decimal places
    
    There are 1 solution(s).


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