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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.
    It supports equations that contain mathematical functions.
    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 3*(1+a/100)+2*(1+3*a/100)*(1-a/100) = 5*(1+29*a/2500) .
    Question type: Equation
    Solution:Original question:
     3(1 + a ÷ 100) + 2(1 + 3 a ÷ 100)(1 a ÷ 100) = 5(1 + 29 a ÷ 2500)
    Remove the bracket on the left of the equation:
     Left side of the equation = 3 × 1 + 3 a ÷ 100 + 2(1 + 3 a ÷ 100)(1 a ÷ 100)
                                             = 3 +
3
100
a + 2(1 + 3 a ÷ 100)(1 a ÷ 100)
                                             = 3 +
3
100
a + 2 × 1(1 a ÷ 100) + 2 × 3 a ÷ 100 × (1 a ÷ 100)
                                             = 3 +
3
100
a + 2(1 a ÷ 100) +
3
50
a (1 a ÷ 100)
                                             = 3 +
3
100
a + 2 × 12 a ÷ 100 +
3
50
a (1 a ÷ 100)
                                             = 3 +
3
100
a + 2
1
50
a +
3
50
a (1 a ÷ 100)
                                             = 5 +
1
100
a +
3
50
a (1 a ÷ 100)
                                             = 5 +
1
100
a +
3
50
a × 1
3
50
a a ÷ 100
                                             = 5 +
1
100
a +
3
50
a
3
5000
a a
                                             = 5 +
7
100
a
3
5000
a a
    The equation is transformed into :
     5 +
7
100
a
3
5000
a a = 5(1 + 29 a ÷ 2500)
    Remove the bracket on the right of the equation:
     Right side of the equation = 5 × 1 + 5 × 29 a ÷ 2500
                                               = 5 +
29
500
a
    The equation is transformed into :
     5 +
7
100
a
3
5000
a a = 5 +
29
500
a

    After the equation is converted into a general formula, it is converted into:
    ( a +0 )( a - 20 )=0
    From
        a + 0 = 0
        a - 20 = 0

    it is concluded that::
        a1=0
        a2=20
    
    There are 2 solution(s).


解一元二次方程的详细方法请参阅:《一元二次方程的解法》



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