Mathematics
         
语言：中文    Language:English
Equations   
Fold
Unary equation
Multivariate equation
Math OP  
Unfold
Linear algebra      
Unfold
Derivative function
Function image
Hot issues
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+1/5)(5+1/3)(X+2) = (4X+16)(48/15) .
    Question type: Equation
    Solution:Original question:
     (3 + 1 ÷ 5)(5 + 1 ÷ 3)( X + 2) = (4 X + 16)(48 ÷ 15)
    Remove the bracket on the left of the equation:
     Left side of the equation = 3(5 + 1 ÷ 3)( X + 2) + 1 ÷ 5 × (5 + 1 ÷ 3)( X + 2)
                                             = 3(5 + 1 ÷ 3)( X + 2) +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 3 × 5( X + 2) + 3 × 1 ÷ 3 × ( X + 2) +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 15( X + 2) + 1( X + 2) +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 15 X + 15 × 2 + 1( X + 2) +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 15 X + 30 + 1( X + 2) +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 15 X + 30 + 1 X + 1 × 2 +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 15 X + 30 + 1 X + 2 +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 16 X + 32 +
1
5
(5 + 1 ÷ 3)( X + 2)
                                             = 16 X + 32 +
1
5
 × 5( X + 2) +
1
5
 × 1 ÷ 3 × ( X + 2)
                                             = 16 X + 32 + 1( X + 2) +
1
15
( X + 2)
                                             = 16 X + 32 + 1 X + 1 × 2 +
1
15
( X + 2)
                                             = 16 X + 32 + 1 X + 2 +
1
15
( X + 2)
                                             = 17 X + 34 +
1
15
( X + 2)
                                             = 17 X + 34 +
1
15
 X +
1
15
 × 2
                                             = 17 X + 34 +
1
15
 X +
2
15
                                             =
256
15
 X +
512
15
    The equation is transformed into :
     
256
15
 X +
512
15
 = (4 X + 16)(48 ÷ 15)
    Remove the bracket on the right of the equation:
     Right side of the equation = 4 X (48 ÷ 15) + 16(48 ÷ 15)
                                               = 4 X × 48 ÷ 15 + 16(48 ÷ 15)
                                               =
64
5
 X + 16(48 ÷ 15)
                                               =
64
5
 X + 16 × 48 ÷ 15
                                               =
64
5
 X +
256
5
    The equation is transformed into :
     
256
15
 X +
512
15
 =
64
5
 X +
256
5

    Transposition :
     
256
15
 X −
64
5
 X =
256
5
 −
512
15

    Combine the items on the left of the equation:
     
64
15
 X =
256
5
 −
512
15

    Combine the items on the right of the equation:
     
64
15
 X =
256
15

    The coefficient of the unknown number is reduced to 1 :
      X =
256
15
 ÷
64
15
        =
256
15
 ×
15
64
        = 4 × 1

    We obtained :
      X = 4
    This is the solution of the equation.



Your problem has not been solved here? Please go to the Hot Problems section！


Return



  New addition:Lenders ToolBox module（Specific location：Math OP > Lenders ToolBox ），welcome。