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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.
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 12×（2/3a×3/5）+16×（1/3a×1/2） = 16×（2/3a×2/5）+（1/3a×1/2） .
Question type: Equation
Solution:Original question:
Remove the bracket on the left of the equation:
The equation is transformed into :
Remove the bracket on the right of the equation:
The equation is transformed into :
Transposition :
Combine the items on the left of the equation:
The coefficient of the unknown number is reduced to 1 :
We obtained :
This is the solution of the equation.
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation 12×（2/3a×3/5）+16×（1/3a×1/2） = 16×（2/3a×2/5）+（1/3a×1/2） .
Question type: Equation
Solution:Original question:
| 12 | ( | 2 | ÷ | 3 | × | a | × | 3 | ÷ | 5 | ) | + | 16 | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) | = | 16 | ( | 2 | ÷ | 3 | × | a | × | 2 | ÷ | 5 | ) | + | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) |
| Left side of the equation = | 12 | × | 2 | ÷ | 3 | × | a | × | 3 | ÷ | 5 | + | 16 | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) |
| = | 24 5 | a | + | 16 | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) |
| = | 24 5 | a | + | 16 | × | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 |
| = | 24 5 | a | + | 8 3 | a |
| = | 112 15 | a |
112 15 | a | = | 16 | ( | 2 | ÷ | 3 | × | a | × | 2 | ÷ | 5 | ) | + | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) |
| Right side of the equation = | 16 | × | 2 | ÷ | 3 | × | a | × | 2 | ÷ | 5 | + | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) |
| = | 64 15 | a | + | ( | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 | ) |
| = | 64 15 | a | + | 1 | ÷ | 3 | × | a | × | 1 | ÷ | 2 |
| = | 64 15 | a | + | 1 6 | a |
| = | 133 30 | a |
112 15 | a | = | 133 30 | a |
Transposition :
112 15 | a | − | 133 30 | a | = | 0 |
Combine the items on the left of the equation:
91 30 | a | = | 0 |
The coefficient of the unknown number is reduced to 1 :
| a | = | 0 | ÷ | 91 30 |
| = | 0 | × | 30 91 |
We obtained :
| a | = | 0 |