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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 4t+33 = 10+20（t-0.5）-1/2×5×（t-0.5） .
Question type: Equation
Solution:Original question:
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:
Combine the items on the right of the equation:
By shifting the terms and changing the symbols on toth sides of the equation, we obtain :
If the left side of the equation is equal to the right side, then the right side must also be equal to the left side, that is :
The coefficient of the unknown number is reduced to 1 :
We obtained :
This is the solution of the equation.
Convert the result to decimal form :
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation 4t+33 = 10+20（t-0.5）-1/2×5×（t-0.5） .
Question type: Equation
Solution:Original question:
| 4 | t | + | 33 | = | 10 | + | 20 | ( | t | − | 1 2 | ) | − | 1 | ÷ | 2 | × | 5 | ( | t | − | 1 2 | ) |
| Right side of the equation = | 10 | + | 20 | ( | t | − | 1 2 | ) | − | 5 2 | ( | t | − | 1 2 | ) |
| 4 | t | + | 33 | = | 10 | + | 20 | ( | t | − | 1 2 | ) | − | 5 2 | ( | t | − | 1 2 | ) |
| Right side of the equation = | 10 | + | 20 | t | − | 20 | × | 1 2 | − | 5 2 | ( | t | − | 1 2 | ) |
| = | 10 | + | 20 | t | − | 10 | − | 5 2 | ( | t | − | 1 2 | ) |
| = | 0 | + | 20 | t | − | 5 2 | ( | t | − | 1 2 | ) |
| = | 0 | + | 20 | t | − | 5 2 | t | + | 5 2 | × | 1 2 |
| = | 0 | + | 20 | t | − | 5 2 | t | + | 5 4 |
| = | 5 4 | + | 35 2 | t |
| 4 | t | + | 33 | = | 5 4 | + | 35 2 | t |
Transposition :
| 4 | t | − | 35 2 | t | = | 5 4 | − | 33 |
Combine the items on the left of the equation:
| - | 27 2 | t | = | 5 4 | − | 33 |
Combine the items on the right of the equation:
| - | 27 2 | t | = | - | 127 4 |
By shifting the terms and changing the symbols on toth sides of the equation, we obtain :
127 4 | = | 27 2 | t |
If the left side of the equation is equal to the right side, then the right side must also be equal to the left side, that is :
27 2 | t | = | 127 4 |
The coefficient of the unknown number is reduced to 1 :
| t | = | 127 4 | ÷ | 27 2 |
| = | 127 4 | × | 2 27 |
| = | 127 2 | × | 1 27 |
We obtained :
| t | = | 127 54 |
Convert the result to decimal form :
| t | = | 2.351852 |