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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 （6000*31.25+4000*125/3）（1-12/5a） = 31.25（1-a）*6000（1-3/2a）+125/3*4000（1-5/2a） .
Question type: Equation
Solution:Original question:
Remove the bracket on the left of the equation:
The equation is transformed into :
The equation is transformed into :
Remove the bracket on the right of the equation:
The equation is transformed into :
After the equation is converted into a general formula, it is converted into:
( a +0 )( 135a - 17 )=0
From
a + 0 = 0
135a - 17 = 0
it is concluded that:：
a1=0
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》
☆1 equations
[ 1/1 Equation]
Work: Find the solution of equation （6000*31.25+4000*125/3）（1-12/5a） = 31.25（1-a）*6000（1-3/2a）+125/3*4000（1-5/2a） .
Question type: Equation
Solution:Original question:
| ( | 6000 | × | 125 4 | + | 4000 | × | 125 | ÷ | 3 | ) | ( | 1 | − | 12 | ÷ | 5 | × | a | ) | = | 125 4 | ( | 1 | − | a | ) | × | 6000 | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 125 | ÷ | 3 | × | 4000 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| Left side of the equation = | 6000 | × | 125 4 | ( | 1 | − | 12 | ÷ | 5 | × | a | ) | + | 4000 | × | 125 | ÷ | 3 | × | ( | 1 | − | 12 | ÷ | 5 | × | a | ) |
| = | 187500 | ( | 1 | − | 12 | ÷ | 5 | × | a | ) | + | 500000 3 | ( | 1 | − | 12 | ÷ | 5 | × | a | ) |
| = | 187500 | × | 1 | − | 187500 | × | 12 | ÷ | 5 | × | a | + | 500000 3 | ( | 1 | − | 12 | ÷ | 5 | × | a | ) |
| = | 187500 | − | 450000 | a | + | 500000 3 | ( | 1 | − | 12 | ÷ | 5 | × | a | ) |
| = | 187500 | − | 450000 | a | + | 500000 3 | × | 1 | − | 500000 3 | × | 12 | ÷ | 5 | × | a |
| = | 187500 | − | 450000 | a | + | 500000 3 | − | 400000 | a |
| = | 1062500 3 | − | 850000 | a |
1062500 3 | − | 850000 | a | = | 125 4 | ( | 1 | − | a | ) | × | 6000 | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 125 | ÷ | 3 | × | 4000 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| Right side of the equation = | 187500 | ( | 1 | − | a | ) | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
1062500 3 | − | 850000 | a | = | 187500 | ( | 1 | − | a | ) | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| Right side of the equation = | 187500 | × | 1 | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | − | 187500 | a | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| = | 187500 | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | − | 187500 | a | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| = | 187500 | × | 1 | − | 187500 | × | 3 | ÷ | 2 | × | a | − | 187500 | a | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| = | 187500 | − | 281250 | a | − | 187500 | a | ( | 1 | − | 3 | ÷ | 2 | × | a | ) | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| = | 187500 | − | 281250 | a | − | 187500 | a | × | 1 | + | 187500 | a | × | 3 | ÷ | 2 | × | a | + | 500000 3 |
| = | 187500 | − | 281250 | a | − | 187500 | a | + | 281250 | a | a | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| = | 187500 | − | 468750 | a | + | 281250 | a | a | + | 500000 3 | ( | 1 | − | 5 | ÷ | 2 | × | a | ) |
| = | 187500 | − | 468750 | a | + | 281250 | a | a | + | 500000 3 | × | 1 | − | 500000 3 | × | 5 | ÷ | 2 | × | a |
| = | 187500 | − | 468750 | a | + | 281250 | a | a | + | 500000 3 | − | 1250000 3 | a |
| = | 1062500 3 | − | 2656250 3 | a | + | 281250 | a | a |
1062500 3 | − | 850000 | a | = | 1062500 3 | − | 2656250 3 | a | + | 281250 | a | a |
After the equation is converted into a general formula, it is converted into:
( a +0 )( 135a - 17 )=0
From
a + 0 = 0
135a - 17 = 0
it is concluded that:：
a1=0
| a2= | 17 135 |
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》