Equations
Fold
Math OP
Unfold
Linear algebra
Unfold
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 (a×0.2+a)×0.2+(a×0.2)+a+((a×0.2+a)×0.2+(a×0.2)+a)×0.6 = 171000 .
Question type: Equation
Solution:Original question:
| ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | + | ( | ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | ) | × | 3 5 | = | 171000 |
| Left side of the equation = | a | × | 1 5 | × | 1 5 | + | a | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | + | ( | ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | ) | × | 3 5 |
| = | a | × | 1 25 | + | a | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | + | ( | ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | ) | × | 3 5 |
| = | 31 25 | a | + | ( | a | × | 1 5 | ) | + | ( | ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | ) | × | 3 5 |
| = | 31 25 | a | + | a | × | 1 5 | + | ( | ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | ) | × | 3 5 |
| = | 36 25 | a | + | ( | ( | a | × | 1 5 | + | a | ) | × | 1 5 | + | ( | a | × | 1 5 | ) | + | a | ) | × | 3 5 |
| = | 36 25 | a | + | ( | a | × | 1 5 | + | a | ) | × | 1 5 | × | 3 5 | + | ( | a | × | 1 5 | ) | × | 3 5 | + | a | × | 3 5 |
| = | 36 25 | a | + | ( | a | × | 1 5 | + | a | ) | × | 3 25 | + | ( | a | × | 1 5 | ) | × | 3 5 | + | a | × | 3 5 |
| = | 51 25 | a | + | ( | a | × | 1 5 | + | a | ) | × | 3 25 | + | ( | a | × | 1 5 | ) | × | 3 5 |
| = | 51 25 | a | + | a | × | 1 5 | × | 3 25 | + | a | × | 3 25 | + | ( | a | × | 1 5 | ) | × | 3 5 |
| = | 51 25 | a | + | a | × | 3 125 | + | a | × | 3 25 | + | ( | a | × | 1 5 | ) | × | 3 5 |
| = | 273 125 | a | + | ( | a | × | 1 5 | ) | × | 3 5 |
| = | 273 125 | a | + | a | × | 1 5 | × | 3 5 |
| = | 273 125 | a | + | a | × | 3 25 |
| = | 288 125 | a |
288 125 | a | = | 171000 |
The coefficient of the unknown number is reduced to 1 :
| a | = | 171000 | ÷ | 288 125 |
| = | 171000 | × | 125 288 |
| = | 2375 | × | 125 4 |
We obtained :
| a | = | 296875 4 |
Convert the result to decimal form :
| a | = | 74218.75 |