To solve the system of equations given, we first rewrite the equations clearly. The system appears to consist of three equations involving the currents \(I_1\), \(I_2\), and \(I_3\):
- \( I_1 = I_2 + I_3 \) (Equation 1)
- \( 10 - 5I_1 - 5I_2 = 0 \) (Equation 2)
- \( 15 - 5I_3 - 5I_2 = 0 \) (Equation 3)
We can rewrite Equations 2 and 3 in a more standard form:
From Equation 2: \[ 5I_1 + 5I_2 = 10 \implies I_1 + I_2 = 2 \quad (Equation 2') \]
From Equation 3: \[ 5I_3 + 5I_2 = 15 \implies I_3 + I_2 = 3 \quad (Equation 3') \]
Now, we have the following system of equations:
- \( I_1 = I_2 + I_3 \) (Equation 1)
- \( I_1 + I_2 = 2 \) (Equation 2')
- \( I_3 + I_2 = 3 \) (Equation 3')
Step 1: Solve for \(I_1\)
From Equation 2': \[ I_1 = 2 - I_2 \]
Step 2: Solve for \(I_3\)
From Equation 3': \[ I_3 = 3 - I_2 \]
Step 3: Substitute into Equation 1
Now substitute \(I_1\) and \(I_3\) into Equation 1: \[ 2 - I_2 = I_2 + (3 - I_2) \]
This simplifies to: \[ 2 - I_2 = 3 \]
Step 4: Solve for \(I_2\)
Solving this equation: \[ 2 - I_2 = 3 \implies -I_2 = 3 - 2 \implies -I_2 = 1 \implies I_2 = -1 \]
Step 5: Substitute to find \(I_1\) and \(I_3\)
Now we can find \(I_1\) and \(I_3\): \[ I_1 = 2 - I_2 = 2 - (-1) = 2 + 1 = 3 \]
\[ I_3 = 3 - I_2 = 3 - (-1) = 3 + 1 = 4 \]
Conclusion
The solutions are: \[ I_1 = 3, \quad I_2 = -1, \quad I_3 = 4 \]
Thus, the solution to the system \( (I_1, I_2, I_3) \) is \( (3, -1, 4) \).