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solve the system

I1 = I2 + I3
-10 - 5I1 - 5I2
-15 - 5I3 - 5I2

1 answer

To solve the system of equations given by:

  1. \( I_1 = I_2 + I_3 \)
  2. \( -10 - 5I_1 - 5I_2 = 0 \) (let's rewrite this as \( -5I_1 - 5I_2 = 10 \), or \( I_1 + I_2 = -2 \))
  3. \( -15 - 5I_3 - 5I_2 = 0 \) (let's rewrite this as \( -5I_3 - 5I_2 = 15 \), or \( I_3 + I_2 = -3 \))

Now we have the following system of equations to work with:

  1. \( I_1 = I_2 + I_3 \)
  2. \( I_1 + I_2 = -2 \)
  3. \( I_3 + I_2 = -3 \)

Now, we can substitute \( I_1 \) in terms of \( I_2 \) and \( I_3 \) into the second equation:

From equation (1), we know that: \[ I_1 = I_2 + I_3 \]

Substituting \( I_1 \) into equation (2): \[ I_2 + I_3 + I_2 = -2 \] \[ 2I_2 + I_3 = -2 \] (equation 4)

Now, we have two equations involving \( I_2 \) and \( I_3 \):

  1. \( 2I_2 + I_3 = -2 \) (equation 4)
  2. \( I_3 + I_2 = -3 \) (equation 3)

From equation (3), we can express \( I_3 \) in terms of \( I_2 \): \[ I_3 = -3 - I_2 \]

Now, substitute this expression back into equation (4): \[ 2I_2 + (-3 - I_2) = -2 \] \[ 2I_2 - 3 - I_2 = -2 \] \[ I_2 - 3 = -2 \] \[ I_2 = 1 \]

Now that we have \( I_2 \), we can substitute it back to find \( I_3 \): \[ I_3 = -3 - I_2 = -3 - 1 = -4 \]

Now substitute \( I_2 \) into the equation for \( I_1 \): \[ I_1 = I_2 + I_3 = 1 - 4 = -3 \]

We have found all the currents:

  • \( I_1 = -3 \)
  • \( I_2 = 1 \)
  • \( I_3 = -4 \)

Thus, the solution to the system is: \[ I_1 = -3, \quad I_2 = 1, \quad I_3 = -4 \]