Two batteries, E 1 = 6 V and E 2 = 9 V, and three resistors, R 1 = 2.04 ?, R 2 =

Question

Two batteries, E1 = 6 V and E2 = 9 V, and three resistors, R1 = 2.04 ?, R2 = 3.90 ?, and R3 = 4.97 ?, are connected as shown in the figure below.

How much current flows through the E2 = 9 V battery when the switch is closed?

How much current flows through the E1 = 6 V battery when the switch is closed?

PLEASE READ! - I have seen similar questions like this but it uses whole numbers which seems to me to be way easier than this problem. I have gotten the following equations, I just do not understand how to combine the two equations to give me the answers. Please explain how to combine the following equations, also correct me if I am incorrect about these equations.

EQ 1: 2.09I1 - 3.90I2 = -3

EQ 2: 8.87I2 + 4.97I1 = 9

Explanation / Answer

Let the current flowing from Battery E1 is I1.
Let the current flowing from Battery E2 is I2.
Let the Current flowing from Resistor R3 is I.

Now,
I = I1+I2 ----------1

Using Kirchoff Voltage Law -

I* 4.97 + 2.04 * I1 = 6.0 -----------2
I* 4.97 + 3.90 * I2 = 9.0 -----------3

solving these equations we get,
I1 = 0.23 A
I2 = 0.89 A

Current flowing from Battery E1. I1 = 0.23 A
Current flowing from Battery E2, I2 = 0.89 A

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