The current sources in the circuit below are I$1=612$I$1=612$ and$$I$2=4+25$I$2=4+25.$Use the node voltage method to determine IBIB in Amps.$$

a$.$ What are the node voltages $($in Volts$)?$

V$1=($x$1)+$j$($y$1)=($V$1=($x$1)+$j$($y$1)=($Answer $1$ Question $6)+$j$()+$j$($Answer $2$ Question $6)$V$)$V

$($note: both x$1$x$1$ and$$y$1$y$1$ can be positive or negative$)$

V$2=($x$2)+$j$($y$2)=($V$2=($x$2)+$j$($y$2)=($Answer $3$ Question $6)+$j$()+$j$($Answer $4$ Question $6)$V$)$V

$($note: both x$2$x$2$ and$$y$2$y$2$ can be positive or negative$)$

b$.$ What is the current IBIB $($in Amps$)?$

IB$=($x$3)+$j$($y$3)=($IB$=($x$3)+$j$($y$3)=($Answer $5$ Question $6)+$j$()+$j$($Answer $6$ Question $6)$A$)$A

$($note: both x$3$x$3$ and$$y$3$y$3$ can be positive or negative$)$