Let #x# be the first number.
Let #y# be the second number.
First Equation:
The first number is 2 less than the second number.
#color(blue)(x=y-2)#
Second Equation:
Twice the first number is 4 more than 3 times the second.
#color(red)(2x=4+3y)#
System:
#color(blue)(x=y-2)#
#color(red)(2x=4+3y)#
We can solve this by substitution. Let's get the value of #x# in the first equation and substitute it into #x# in the second equation. Then, we can solve for #y#.
#2color(blue)(x)=4+3y#
#2color(blue)((y-2))=4+3y#
#2y-4=4+3y#
#2y-4-3y=4+3y-3y#
#-4-y=4#
#-4-y+4=4+4#
#-y=8#
#color(magenta)(y=-8)#
To solve for #x#, just substitute the value of #y# into either of the two equations.
#x=color(magenta)(y)-2#
#x=color(magenta)((-8))-2#
#color(orange)(x=-10)#
