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)#