Ben's boat will take 1½ hours longer to go 12 miles up a river than to return. What is the rate of his boat if the rate of the current is 2 miles an hour?

From the given data:
Let #d=12# miles
Let #t=#time going down with the river current
Let #v =#the rate of Ben's boat
Let #v_c=2# miles/hour-rate of the current

Let #v-v_c=#the rate going up the river
Let #v+v_c=#the rate going down the river

#(v+v_c)(t)=12" "# the equation for distance going down
#(v-v_c)(t+3/2)=12" "#the equation for distance going up

Let us do the simultaneous solution of these equations

From #(v+v_c)(t)=12" "# the equation for distance going down

#(v+v_c)(t)=12" "#

#t=color(red)(12/(v+v_c)" ")#substitute this in the equation for distance going up

#(v-v_c)(t+3/2)=12" "#equation for distance going up
#(v-v_c)(color(red)(12/(v+v_c))+3/2)=12" "#
Now change #v_c=2#

#(v-2)(color(red)(12/(v+2))+3/2)=12" "#

We can now solve for v:
Multiply both sides by (v+2)

#(v-2)(12/(v+2)+3/2)(v+2)=12(v+2)#

#(12(v-2)(v+2))/(v+2)+3/2*(v-2)(v+2)=12v+24#

Simplify

#(12(v-2)cancel(v+2))/cancel(v+2)+3/2*(v-2)(v+2)=12v+24#

#12(v-2)+3/2*(v-2)(v+2)=12v+24#
Expand
#12v-24+3/2(v^2-4)=12v+24#

Simplify

#3/2(v^2-4)=48#

#v^2-4=32#

#v^2=32+4#

#v^2=36#

#v=6" "#miles/hour