I'm first going to solve this assuming that the first #x# term should really be a #t#:

#7t+6-2(5+3t/2)=5t-11#

Distribute the 2 across the brackets:

#7t+6-10+3t=5t-11#

Let's move all the #t# terms to the left and all the non-#t# terms to the right:

#7t+6-10+3tcolor(red)(-6+10-5t)=5t-11color(red)(-6+10-5t)#

#7t+3t-5t=-11-6+10#

And now simplify:

#5t=-7#

#(5t)/color(red)(5)=-7/color(red)(5)#

#t=-7/5#

~~~~~

Now let's assume that we want to solve for #x# in terms of #t#:

#7x+6-2(5+3t/2)=5t-11#

#7x+6-10+3t=5t-11#

Let's move all non-#x# terms to the right:

#7x+6-10+3tcolor(red)(-6+10-3t)=5t-11color(red)(-6+10-3t)#

#7x=2t-7#

#x=(2t-7)/7=2/7t-1#

~~~~~

Now let's assume we want to solve for #t# in terms of #x#:

Start with

#7x=2t-7#

#2t=7x+7#

#t=(7x+7)/2=(7(x+1))/2=7/2(x+1)#