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)#
