There are a couple of ways to start solving this problem, either expand the terms within the parenthesis or add #color(red)(7)# to each side of the equation to eliminate the constant on the right side of the equation. I will start with expanding the terms within parenthesis by multiplying each term within the parenthesis by #color(red)(2)#. Be careful to handle the signs of the individual terms correctly:
#-119 = -7 - color(red)(2)(-7v - 1)#
#-119 = -7 - (color(red)(2) xx -7v) - (color(red)(2) xx -1)#
#-119 = -7 - (-14v) - (-2)#
#-119 = -7 + 14v + 2#
Next, combine the constants on the right side of the equation:
#-119 = 14v + 2 - 7#
#-119 = 14v - 5#
Then, add #color(red)(5)# to each side of the equation to isolate the #v# term while keeping the equation balanced:
#-119 + color(red)(5) = 14v - 5 + color(red)(5)#
#-114 = 14v - 0#
#-114 = 14v#
Now, divide each side of the equation by #color(red)(14)# to solve for #v# while keeping the equation balanced:
#-114/color(red)(14) = (14v)/color(red)(14)#
#-114/14 = (color(red)(cancel(color(black)(14)))v)/cancel(color(red)(14))#
#-(2 xx 57)/(2 xx 7) = v#
#-(color(red)(cancel(color(black)(2))) xx 57)/(color(red)(cancel(color(black)(2))) xx 7) = v#
#-57/7 = v#
#v = -57/7#
