How do you solve #17= - 2x - 3+ 7x#?

First, lets combine like terms on the Right-Hand Side (RHS). We have two numbers that are both multiplied by #x#, so we can group them like so:

#17=-2x-3+7x#

#17=-3+7x-2x#

#17=-3+x(7-2)#

#17=-3+5x#

Next, we'll re-arrange the equation to get #x# and its coefficient by itself, on the RHS. We'll achieve this by adding 3 to both sides, which will eliminate the -3 on the RHS, and effectively moving it to the Left-Hand Side (LHS):

#17color(red)(+3)=cancel(-3)+5xcolor(red)(cancel(+3))#

#20=5x#

Finally, we'll divide both sides by the coefficient that #x# is multiplied by, which should give us our answer:

#20/color(red)(5)=(cancel(5)x)/color(red)(cancel(5))#

#color(green)(x=4)#

#17=-2x-3+7x#

can also be written as

#17=5x-3#

To solve this, you want all of the #x#'s on one side of the equation: Therefore, you can add #3# to both sides to make;

#20=5x#

To get #x# on its own, you then divide both sides by #5# to get:

#4=x#

Given:

#17 = -2x-3+7x#

Adding 3 to both sides,
#17+3 = cancel(3)-cancel(3)+ 5x#
#20 = 5x#

Dividing both sides with 5, we get,
#cancel(20)/cancel(5) = cancel(5)/cancel(5)x#
#:.x = color(red)(4)#