We have only 1 unknown and 1 equation. So it is solvable. The equation count has to be the same as the count of 'different' unknowns.
Given: #2x-3+4=3#
Method:
Step 1:
Simplify as far as you are able.
Step 2:
Isolate (get on its own) the term with #x# in it. That is, have #2x# on one side of the = and everything else on the other side.
Step 3:
Get #x# on its own on one side of the = and everything else on the other side.
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#color(blue)("Step 1")#
Deal with the -3+4
#color(green)(2xcolor(red)(-3+4)=3color(white)("dd")-> color(white)("ddd") 2xcolor(red)(+1)=3#
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#color(blue)("Step 2")#
Subtract #color(red)(1)# from both sides. This turns the +1 into 0
#color(green)(2x+1=3color(white)("dddd") ->color(white)("dddd") 2x+1color(red)(-1)color(white)("d")=color(white)("d")3color(red)(-1))#
#color(green)(color(white)("dddddddddddd.")->color(white)("dddd")2x+color(white)("d.")0color(white)("d.d")=color(white)("dd")2 )#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#
Divide both sides by #color(red)(2)#. This turns the 2 from #2x# into 1 and 1 times anything does not change its value.
#color(green)(2x=2 color(white)("ddddddd")->color(white)("ddd")2/color(red)(2)x=2/color(red)(2)#
#color(green)(color(white)("ddddddddddd.d")->color(white)("ddd")x=1)#
