Given: #x+2(x+2)=16#

#xcolor(red)(+2)color(green)((x+2))=16#

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#color(blue)("Dealing with just the brackets")#

You have two options of approach for dealing with the brackets. Actually they are both the same thing but look different.

#color(brown)("Option 1:")#

The #color(red)(2)# in front of the brackets means we have 2 of them so you could do this.

#color(green)(x+2)#

#ul(color(green)(x+2) larr" Add")#

#2x+4#

#color(brown)("Option 2:")#

Multiply everything inside the brackets by the #color(red)(2)# outside it.

#color(green)( [color(red)(2xx)x]+[color(red)(2xx)2]#

#color(white)("dd")2xcolor(white)(".d")+color(white)("dd")4#

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#color(blue)("Putting it all together")#

#color(white)("ddddddddddd")xcolor(red)(+2)color(green)((x+2))color(white)("d")=16#

#color(white)("ddddddddddd")x+color(white)("d")2x+4color(white)("d")=16#

Adding all the #x's#

#color(white)("ddddddddddd")3x+4=16#

Getting the #3x# on its own: Subtract #color(red)(4)# from both sides

#color(white)("ddddddddddd")color(green)(3x+4color(red)(-4)=16color(red)(-4))#

#color(white)("ddddddddddd")3x+color(white)("d")0color(white)(".d")=color(white)("d")12#

Getting the #x# on its own: multiply both sides by #color(red)(1/3)#

#color(white)("ddddddddddd")color(green)(3x color(red)(xx1/3)=12color(red)(xx1/3))#

#color(white)("ddddddddddddd")color(green)(3/color(red)(3) x color(white)("d.")= color(white)("d")12/color(red)(3))#

But #3/3# is the same as 1 and anything multiplied by 1 does not change.

#color(white)("ddddddddddd")1xx x =4#

#color(white)("dddddddddddddd")x=4#