Given: #5(x+17)=100#
#color(blue)("Approach 1 - shows usfull methods")#
Multiply everything inside the brackets by the 5 that is outside
#5x+85=100#
To 'get rid' of the 85 on the left subtract #color(red)(85)# from BOTH sides
#color(green)(5x+ubrace(85color(red)(-85))=100color(red)(-85))#
#color(white)("dddddd")darr#
#color(green)(5x+color(white)("d.d")0color(white)("d.d")=color(white)("dd")15)#
To get rid of the #5# from #5x# divide both sides by #color(red)(5)#
#color(green)(5/color(red)(5) xcolor(white)("d")=color(white)("d")15/color(red)(5)#
But #5/5=1 and 1xx x = x# and #15/5=3# giving:
#color(green)(x=3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Approach 2 - quicker as it uses less steps")#
#5(x+17)=100#
Divide both sides by 5
#x+17=20#
Subtract 17 from both sides
#x=3#
