#color(blue)(underline(Stepcolor(white)(x)1))#
Add #color(blue)(5y)# to both sides
#color(brown)((8x-5y) color(blue)(+5y) =( 2)color(blue)(+5y)#
I am using the brackets to show what is being altered or grouping to make understanding easier. They serve no other purpose!
#8x +(color(blue)(5y)-5y)=2+color(blue)(5y)#
#8x +0 =2+5y color(green)(" This action has made the y-term positive")#
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#color(blue)(underline(Stepcolor(white)(x)2)#
Subtract 2 from both sides
#color(brown)((8x)color(blue)( -2)=(5y+2)color(blue)(-2)) #
#8x-2 = 5y +(2-2)#
#8x-2 =5y +0 color(white)(xx)color(green)("This action moved the 2 to the other side of =")#
Due to convention rewrite with the target variable on the left:
#5y=8x-2#
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#color(blue)(underline(Stepcolor(white)(x)3)#
Divide both sides by #color(blue)(5)#. Note that this is the same as #times 1/5#
#(color(brown)(5y))/(color(blue)(5)) =(color(brown)(8x-2))/(color(blue)(5))#
#5/5 times y=8/5 x-2/5#
But #5/5=1# giving
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#color(blue)(underline(Solution)#
#color(green)(y=8/5x-2/5)#
