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)
