x+3y-3=0; add 3 to both sides

x + 3y = 3; subtract x from both sides

3y = 3 – x; change 3 – x to -x + 3

3y = -x + 3; divide by 3

y = -x/3 + 1

in front of the x is a negative 1 but we don’t always write the one. We have the fraction

-1x/3 which we can write as -1/3 x

so the equation is

y = -1/3 x + 1

solve for y

first add the 3 from both sides

then subtract the X from both sides

you are left with 3y= – x+3 right now m= -1

then divide 3 on both sides and then you get y= -1/3x+3

remember the coeffcient of x is -1 and when divide it by 3 you get -1/3

-x -x

3y-3=-x

+3 +3

3y/3= (-x+3)/3

y= -1/3x+1

First You Subtract x on both sides and you get: 3y-3=-x because positives and negatives cancel.

Then You Add 3 on both sides and you get: 3y= -x+3 because positives and negatives cancel.

Then You Divide by 3 on both sides to get y by itself and you get: y= -1/3x+1. I hope this gets the job done!

Peace Luck Love Life.

add 3 to each side, then

subtract x from each side, then

divide both sides by 3 (remember to divide each term by 3)

It’s just solving for y.

3y-3=-x+0

3y=-x+3

y=-x/3+1

y=-1/3(x)+1