I figured it out. Nevermind. Sorry to bother y'all.
Had to do y=5/3x + 0
then do 3*y=3*5/3x + 0
to get,3y=15/3x + 0
3y=5x + 0
3y-5x=5x-5x + 0
Answer: 5x-3y=0
Write the standard form of the equation of the line through the given point with the given slope.
through: (3, 5), slope = 5/3
I'm using the point-slope formula of y-y1=m(x-x1)
y-y1=m(x-x1)
y-5=5/3(x-3)
y-5=5/3x-15/3
y-5=5/3x - 5
y-5+5=5/3x -5+5
y=5/3x
This is where I am stuck. Not sure what to do from here. Did I plug in the xs and ys incorrectly?
2 answers
(3, 5), m = 5/3.
Y = mx + b = 5.
(5/3)3 + b = 5,
b = 0.
Y = (5/3)x + 0.
Standard form: (5/3)x - Y = 0.
Y = mx + b = 5.
(5/3)3 + b = 5,
b = 0.
Y = (5/3)x + 0.
Standard form: (5/3)x - Y = 0.