Which equation shows y+1/5=3x in standard form? a. -15x+5y=-1 b. -5x+15y=-1 c. -15x+5y=1 d. 15x+5y=-1

You always begin by the Given equation, which is in this case:
y+ = 3x
To remove the fraction, you multiply the whole equation by 5:
5y + 1 = 15x
Then, you move the "15x" the left-side of the equation, changing it's sign
-15x + 5y +1 = 0
You move the "1" to the right side, changing it's sign
-15x + 5y = -1
So the correct answer is A :)
Hope it helped! ^^

Standard form is ax+by=c where a and b are integers and a is normally positive

y+1/5=3x
minus 3x both sides
-3x+y+1/5=
mnus 1/5 both sides
-3x+y=-1/5
times both sides by -5
15x-5y=1
seems they want x to be negative
multiply by -1
-15x+5y=-1

A

Jjutsclopyrsvnkiu it’s c