Question 1112005
a. 
For {{{y=0}}} , {{{-3x = 15}}} , and solving for {{{x}}} we find
{{{x=15/(-3)}}} and {{{x=highlight(-5)}}} for the x-intercept.
For {{{x=0}}} , {{{5y = 15}}} , and solving for {{{y}}} we find
{{{5=15/5}}} and {{{y=highlight(3)}}} for the y-intercept.
The points called x-intercept and y-intercept are
{{{highlight(A(-5,0))}}} and {{{highlight(B(0,3))}}} respectively.
 
EXPLANATION:
The x-intercept sometimes means the point where the line intersects the x-axis,
but as {{{y=0}}} for all points of the x-axis, that intersection point has {{{y=0}}}
so its x-coordinate is also called the x-intercept of the line.
Similarly, the point (with {{{x=0}}} ) , or just its y-coordinate is called the y-intercept.
 
b.
Maybe you were taught that the slope of a line going through points {{{A(x[A],y[A])}}} and {{{B(x[B],y[B])}}} is
{{{m=(y[B]-y[A])/(x[B]-x[A])}}} .
Applied to the points found in part a,
{{{m=(3-0)/(0-(-5))=3/(0+5)=3/5}}} .
So, {{{highlight(m=3/5)}}} .
 
c.
There is an infinite number of equivalent equations that represent the same line as {{{-3x+5y=15}}} ,
but there is only one that starts with "y=", and that is the slope-intercept form.
To get it, we just solve for y:
{{{-3x+5y=15}}}
{{{5y=3x+15}}}
{{{y=(3x+15)/5}}}
{{{y=3x/5+15/5}}}
{{{highlight(y=(3/5)x+3)}}} .
 
d.
The slope-intercept form of the equation of a line is an equation of the form
{{{y=m*x+b}}} where the constants{{{m}}} and {{{b}}}
are the slope and y-intercept of the line respectively.
The equation {{{y=(3/5)x+3}}} found in part c,
tells us That {{{m=3/5}}} is the slope (as found in part b),
and that {{{b=3}}} is the y-intercept (as found in part a).