the x-intercept is the value of x when y is equal to 0.
your equation is 5y = -4.
put this in slope intercept form and the equation becomes y = -4/5.
slope intercept form is y = mx + b
m is the slope
b is the y-intercept
the slope is the change in the value of y divided by the change in the value of x.
when the slope is equal to 0, as it is in the equation of y = -4/5, you will not see the x in the equation.
if it was there, it would show up as y = 0x - 4/5.
when the slope is equal to 0, you have a horizontal line.
when the slope is undefined, you have a vertical line.
an equation of a vertical line will show up as x = a constant.
for example:
y = -4/5 is a horizontal line at y = -4/5.
x = -4/5 is a vertical line at x = -4/5.
outside of vertical and horizontal lines, you will see equations in the form of y = mx + b.
m is the slope.
b is the y-intercept.
when you want to find the y-intercept, you replace x with 0 and solve the equation for y.
when you want to find the x-intercept, you set y = 0 and solve the equation for x.
your equation is y = -4/5.
since there is no x in the equation, this means that x can be any value and y will always be -4/5.
that's a horizontal line.
it's always at y = -4/5 regardless of what the value of x is.
this includes when x = 0, therefore, your y-intercept is equal to -4/5.
graph it and you see a horizontal line on the graph at y = -4/5.
if you want to find the x-intercept in this equation, you will find that there is no x-intercept.
since y = -4/5 regardless of the value of x, y will always be -4/5 which means that y can never be equal to 0.
since y can never be equal to 0, then there is no x-intercept for this equation.
now let's look at a line that is vertical.
an equation for a vertical line could be x = -4/5.
if you look for the y-intercept, there won't be any because the value of x is equal to -4/5 regardless of the value of y.
since x is always -4/5, it can never cross the y-axis and so you never have a y-intercept.
if you look for the x-intercept, it will be at x = -4/5 because x is always at -4/5 regardless of the value of y, including y = 0.
outside of vertical or horizontal lines, you would solve for y-intercept and x-intercept as follows.
assume the equation is y = -5x + 3.
to find the x-intercept, set y equal to 0 and solve for x.
you will get 0 = -5x + 3.
subtract 3 from both sides of this equation to get -3 = -5x.
divide both sides of this equation by -5 to get x = -3/-5 which becomes y = 3/5.
that's your x-intercept.
the coordinate point of your x-intercept is (3/5,0)
to find the y-intercept, set x equal to 0 and solve for y.
you will get y = -5*0 + 3 which becomes y = 3.
that's your y-intercept.
the coordinate point of your y-intercept is (0,3).
the following graph shows you a horizontal line at y = -4/5 and a vertical line at x = -4/5 and the equation of y = -5x + 3 that i just showed you above.
look below the graph for further comments.
in this graph:
blue line is x = -4/5.
it's a vertical line that intersects the x-axis at x = -4/5 and doesn't intersect the y-axis because it is parallel to the y-axis and will never touch it.
red line is y = -4/5.
it's a horizontal line that intersects the y-axis at y = -4/5 and doesn't intersect the x-axis because it is parallel to the x-axis and will never touch it.
orange line is y = -5x + 3.
this is the more traditional line.
it intersects the y-axis at y = 3 and it intersects the x-axis at x = 3/5.
on the graph, -4/5 shows up as -.8 and x = 3/5 shows up as .6.
the slope of y = 5x + 3 can be calculated from the x-intercept and the y-intercept as follows:
the y-intercept is at (0,3)
the x-intercept is at (.6,0)
the slope is (y2-y1) / (x2-x1).
you can assign either of those coordinates to (x1,y1) or (x2,y2).
the slope will be the same.
for example:
x1 = 0 and y1 = 3, while x2 = .6 and y2 = 0.
(y2-y1)/(x2-x1) becomes (0-3)/.6-0 which becomes -3/.6 which becomes -5.
if you assign x1 = .6 and y1 = 0, while x2 = 0 and y2 = 3, you will get:
(y2-y1)/(x2-x1) becomes (3-0)/(0-.6) = 3/-.6 which becomes -5.
you get the same slope either way.
you could not find the slope using x and y intercepts for y = -4/5 because that equation has no x-intercept.
you could not find the slope using x and y intercepts for x = -4/5 because that equation has no y-intercept.
in those cases, you would need to find another point on the line.
in y = -4/5, all you need to do is assume any value of x and y will be -4/5 for that value of x.
in x = -4/5, all you need to do is assume any value of y and x will be at -4/5 for that value of y.
to find any point on a more standard line, just assume any value for x and then solve for the corresponding value of y.
here's some references that should be helpful.
http://www.purplemath.com/modules/strtlneq.htm
http://www.purplemath.com/modules/strtlneq3.htm
http://www.purplemath.com/modules/strtlneq2.htm
http://www.purplemath.com/modules/slope.htm
purplemath has more.
here's the page that i did the search on.
https://www.google.com/?gws_rd=ssl#q=solving+equation+of+straight+lines+purplemath
my search was for "solving equation of straight lines purplemath"
take out the "purplemath" and you'll get lots of other websites as well.