How do you graph #-8x+10y=40# using intercepts?

To find the #color(blue)" x and y intercepts"#

#• " let x = 0, in the equation, to find y-intercept"#

#• " let y = 0, in the equation, to find x-intercept"#

#x=0to10y=40rArry=4larrcolor(red)" y-intercept"#

#y=0to-8x=40rArrx=-5larrcolor(red)" x-intercept"#

Plot the points (0 ,4), (-5 ,0) and draw a straight line through them for the graph.
graph{4/5x+4 [-11.25, 11.25, -5.625, 5.625]}

Keep in mind that the #x#-intercept is the point where #y# is #0#, and the #y#-intercept is the point where #x# is #0#.

It's easier to remember this if you think about the intercepts on a graph #-# any point on the #y#-axis has to have an #x# coordinate of #0#, or else it wouldn't be on the #y#-axis (and vice versa).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First, find the #x#-intercept by plugging in #0# for #y#.

#-8x + 10(0) = 40#
#color(white)"XXXX--"-8x = 40#
#color(white)"XXXXXX.." x = -5#

The #x#-intercept, then, is #color(red)("(-5, 0)#

Next, find the #y#-intercept by plugging in #0# for #x#.

#-8(0) + 10y = 40#
#color(white)"XXXXX-"10y=40#
#color(white)"XXXXXX.."y=4#

The #y#-intercept, then, is #color(blue)("(0, 4))#

Finally, to graph the line, just graph these two points, and connect them, as shown below.