To find the #x#-intercept, substitute #0# for #y# and solve for #x#:
#8y = 3x - 9# becomes:
#8 * 0 = 3x - 9#
#0 = 3x - 9#
#0 + color(red)(9) = 3x - 9 + color(red)(9)#
#9 = 3x - 0#
#9 = 3x#
#9/color(red)(3) = (3x)/color(red)(3)#
#3 = (color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3))#
#3 = x#
#x = 3#
The #x#-intercept is #3# or #(3, 0)#
To find the #y#-intercept, substitute #0# for #x# and solve for #y#:
#8y = 3x - 9# becomes:
#8y = (3 * 0) - 9#
#8y = 0 - 9#
#8y = -9#
#(8y)/color(red)(8) = -9/color(red)(8)#
#(color(red)(cancel(color(black)(8)))y)/cancel(color(red)(8)) = -9/8#
#y = -9/8#
The #y#-intercept is #-9/8# or #(0, -9/8)#