How do you find the x and y intercept of #4y=3x+12#?

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Answer 1 · 2016-12-17T22:10:52

The x-intercept is: #x = -4# or #(-4, 0#).

The y-intercept is: #y = 3# or (0, 3)

To find the x-intercept you set #y# equal to #0# and solve for #x#:

#4*0 = 3x + 12#

#0 = 3x + 12#

#0 - 12 = 3x + 12 - 12#

#-12 = 3x + 0#

#-12 = 3x#

#(-12)/3 = (3x)/3#

#-4 = x# or #x = -4# or #(-4, 0#).

To find the x-intercept you set #x# equal to #0# and solve for #y#:

#4y = 3*0 + 12#

#4y = 0 + 12#

#4y = 12#

#(4y)/4 = 12/4#

#y = 3# or (0, 3)