How do you find the x and y intercept of #6x + 4y = 12#?

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Answer 1 · 2017-02-27T13:24:33

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

#6x + 4y = 12# becomes:

#6x + (4 * 0) = 12#

#6x + 0 = 12#

#6x = 12#

#(6x)/color(red)(6) = 12/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 2#

#x = 2# therefore the x-intercept is #2# or #(2, 0)#

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

#6x + 4y = 12# becomes:

#(6 xx 0) + 4y = 12#

#0 + 4y = 12#

#4y = 12#

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

#(color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4)) = 3#

#y = 3# therefore the y-intercept is #3# or #(0, 3)#