How do you find the intercepts for #x-3y+7=0?

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Answer 1 · 2015-06-19T18:37:37

To find the intercept with the #x# axis, let #y=0# and solve to find #x = -7#

To find the intercept with the #y# axis, let #x=0# and solve to find #y = 7/3#

This line will intercept the #x# axis where #y=0#, so substitute #y=0# into the equation to get:

#x+7 = 0#

Subtract #7# from both sides to get #x = -7#.

So the intercept is at #(-7, 0)#

The line will intercept the #y# axis where #x=0#, so substitute #x=0# into the equation to get:

#-3y+7 = 0#

Add #3y# to both sides to get:

#7=3y#

Divide both sides by #3# to get:

#y = 7/3#

So the intercept is at #(0, 7/3)#