What are the intercepts of #-x-2y=6#?

1 Answer
Aug 5, 2017

See a solution process below:

Explanation:

y-intercept

First, cover up the #-x# term leaving:

#-2y = 6#

Solving for #y# gives:

#(-2y)/color(red)(-2) = 6/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = -3#

#y = -3#

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

x-intercept

Now, cover up the #-2y# term leaving:

#-x = 6#

Solving for #x# gives:

#color(red)(-1) xx -x = color(red)(-1) xx 6#

#x = -6#

The #x#-intercept is: #-6# or #(-6, 0)#