How do you find the point of intersection for # x+y= -3# and #x+y=3#?

1 Answer
Jul 8, 2015

Answer:

From #x+y=3->y=-x+3=3-x#

Explanation:

Now substitute that in the other equation:
#x+y=x+(3-x)=-3->3=-3#
This cannot be. Clearly there is no intersection.

We could have seen this, because (as we have seen before), the second equation translates to a slope-intercept form of
#y=-x+3#
While the first will translate to:
#y=-x-3#
Same slope (#-1#), but different intercepts (#+3and-3#), in other words, these are parallel lines.
Here is the graph for #y=-x+3# the other one runs parallel and #6# units below it:
graph{-x+3 [-10, 10, -5, 5]}