How do determine if #x+y=-2,y=x+2# has no solution, one solution, or an Infinite number of solutions and find the solution?

1 Answer
May 10, 2018

There is a single solution. As an #(x, y)# pair, the solution is #(-2, 0)#.

Explanation:

There are a few things we can do to determine if the system is inconsistent (has no solutions) or is consistent (has one or many solutions). The easiest, in this case, is to eliminate a variable and see if the remaining equation can ever be true.

Consider that our second equation is #y = x + 2#. It is already solved for #y#, so we should solve our first equation for #y#:

#x + y = -2#
#y = -2 - x#

We now have two expressions for #y#. Since #y = y#, it follows that the two expressions should be equal. That is, #x + 2 = -2 - x#. We can manipulate this as follows:

#x + 2 = -2 - x#
#x = -x - 4#
#2x = -4#
#x = -2#

Remember that #y = x + 2#, so #y# must equal #0#. Plugging these in reveals that, indeed, there is a single solution for the system, represented by the #(x, y)# pair #(-2, 0)#.