Question 324016
{{{3x+6y=-15}}}
{{{x+2y=-5}}}
From the second equation,
{{{x=-2y-5}}}
Substitute into the first,
{{{3(-2y-5)+6y=-15}}}
{{{-6y-15+6y=-15}}}
{{{0=0}}}
True, now what?
The second equation and the first equation are the same equation (multiply the second by 3 to get the first).
So every solution of eq. 1 is a solution of eq. 2, there are an infinite number of solutions.