What are the solutions to #y = -2x + 2# and #2y = -4x +4#?

2 Answers
Jul 14, 2016

Let's start by putting the two equations in the same form-#y = mx + b#.

This can be done by dividing both sides of the first equation by #2#.

#2y = -4x + 4 -> y = (-4x + 4)/2 -> y = (2(-2x + 2))/2 -> y = -2x + 2#

The two lines have the same equation, hence there are infinite solutions (since the two lines lie one on top of one another). The previous statement can be confirmed by selecting any x point and substituting it into both equations, and getting the same result.

For example:

Let #x = -27#

#2y = -4(-27) + 4" and "y = -2(-27) +2#

#y = 56 " and "y = 56#

Let #x = 3195#

#2y = -4(3195) + 4" and "y = -2(3195) + 2#

#y = -6388" and "y = -6388#

Hopefully this helps!

Jul 14, 2016

Infinitely many solutions. See below.

Explanation:

The first thing we need to do in order to classify this system is to solve for #y#. This is done for us in one of the equations already:
#y=-2x+2#

We just have to focus on this one:
#2y=-4x+4#

We want to get #y# by itself, which means removing the #2#; and we do that by dividing by #2#:
#2y=-4x+4#
#->ul(cancel(2)y)=ul(-cancel(4)^2x+cancel(4)^2)#
#color(white)(XX)cancel2color(white)(XXXII)cancel(2)#
#->y=-2x+2#

Our system now looks like this:
#y=-2x+2#
#y=-2x+2#

Hm...they're the same equation! That means there are infinitely many solutions. An infinite number of points satisfy the system, because every #x# value will produce the same #y# value in both of the equations. When #x=5#, #y# equals #-8# in both equations; when #x=-2#, #y# equals #6# in both equations; and so on to infinity (and beyond).

Protip: Whenever the equations are the same, there will be infinitely many solutions. Whenever the slopes are the same, but the intercepts are different (for example, #2x+2# and #2x+4#), there are no solutions because the lines never intersect (they are parallel). Other than that, there will generally be one solution.