How do you solve the system of equations: 2x + y = 1 and 4x + 2y = −1?

1 Answer
Aug 15, 2015

Answer:

#{(x = O/), (y = O/) :}#

Explanation:

Start by writing the system as given to you

#{(2x + y = 1), (4x + 2y = -1) :}#

Notice that you can simplify the second equation by dividing all the terms by #2# to get

#4/2 * x + 2/2 * y = -1/2#

This is equivalent to

#2x + y = -1/2#

Notice that the left side of the second equation is identical to the left side of the first equation, but that this expression equals two different values, #1# and #-1/2#, respectively.

In other words, you have

#{(2x + y = 1), (2x + y = -1/2) :}#

Let's say that you wanted to solve this system by substitution

#y = 1 -2 x#

#2x + (1 - 2x) = -1/2#

#color(blue)(cancel(color(black)(2x))) + 1 - color(blue)(cancel(color(black)(2x))) = -1/2#

#1 color(red)(!=) -1/2#

Since #1!=-1/2# for any value of #x# and of #y#, the system of equations has no real solution. You're essentially dealing with two parralel lines that will never intersect to produce a valid solution.