How do you solve # y=x-2# and #y=4x+1#?

1 Answer
Jun 3, 2016

This is called system of equations because you are searching a pair of numbers that inserted as #x# and #y# verify both equations at the same time.

To solve a system of equation you have to use one equation to express one variable as a function of the other, and substitute the result in the second equation.

In your case you have

#y=x-2#

So your #y# variable will be always equal to #x-2#, even in the second equation. Then we can write

#x-2=4x+1#

Here I replaced the "value" of #y# in the second equation.

Now you have an equation in one variable that you can solve as usual

#x-2=4x+1#
#4x-x=-2-1#
#3x=-3#
#x=-1#

With a value for #x# you can substitute in one of the two equations and obtain #y#. Because both equations has to be valid at the same time, both equations must give you the same number for #y#. Let's try:

#y=x-2#
#y=-1-2#
#y=-3#

and, with the second equation

#y=4x+1#
#y=4(-1)+1#
#y=-4+1#
#y=-3#.

It works! Then our solution is given by #x=-1# and #y=-3#.