How do you find the solution of the system of equations #y= x^2-4# and #y= x-2#?

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Answer 1 · 2015-05-17T13:27:42

You can isolate one variable in one equation and substitute its value in the other one. Let's do it.

Let's get the value of #y# in the first equation and substitute it in the second one and, then, find #x#:

#x^2-4=x-2#
#x^2-x-2=0#

Using Bhaskara:

#(1+-sqrt(1^2-4(1)(-2)))/2*1#
#(1+-3)/2#
#x_1=2# and #x_2=-1#

Now we have these values for #x#, let's find the values of #y# - using the first equation:

#y_1=(2)^2-4=0#
#y_2=(-1)^2-4=-3#

So, your answers are:

#y=0# when #x=2#
#y=-3# when #x=-1#