quick algebra question

This involves systems of equations. To do this, let's first set up our equation:

x + y = 6

xy = 3

We now can see that x in terms of y is 6-y

Plug that in for x in the second equation, and you get 6y-y^2=3. We can turn this into a quadratic and solve.

-y^2+6y-3=0.

Divide both sides by -1

y^2-6y+3=0

Now, can you solve the quadratic? 

Once you have done that, find x, and then you are all set!

You can solve this one way easier.

X+Y=6

XY=3

|x-y|=?

 

(x-y)^2=(x+y)^2 -4xy

(x-y)^2=36-12

(x-y)^2=24

x-y = + - sqr rt 24