Question 289082
Which ordered pair is a solution of the system of equations y=x^2-x-20 and 
y=3x-15? 
I did do some work i did 
y=x^2-x-20 
1.) y=(x^2-5x)(4x-20)
y= x(x-5) 4(x-5)= y=(x+4)(x-5) 
and 
y=3x-15
y=3(x-5) 
but i don't know what to do from there, i would appreciate the help

Your factoring at 1.) is not correct.

Here's another approach:

2.) y = x^2 - x - 20
3.) y = 3x - 15 

From 2.) we know we can substitute 3x - 15 for y in equation 1.)

3x - 15 = x^2 - x - 20
x^2 - 4x - 5 = 0
(x - 5)*(x + 1) = 0

x = 5 and x = -1 are solutions for x. 

Substituting x = 5 in 3.) we have:

y = 3*5 - 15 
y = 15 - 15 = 0

So the point (5,0) is the first "possible" solution to the system.

The other solution, substituing -1 for x:

y =  3*(-1) - 15
y = -3 - 15 = -18

So the point (-18, -1) is the seocnd possible solution to the system.

Let's check each soluiton in 2.):

First (5,0):

Does  0 = 5^2 - 5 - 20?
5^2 - 5 - 20 = 25 - 5 - 20 = 0
Yes, so (5,0) is a solution to both equations.
 
How about (-18,-1)?

Does -1 = (-18)^2 - 18 - 20
(-18)^2 - 18 - 20 = 324 - 38 = 286
286 is not equal to -1 so (18,-1) is not a solution to the system.